Compute the product of 3 dictionaries and concatenate keys and values Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) The Ask Question Wizard is Live! Data science time! April 2019 and salary with experience Should we burninate the [wrap] tag?“Least Astonishment” and the Mutable Default ArgumentHow to merge two dictionaries in a single expression?How do I sort a list of dictionaries by a value of the dictionary?How do I sort a dictionary by value?Add new keys to a dictionary?Check if a given key already exists in a dictionaryHow do I concatenate two lists in Python?Iterating over dictionaries using 'for' loopsHow to remove a key from a Python dictionary?check two dictionaries that have similar keys but different valueshow to compare two dictionaries to check if a key is present in both of them

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Compute the product of 3 dictionaries and concatenate keys and values



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experience
Should we burninate the [wrap] tag?“Least Astonishment” and the Mutable Default ArgumentHow to merge two dictionaries in a single expression?How do I sort a list of dictionaries by a value of the dictionary?How do I sort a dictionary by value?Add new keys to a dictionary?Check if a given key already exists in a dictionaryHow do I concatenate two lists in Python?Iterating over dictionaries using 'for' loopsHow to remove a key from a Python dictionary?check two dictionaries that have similar keys but different valueshow to compare two dictionaries to check if a key is present in both of them



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9















Assuming that I have 3 different dictionaries:



dict1 = 
"A": "a"


dict2 =
"B": "b",
"C": "c",
"D": "d",
"E": "e"


dict3 =
"F": "f",
"G": "g"



I want to compute the product of these dictionaries (excluding the product between dict2 and dict3) and combine both the keys and values where the keys are concatenated with _ and values with ' and '



The desired output would be a single dictionary:




# dict1 x dict2
"A_B": "a and b",
"A_C": "a and c",
"A_D": "a and d",
"A_E": "a and e",

# dict1 x dict3
"A_F": "a and f",
"A_G": "a and g",

# dict1 x dict2 x dict3
"A_B_F": "a and b and f",
"A_B_G": "a and b and g",
"A_C_F": "a and c and f",
"A_C_G": "a and c and g",
"A_D_F": "a and d and f",
"A_D_G": "a and d and g",
"A_E_F": "a and e and f",
"A_E_G": "a and e and g"



I had a look at the documentation for itertools but I was not able to understand how I can achieve the desired output.










share|improve this question




























    9















    Assuming that I have 3 different dictionaries:



    dict1 = 
    "A": "a"


    dict2 =
    "B": "b",
    "C": "c",
    "D": "d",
    "E": "e"


    dict3 =
    "F": "f",
    "G": "g"



    I want to compute the product of these dictionaries (excluding the product between dict2 and dict3) and combine both the keys and values where the keys are concatenated with _ and values with ' and '



    The desired output would be a single dictionary:




    # dict1 x dict2
    "A_B": "a and b",
    "A_C": "a and c",
    "A_D": "a and d",
    "A_E": "a and e",

    # dict1 x dict3
    "A_F": "a and f",
    "A_G": "a and g",

    # dict1 x dict2 x dict3
    "A_B_F": "a and b and f",
    "A_B_G": "a and b and g",
    "A_C_F": "a and c and f",
    "A_C_G": "a and c and g",
    "A_D_F": "a and d and f",
    "A_D_G": "a and d and g",
    "A_E_F": "a and e and f",
    "A_E_G": "a and e and g"



    I had a look at the documentation for itertools but I was not able to understand how I can achieve the desired output.










    share|improve this question
























      9












      9








      9


      2






      Assuming that I have 3 different dictionaries:



      dict1 = 
      "A": "a"


      dict2 =
      "B": "b",
      "C": "c",
      "D": "d",
      "E": "e"


      dict3 =
      "F": "f",
      "G": "g"



      I want to compute the product of these dictionaries (excluding the product between dict2 and dict3) and combine both the keys and values where the keys are concatenated with _ and values with ' and '



      The desired output would be a single dictionary:




      # dict1 x dict2
      "A_B": "a and b",
      "A_C": "a and c",
      "A_D": "a and d",
      "A_E": "a and e",

      # dict1 x dict3
      "A_F": "a and f",
      "A_G": "a and g",

      # dict1 x dict2 x dict3
      "A_B_F": "a and b and f",
      "A_B_G": "a and b and g",
      "A_C_F": "a and c and f",
      "A_C_G": "a and c and g",
      "A_D_F": "a and d and f",
      "A_D_G": "a and d and g",
      "A_E_F": "a and e and f",
      "A_E_G": "a and e and g"



      I had a look at the documentation for itertools but I was not able to understand how I can achieve the desired output.










      share|improve this question














      Assuming that I have 3 different dictionaries:



      dict1 = 
      "A": "a"


      dict2 =
      "B": "b",
      "C": "c",
      "D": "d",
      "E": "e"


      dict3 =
      "F": "f",
      "G": "g"



      I want to compute the product of these dictionaries (excluding the product between dict2 and dict3) and combine both the keys and values where the keys are concatenated with _ and values with ' and '



      The desired output would be a single dictionary:




      # dict1 x dict2
      "A_B": "a and b",
      "A_C": "a and c",
      "A_D": "a and d",
      "A_E": "a and e",

      # dict1 x dict3
      "A_F": "a and f",
      "A_G": "a and g",

      # dict1 x dict2 x dict3
      "A_B_F": "a and b and f",
      "A_B_G": "a and b and g",
      "A_C_F": "a and c and f",
      "A_C_G": "a and c and g",
      "A_D_F": "a and d and f",
      "A_D_G": "a and d and g",
      "A_E_F": "a and e and f",
      "A_E_G": "a and e and g"



      I had a look at the documentation for itertools but I was not able to understand how I can achieve the desired output.







      python






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Apr 11 at 16:08









      Old-SchoolOld-School

      726




      726






















          4 Answers
          4






          active

          oldest

          votes


















          8














          The function that will do the job is itertools.product.
          First, here is how you can print out the product dict1 x dict2 x dict3:



          for t in product(dict1.items(), dict2.items(), dict3.items()): 
          k, v = zip(*t)
          print("_".join(k), "-", " and ".join(v))


          Output:



          A_B_F - a and b and f
          A_B_G - a and b and g
          A_C_F - a and c and f
          A_C_G - a and c and g
          A_D_F - a and d and f
          A_D_G - a and d and g
          A_E_F - a and e and f
          A_E_G - a and e and g


          Now, just populate a result dictionary:



          result = 
          for t in product(dict1.items(), dict2.items(), dict3.items()):
          k, v = zip(*t)
          result["_".join(k)] = " and ".join(v)


          You can now add to this dictionary the dict1 x dict2 and dict1 x dict3 products, that are even simpler to compute.




          Based on @ShadowRanger's comment, here is a complete snippet:



          import itertools
          import pprint


          dict1 =
          "A": "a"


          dict2 =
          "B": "b",
          "C": "c",
          "D": "d",
          "E": "e"


          dict3 =
          "F": "f",
          "G": "g"



          result =
          for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)):
          for t in itertools.product(*(d.items() for d in dicts)):
          k, v = zip(*t)
          result["_".join(k)] = " and ".join(v)

          pprint.pprint(result)


          Output:



          'A_B': 'a and b',
          'A_B_F': 'a and b and f',
          'A_B_G': 'a and b and g',
          'A_C': 'a and c',
          'A_C_F': 'a and c and f',
          'A_C_G': 'a and c and g',
          'A_D': 'a and d',
          'A_D_F': 'a and d and f',
          'A_D_G': 'a and d and g',
          'A_E': 'a and e',
          'A_E_F': 'a and e and f',
          'A_E_G': 'a and e and g',
          'A_F': 'a and f',
          'A_G': 'a and g'





          share|improve this answer




















          • 1





            is functools supposed to be itertools?

            – Ben Jones
            Apr 11 at 16:23







          • 1





            @BenJones Yeah sure my bad, I always mix them up...

            – Right leg
            Apr 11 at 16:24











          • No worries. Now I know about functools!

            – Ben Jones
            Apr 11 at 16:25






          • 1





            @BenJones Wanna learn about some more magic? Check out more_itertools :)

            – Right leg
            Apr 11 at 16:27











          • Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

            – ShadowRanger
            Apr 11 at 16:27


















          1














          To produce all pairings, you can use two recursive generator functions: one to find the overall combinations of dictionaries, and the other to pair the keys and values:



          def pair_dicts(data, c):
          if not data:
          keys, values = zip(*c)
          yield ('_'.join(keys), ' and '.join(values))
          else:
          for i in data[0]:
          yield from pair_dicts(data[1:], c+[i])

          def combos(d, c = []):
          if len(c) == len(d):
          yield c
          else:
          if len(c) > 1:
          yield c
          for i in d:
          if all(h != i for h in c):
          yield from combos(d, c+[i])

          new_d = [[list(c.items()) for c in i] for i in combos([dict1, dict2, dict3])]
          final_result = dict(i for b in new_d for i in pair_dicts(b, []))


          Output:



          'A_B': 'a and b', 'A_C': 'a and c', 'A_D': 'a and d', 'A_E': 'a and e', 'A_B_F': 'a and b and f', 'A_B_G': 'a and b and g', 'A_C_F': 'a and c and f', 'A_C_G': 'a and c and g', 'A_D_F': 'a and d and f', 'A_D_G': 'a and d and g', 'A_E_F': 'a and e and f', 'A_E_G': 'a and e and g', 'A_F': 'a and f', 'A_G': 'a and g', 'A_F_B': 'a and f and b', 'A_F_C': 'a and f and c', 'A_F_D': 'a and f and d', 'A_F_E': 'a and f and e', 'A_G_B': 'a and g and b', 'A_G_C': 'a and g and c', 'A_G_D': 'a and g and d', 'A_G_E': 'a and g and e', 'B_A': 'b and a', 'C_A': 'c and a', 'D_A': 'd and a', 'E_A': 'e and a', 'B_A_F': 'b and a and f', 'B_A_G': 'b and a and g', 'C_A_F': 'c and a and f', 'C_A_G': 'c and a and g', 'D_A_F': 'd and a and f', 'D_A_G': 'd and a and g', 'E_A_F': 'e and a and f', 'E_A_G': 'e and a and g', 'B_F': 'b and f', 'B_G': 'b and g', 'C_F': 'c and f', 'C_G': 'c and g', 'D_F': 'd and f', 'D_G': 'd and g', 'E_F': 'e and f', 'E_G': 'e and g', 'B_F_A': 'b and f and a', 'B_G_A': 'b and g and a', 'C_F_A': 'c and f and a', 'C_G_A': 'c and g and a', 'D_F_A': 'd and f and a', 'D_G_A': 'd and g and a', 'E_F_A': 'e and f and a', 'E_G_A': 'e and g and a', 'F_A': 'f and a', 'G_A': 'g and a', 'F_A_B': 'f and a and b', 'F_A_C': 'f and a and c', 'F_A_D': 'f and a and d', 'F_A_E': 'f and a and e', 'G_A_B': 'g and a and b', 'G_A_C': 'g and a and c', 'G_A_D': 'g and a and d', 'G_A_E': 'g and a and e', 'F_B': 'f and b', 'F_C': 'f and c', 'F_D': 'f and d', 'F_E': 'f and e', 'G_B': 'g and b', 'G_C': 'g and c', 'G_D': 'g and d', 'G_E': 'g and e', 'F_B_A': 'f and b and a', 'F_C_A': 'f and c and a', 'F_D_A': 'f and d and a', 'F_E_A': 'f and e and a', 'G_B_A': 'g and b and a', 'G_C_A': 'g and c and a', 'G_D_A': 'g and d and a', 'G_E_A': 'g and e and a'





          share|improve this answer























          • Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

            – Right leg
            Apr 11 at 16:59


















          0














          I created a (not so nice) function to do your task with arbitrary number of dictionaries.



          (Explanation below)



          import itertools as it

          dict1 =
          "A": "a"


          dict2 =
          "B": "b",
          "C": "c",
          "D": "d",
          "E": "e"


          dict3 =
          "F": "f",
          "G": "g"




          def custom_dict_product(dictionaries):
          return dict(zip(map("_".join, it.product(*map(dict.keys, dictionaries))),
          map(" and ".join, it.product(*map(dict.values, dictionaries)))))

          result = custom_dict_product([dict1,dict2])
          result.update(custom_dict_product([dict1,dict3]))
          result.update(custom_dict_product([dict1,dict2,dict3]))
          result
          #'A_B': 'a and b',
          # 'A_B_F': 'a and b and f',
          # 'A_B_G': 'a and b and g',
          # 'A_C': 'a and c',
          # 'A_C_F': 'a and c and f',
          # 'A_C_G': 'a and c and g',
          # 'A_D': 'a and d',
          # 'A_D_F': 'a and d and f',
          # 'A_D_G': 'a and d and g',
          # 'A_E': 'a and e',
          # 'A_E_F': 'a and e and f',
          # 'A_E_G': 'a and e and g',
          # 'A_F': 'a and f',
          # 'A_G': 'a and g'


          The function takes the given dictionaries and gets their keys and values, which is done by map(dict.keys, dictionaries))and map(dict.values, dictionaries)). The results of the first call



          list(it.product(*map(dict.keys, [dict1,dict2])))
          # [('A', 'C'), ('A', 'E'), ('A', 'B'), ('A', 'D')]


          The tuples insides this list are then forced to your desired structure with join(and again an map call to do this for every element):



          "_".join(('A', 'C'))
          # 'A_C'
          list(map("_".join, it.product(*map(dict.keys, [dict1,dict2]))))
          # ['A_C', 'A_E', 'A_B', 'A_D']


          Finally the two resulting lists are transformed to tuples of (keys, values) with the call of zip and handed to the dictionary creation.






          share|improve this answer










          New contributor




          Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.



























            0














            Here a dirty, but working, solution that makes use of itertools



            from itertools import product, combinations


            # create a list and sum dict to be used later
            t = [dict1, dict2, dict3]
            k =
            for d in t:
            k.update(d)


            # iterate over "i" order of combinations ("dict1_X" or "dict1_X_Y") and
            # the cartesian product of keys for each combination

            results =
            for i in range(2, 4):
            a = [
            [
            results.update("_".join(y): " and ".join([k[j] for j in y]))
            for y in product(*x)
            ]
            for x in combinations(t, i)
            if dict1 in x
            ]

            results


            Output:



            'A_B': 'a and b',
            'A_B_F': 'a and b and f',
            'A_B_G': 'a and b and g',
            'A_C': 'a and c',
            'A_C_F': 'a and c and f',
            'A_C_G': 'a and c and g',
            'A_D': 'a and d',
            'A_D_F': 'a and d and f',
            'A_D_G': 'a and d and g',
            'A_E': 'a and e',
            'A_E_F': 'a and e and f',
            'A_E_G': 'a and e and g',
            'A_F': 'a and f',
            'A_G': 'a and g'





            share|improve this answer










            New contributor




            Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.




















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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              8














              The function that will do the job is itertools.product.
              First, here is how you can print out the product dict1 x dict2 x dict3:



              for t in product(dict1.items(), dict2.items(), dict3.items()): 
              k, v = zip(*t)
              print("_".join(k), "-", " and ".join(v))


              Output:



              A_B_F - a and b and f
              A_B_G - a and b and g
              A_C_F - a and c and f
              A_C_G - a and c and g
              A_D_F - a and d and f
              A_D_G - a and d and g
              A_E_F - a and e and f
              A_E_G - a and e and g


              Now, just populate a result dictionary:



              result = 
              for t in product(dict1.items(), dict2.items(), dict3.items()):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)


              You can now add to this dictionary the dict1 x dict2 and dict1 x dict3 products, that are even simpler to compute.




              Based on @ShadowRanger's comment, here is a complete snippet:



              import itertools
              import pprint


              dict1 =
              "A": "a"


              dict2 =
              "B": "b",
              "C": "c",
              "D": "d",
              "E": "e"


              dict3 =
              "F": "f",
              "G": "g"



              result =
              for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)):
              for t in itertools.product(*(d.items() for d in dicts)):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)

              pprint.pprint(result)


              Output:



              'A_B': 'a and b',
              'A_B_F': 'a and b and f',
              'A_B_G': 'a and b and g',
              'A_C': 'a and c',
              'A_C_F': 'a and c and f',
              'A_C_G': 'a and c and g',
              'A_D': 'a and d',
              'A_D_F': 'a and d and f',
              'A_D_G': 'a and d and g',
              'A_E': 'a and e',
              'A_E_F': 'a and e and f',
              'A_E_G': 'a and e and g',
              'A_F': 'a and f',
              'A_G': 'a and g'





              share|improve this answer




















              • 1





                is functools supposed to be itertools?

                – Ben Jones
                Apr 11 at 16:23







              • 1





                @BenJones Yeah sure my bad, I always mix them up...

                – Right leg
                Apr 11 at 16:24











              • No worries. Now I know about functools!

                – Ben Jones
                Apr 11 at 16:25






              • 1





                @BenJones Wanna learn about some more magic? Check out more_itertools :)

                – Right leg
                Apr 11 at 16:27











              • Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

                – ShadowRanger
                Apr 11 at 16:27















              8














              The function that will do the job is itertools.product.
              First, here is how you can print out the product dict1 x dict2 x dict3:



              for t in product(dict1.items(), dict2.items(), dict3.items()): 
              k, v = zip(*t)
              print("_".join(k), "-", " and ".join(v))


              Output:



              A_B_F - a and b and f
              A_B_G - a and b and g
              A_C_F - a and c and f
              A_C_G - a and c and g
              A_D_F - a and d and f
              A_D_G - a and d and g
              A_E_F - a and e and f
              A_E_G - a and e and g


              Now, just populate a result dictionary:



              result = 
              for t in product(dict1.items(), dict2.items(), dict3.items()):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)


              You can now add to this dictionary the dict1 x dict2 and dict1 x dict3 products, that are even simpler to compute.




              Based on @ShadowRanger's comment, here is a complete snippet:



              import itertools
              import pprint


              dict1 =
              "A": "a"


              dict2 =
              "B": "b",
              "C": "c",
              "D": "d",
              "E": "e"


              dict3 =
              "F": "f",
              "G": "g"



              result =
              for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)):
              for t in itertools.product(*(d.items() for d in dicts)):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)

              pprint.pprint(result)


              Output:



              'A_B': 'a and b',
              'A_B_F': 'a and b and f',
              'A_B_G': 'a and b and g',
              'A_C': 'a and c',
              'A_C_F': 'a and c and f',
              'A_C_G': 'a and c and g',
              'A_D': 'a and d',
              'A_D_F': 'a and d and f',
              'A_D_G': 'a and d and g',
              'A_E': 'a and e',
              'A_E_F': 'a and e and f',
              'A_E_G': 'a and e and g',
              'A_F': 'a and f',
              'A_G': 'a and g'





              share|improve this answer




















              • 1





                is functools supposed to be itertools?

                – Ben Jones
                Apr 11 at 16:23







              • 1





                @BenJones Yeah sure my bad, I always mix them up...

                – Right leg
                Apr 11 at 16:24











              • No worries. Now I know about functools!

                – Ben Jones
                Apr 11 at 16:25






              • 1





                @BenJones Wanna learn about some more magic? Check out more_itertools :)

                – Right leg
                Apr 11 at 16:27











              • Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

                – ShadowRanger
                Apr 11 at 16:27













              8












              8








              8







              The function that will do the job is itertools.product.
              First, here is how you can print out the product dict1 x dict2 x dict3:



              for t in product(dict1.items(), dict2.items(), dict3.items()): 
              k, v = zip(*t)
              print("_".join(k), "-", " and ".join(v))


              Output:



              A_B_F - a and b and f
              A_B_G - a and b and g
              A_C_F - a and c and f
              A_C_G - a and c and g
              A_D_F - a and d and f
              A_D_G - a and d and g
              A_E_F - a and e and f
              A_E_G - a and e and g


              Now, just populate a result dictionary:



              result = 
              for t in product(dict1.items(), dict2.items(), dict3.items()):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)


              You can now add to this dictionary the dict1 x dict2 and dict1 x dict3 products, that are even simpler to compute.




              Based on @ShadowRanger's comment, here is a complete snippet:



              import itertools
              import pprint


              dict1 =
              "A": "a"


              dict2 =
              "B": "b",
              "C": "c",
              "D": "d",
              "E": "e"


              dict3 =
              "F": "f",
              "G": "g"



              result =
              for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)):
              for t in itertools.product(*(d.items() for d in dicts)):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)

              pprint.pprint(result)


              Output:



              'A_B': 'a and b',
              'A_B_F': 'a and b and f',
              'A_B_G': 'a and b and g',
              'A_C': 'a and c',
              'A_C_F': 'a and c and f',
              'A_C_G': 'a and c and g',
              'A_D': 'a and d',
              'A_D_F': 'a and d and f',
              'A_D_G': 'a and d and g',
              'A_E': 'a and e',
              'A_E_F': 'a and e and f',
              'A_E_G': 'a and e and g',
              'A_F': 'a and f',
              'A_G': 'a and g'





              share|improve this answer















              The function that will do the job is itertools.product.
              First, here is how you can print out the product dict1 x dict2 x dict3:



              for t in product(dict1.items(), dict2.items(), dict3.items()): 
              k, v = zip(*t)
              print("_".join(k), "-", " and ".join(v))


              Output:



              A_B_F - a and b and f
              A_B_G - a and b and g
              A_C_F - a and c and f
              A_C_G - a and c and g
              A_D_F - a and d and f
              A_D_G - a and d and g
              A_E_F - a and e and f
              A_E_G - a and e and g


              Now, just populate a result dictionary:



              result = 
              for t in product(dict1.items(), dict2.items(), dict3.items()):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)


              You can now add to this dictionary the dict1 x dict2 and dict1 x dict3 products, that are even simpler to compute.




              Based on @ShadowRanger's comment, here is a complete snippet:



              import itertools
              import pprint


              dict1 =
              "A": "a"


              dict2 =
              "B": "b",
              "C": "c",
              "D": "d",
              "E": "e"


              dict3 =
              "F": "f",
              "G": "g"



              result =
              for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)):
              for t in itertools.product(*(d.items() for d in dicts)):
              k, v = zip(*t)
              result["_".join(k)] = " and ".join(v)

              pprint.pprint(result)


              Output:



              'A_B': 'a and b',
              'A_B_F': 'a and b and f',
              'A_B_G': 'a and b and g',
              'A_C': 'a and c',
              'A_C_F': 'a and c and f',
              'A_C_G': 'a and c and g',
              'A_D': 'a and d',
              'A_D_F': 'a and d and f',
              'A_D_G': 'a and d and g',
              'A_E': 'a and e',
              'A_E_F': 'a and e and f',
              'A_E_G': 'a and e and g',
              'A_F': 'a and f',
              'A_G': 'a and g'






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Apr 11 at 16:39

























              answered Apr 11 at 16:19









              Right legRight leg

              8,56242450




              8,56242450







              • 1





                is functools supposed to be itertools?

                – Ben Jones
                Apr 11 at 16:23







              • 1





                @BenJones Yeah sure my bad, I always mix them up...

                – Right leg
                Apr 11 at 16:24











              • No worries. Now I know about functools!

                – Ben Jones
                Apr 11 at 16:25






              • 1





                @BenJones Wanna learn about some more magic? Check out more_itertools :)

                – Right leg
                Apr 11 at 16:27











              • Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

                – ShadowRanger
                Apr 11 at 16:27












              • 1





                is functools supposed to be itertools?

                – Ben Jones
                Apr 11 at 16:23







              • 1





                @BenJones Yeah sure my bad, I always mix them up...

                – Right leg
                Apr 11 at 16:24











              • No worries. Now I know about functools!

                – Ben Jones
                Apr 11 at 16:25






              • 1





                @BenJones Wanna learn about some more magic? Check out more_itertools :)

                – Right leg
                Apr 11 at 16:27











              • Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

                – ShadowRanger
                Apr 11 at 16:27







              1




              1





              is functools supposed to be itertools?

              – Ben Jones
              Apr 11 at 16:23






              is functools supposed to be itertools?

              – Ben Jones
              Apr 11 at 16:23





              1




              1





              @BenJones Yeah sure my bad, I always mix them up...

              – Right leg
              Apr 11 at 16:24





              @BenJones Yeah sure my bad, I always mix them up...

              – Right leg
              Apr 11 at 16:24













              No worries. Now I know about functools!

              – Ben Jones
              Apr 11 at 16:25





              No worries. Now I know about functools!

              – Ben Jones
              Apr 11 at 16:25




              1




              1





              @BenJones Wanna learn about some more magic? Check out more_itertools :)

              – Right leg
              Apr 11 at 16:27





              @BenJones Wanna learn about some more magic? Check out more_itertools :)

              – Right leg
              Apr 11 at 16:27













              Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

              – ShadowRanger
              Apr 11 at 16:27





              Adding an outer loop of for dicts in ((dict1, dict2), (dict1, dict3), (dict1, dict2, dict3)): and making the inner loop for t in product(*[d.items() for d in dicts]): would let you produce the result with minimal code repetition.

              – ShadowRanger
              Apr 11 at 16:27













              1














              To produce all pairings, you can use two recursive generator functions: one to find the overall combinations of dictionaries, and the other to pair the keys and values:



              def pair_dicts(data, c):
              if not data:
              keys, values = zip(*c)
              yield ('_'.join(keys), ' and '.join(values))
              else:
              for i in data[0]:
              yield from pair_dicts(data[1:], c+[i])

              def combos(d, c = []):
              if len(c) == len(d):
              yield c
              else:
              if len(c) > 1:
              yield c
              for i in d:
              if all(h != i for h in c):
              yield from combos(d, c+[i])

              new_d = [[list(c.items()) for c in i] for i in combos([dict1, dict2, dict3])]
              final_result = dict(i for b in new_d for i in pair_dicts(b, []))


              Output:



              'A_B': 'a and b', 'A_C': 'a and c', 'A_D': 'a and d', 'A_E': 'a and e', 'A_B_F': 'a and b and f', 'A_B_G': 'a and b and g', 'A_C_F': 'a and c and f', 'A_C_G': 'a and c and g', 'A_D_F': 'a and d and f', 'A_D_G': 'a and d and g', 'A_E_F': 'a and e and f', 'A_E_G': 'a and e and g', 'A_F': 'a and f', 'A_G': 'a and g', 'A_F_B': 'a and f and b', 'A_F_C': 'a and f and c', 'A_F_D': 'a and f and d', 'A_F_E': 'a and f and e', 'A_G_B': 'a and g and b', 'A_G_C': 'a and g and c', 'A_G_D': 'a and g and d', 'A_G_E': 'a and g and e', 'B_A': 'b and a', 'C_A': 'c and a', 'D_A': 'd and a', 'E_A': 'e and a', 'B_A_F': 'b and a and f', 'B_A_G': 'b and a and g', 'C_A_F': 'c and a and f', 'C_A_G': 'c and a and g', 'D_A_F': 'd and a and f', 'D_A_G': 'd and a and g', 'E_A_F': 'e and a and f', 'E_A_G': 'e and a and g', 'B_F': 'b and f', 'B_G': 'b and g', 'C_F': 'c and f', 'C_G': 'c and g', 'D_F': 'd and f', 'D_G': 'd and g', 'E_F': 'e and f', 'E_G': 'e and g', 'B_F_A': 'b and f and a', 'B_G_A': 'b and g and a', 'C_F_A': 'c and f and a', 'C_G_A': 'c and g and a', 'D_F_A': 'd and f and a', 'D_G_A': 'd and g and a', 'E_F_A': 'e and f and a', 'E_G_A': 'e and g and a', 'F_A': 'f and a', 'G_A': 'g and a', 'F_A_B': 'f and a and b', 'F_A_C': 'f and a and c', 'F_A_D': 'f and a and d', 'F_A_E': 'f and a and e', 'G_A_B': 'g and a and b', 'G_A_C': 'g and a and c', 'G_A_D': 'g and a and d', 'G_A_E': 'g and a and e', 'F_B': 'f and b', 'F_C': 'f and c', 'F_D': 'f and d', 'F_E': 'f and e', 'G_B': 'g and b', 'G_C': 'g and c', 'G_D': 'g and d', 'G_E': 'g and e', 'F_B_A': 'f and b and a', 'F_C_A': 'f and c and a', 'F_D_A': 'f and d and a', 'F_E_A': 'f and e and a', 'G_B_A': 'g and b and a', 'G_C_A': 'g and c and a', 'G_D_A': 'g and d and a', 'G_E_A': 'g and e and a'





              share|improve this answer























              • Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

                – Right leg
                Apr 11 at 16:59















              1














              To produce all pairings, you can use two recursive generator functions: one to find the overall combinations of dictionaries, and the other to pair the keys and values:



              def pair_dicts(data, c):
              if not data:
              keys, values = zip(*c)
              yield ('_'.join(keys), ' and '.join(values))
              else:
              for i in data[0]:
              yield from pair_dicts(data[1:], c+[i])

              def combos(d, c = []):
              if len(c) == len(d):
              yield c
              else:
              if len(c) > 1:
              yield c
              for i in d:
              if all(h != i for h in c):
              yield from combos(d, c+[i])

              new_d = [[list(c.items()) for c in i] for i in combos([dict1, dict2, dict3])]
              final_result = dict(i for b in new_d for i in pair_dicts(b, []))


              Output:



              'A_B': 'a and b', 'A_C': 'a and c', 'A_D': 'a and d', 'A_E': 'a and e', 'A_B_F': 'a and b and f', 'A_B_G': 'a and b and g', 'A_C_F': 'a and c and f', 'A_C_G': 'a and c and g', 'A_D_F': 'a and d and f', 'A_D_G': 'a and d and g', 'A_E_F': 'a and e and f', 'A_E_G': 'a and e and g', 'A_F': 'a and f', 'A_G': 'a and g', 'A_F_B': 'a and f and b', 'A_F_C': 'a and f and c', 'A_F_D': 'a and f and d', 'A_F_E': 'a and f and e', 'A_G_B': 'a and g and b', 'A_G_C': 'a and g and c', 'A_G_D': 'a and g and d', 'A_G_E': 'a and g and e', 'B_A': 'b and a', 'C_A': 'c and a', 'D_A': 'd and a', 'E_A': 'e and a', 'B_A_F': 'b and a and f', 'B_A_G': 'b and a and g', 'C_A_F': 'c and a and f', 'C_A_G': 'c and a and g', 'D_A_F': 'd and a and f', 'D_A_G': 'd and a and g', 'E_A_F': 'e and a and f', 'E_A_G': 'e and a and g', 'B_F': 'b and f', 'B_G': 'b and g', 'C_F': 'c and f', 'C_G': 'c and g', 'D_F': 'd and f', 'D_G': 'd and g', 'E_F': 'e and f', 'E_G': 'e and g', 'B_F_A': 'b and f and a', 'B_G_A': 'b and g and a', 'C_F_A': 'c and f and a', 'C_G_A': 'c and g and a', 'D_F_A': 'd and f and a', 'D_G_A': 'd and g and a', 'E_F_A': 'e and f and a', 'E_G_A': 'e and g and a', 'F_A': 'f and a', 'G_A': 'g and a', 'F_A_B': 'f and a and b', 'F_A_C': 'f and a and c', 'F_A_D': 'f and a and d', 'F_A_E': 'f and a and e', 'G_A_B': 'g and a and b', 'G_A_C': 'g and a and c', 'G_A_D': 'g and a and d', 'G_A_E': 'g and a and e', 'F_B': 'f and b', 'F_C': 'f and c', 'F_D': 'f and d', 'F_E': 'f and e', 'G_B': 'g and b', 'G_C': 'g and c', 'G_D': 'g and d', 'G_E': 'g and e', 'F_B_A': 'f and b and a', 'F_C_A': 'f and c and a', 'F_D_A': 'f and d and a', 'F_E_A': 'f and e and a', 'G_B_A': 'g and b and a', 'G_C_A': 'g and c and a', 'G_D_A': 'g and d and a', 'G_E_A': 'g and e and a'





              share|improve this answer























              • Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

                – Right leg
                Apr 11 at 16:59













              1












              1








              1







              To produce all pairings, you can use two recursive generator functions: one to find the overall combinations of dictionaries, and the other to pair the keys and values:



              def pair_dicts(data, c):
              if not data:
              keys, values = zip(*c)
              yield ('_'.join(keys), ' and '.join(values))
              else:
              for i in data[0]:
              yield from pair_dicts(data[1:], c+[i])

              def combos(d, c = []):
              if len(c) == len(d):
              yield c
              else:
              if len(c) > 1:
              yield c
              for i in d:
              if all(h != i for h in c):
              yield from combos(d, c+[i])

              new_d = [[list(c.items()) for c in i] for i in combos([dict1, dict2, dict3])]
              final_result = dict(i for b in new_d for i in pair_dicts(b, []))


              Output:



              'A_B': 'a and b', 'A_C': 'a and c', 'A_D': 'a and d', 'A_E': 'a and e', 'A_B_F': 'a and b and f', 'A_B_G': 'a and b and g', 'A_C_F': 'a and c and f', 'A_C_G': 'a and c and g', 'A_D_F': 'a and d and f', 'A_D_G': 'a and d and g', 'A_E_F': 'a and e and f', 'A_E_G': 'a and e and g', 'A_F': 'a and f', 'A_G': 'a and g', 'A_F_B': 'a and f and b', 'A_F_C': 'a and f and c', 'A_F_D': 'a and f and d', 'A_F_E': 'a and f and e', 'A_G_B': 'a and g and b', 'A_G_C': 'a and g and c', 'A_G_D': 'a and g and d', 'A_G_E': 'a and g and e', 'B_A': 'b and a', 'C_A': 'c and a', 'D_A': 'd and a', 'E_A': 'e and a', 'B_A_F': 'b and a and f', 'B_A_G': 'b and a and g', 'C_A_F': 'c and a and f', 'C_A_G': 'c and a and g', 'D_A_F': 'd and a and f', 'D_A_G': 'd and a and g', 'E_A_F': 'e and a and f', 'E_A_G': 'e and a and g', 'B_F': 'b and f', 'B_G': 'b and g', 'C_F': 'c and f', 'C_G': 'c and g', 'D_F': 'd and f', 'D_G': 'd and g', 'E_F': 'e and f', 'E_G': 'e and g', 'B_F_A': 'b and f and a', 'B_G_A': 'b and g and a', 'C_F_A': 'c and f and a', 'C_G_A': 'c and g and a', 'D_F_A': 'd and f and a', 'D_G_A': 'd and g and a', 'E_F_A': 'e and f and a', 'E_G_A': 'e and g and a', 'F_A': 'f and a', 'G_A': 'g and a', 'F_A_B': 'f and a and b', 'F_A_C': 'f and a and c', 'F_A_D': 'f and a and d', 'F_A_E': 'f and a and e', 'G_A_B': 'g and a and b', 'G_A_C': 'g and a and c', 'G_A_D': 'g and a and d', 'G_A_E': 'g and a and e', 'F_B': 'f and b', 'F_C': 'f and c', 'F_D': 'f and d', 'F_E': 'f and e', 'G_B': 'g and b', 'G_C': 'g and c', 'G_D': 'g and d', 'G_E': 'g and e', 'F_B_A': 'f and b and a', 'F_C_A': 'f and c and a', 'F_D_A': 'f and d and a', 'F_E_A': 'f and e and a', 'G_B_A': 'g and b and a', 'G_C_A': 'g and c and a', 'G_D_A': 'g and d and a', 'G_E_A': 'g and e and a'





              share|improve this answer













              To produce all pairings, you can use two recursive generator functions: one to find the overall combinations of dictionaries, and the other to pair the keys and values:



              def pair_dicts(data, c):
              if not data:
              keys, values = zip(*c)
              yield ('_'.join(keys), ' and '.join(values))
              else:
              for i in data[0]:
              yield from pair_dicts(data[1:], c+[i])

              def combos(d, c = []):
              if len(c) == len(d):
              yield c
              else:
              if len(c) > 1:
              yield c
              for i in d:
              if all(h != i for h in c):
              yield from combos(d, c+[i])

              new_d = [[list(c.items()) for c in i] for i in combos([dict1, dict2, dict3])]
              final_result = dict(i for b in new_d for i in pair_dicts(b, []))


              Output:



              'A_B': 'a and b', 'A_C': 'a and c', 'A_D': 'a and d', 'A_E': 'a and e', 'A_B_F': 'a and b and f', 'A_B_G': 'a and b and g', 'A_C_F': 'a and c and f', 'A_C_G': 'a and c and g', 'A_D_F': 'a and d and f', 'A_D_G': 'a and d and g', 'A_E_F': 'a and e and f', 'A_E_G': 'a and e and g', 'A_F': 'a and f', 'A_G': 'a and g', 'A_F_B': 'a and f and b', 'A_F_C': 'a and f and c', 'A_F_D': 'a and f and d', 'A_F_E': 'a and f and e', 'A_G_B': 'a and g and b', 'A_G_C': 'a and g and c', 'A_G_D': 'a and g and d', 'A_G_E': 'a and g and e', 'B_A': 'b and a', 'C_A': 'c and a', 'D_A': 'd and a', 'E_A': 'e and a', 'B_A_F': 'b and a and f', 'B_A_G': 'b and a and g', 'C_A_F': 'c and a and f', 'C_A_G': 'c and a and g', 'D_A_F': 'd and a and f', 'D_A_G': 'd and a and g', 'E_A_F': 'e and a and f', 'E_A_G': 'e and a and g', 'B_F': 'b and f', 'B_G': 'b and g', 'C_F': 'c and f', 'C_G': 'c and g', 'D_F': 'd and f', 'D_G': 'd and g', 'E_F': 'e and f', 'E_G': 'e and g', 'B_F_A': 'b and f and a', 'B_G_A': 'b and g and a', 'C_F_A': 'c and f and a', 'C_G_A': 'c and g and a', 'D_F_A': 'd and f and a', 'D_G_A': 'd and g and a', 'E_F_A': 'e and f and a', 'E_G_A': 'e and g and a', 'F_A': 'f and a', 'G_A': 'g and a', 'F_A_B': 'f and a and b', 'F_A_C': 'f and a and c', 'F_A_D': 'f and a and d', 'F_A_E': 'f and a and e', 'G_A_B': 'g and a and b', 'G_A_C': 'g and a and c', 'G_A_D': 'g and a and d', 'G_A_E': 'g and a and e', 'F_B': 'f and b', 'F_C': 'f and c', 'F_D': 'f and d', 'F_E': 'f and e', 'G_B': 'g and b', 'G_C': 'g and c', 'G_D': 'g and d', 'G_E': 'g and e', 'F_B_A': 'f and b and a', 'F_C_A': 'f and c and a', 'F_D_A': 'f and d and a', 'F_E_A': 'f and e and a', 'G_B_A': 'g and b and a', 'G_C_A': 'g and c and a', 'G_D_A': 'g and d and a', 'G_E_A': 'g and e and a'






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Apr 11 at 16:35









              Ajax1234Ajax1234

              43.2k42954




              43.2k42954












              • Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

                – Right leg
                Apr 11 at 16:59

















              • Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

                – Right leg
                Apr 11 at 16:59
















              Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

              – Right leg
              Apr 11 at 16:59





              Although it's not an issue here, I'd generally advise against using a list or any other mutable value as a default value, and would rather go for def combos(d, c=None): if c is None: c = []. See stackoverflow.com/questions/1132941/…

              – Right leg
              Apr 11 at 16:59











              0














              I created a (not so nice) function to do your task with arbitrary number of dictionaries.



              (Explanation below)



              import itertools as it

              dict1 =
              "A": "a"


              dict2 =
              "B": "b",
              "C": "c",
              "D": "d",
              "E": "e"


              dict3 =
              "F": "f",
              "G": "g"




              def custom_dict_product(dictionaries):
              return dict(zip(map("_".join, it.product(*map(dict.keys, dictionaries))),
              map(" and ".join, it.product(*map(dict.values, dictionaries)))))

              result = custom_dict_product([dict1,dict2])
              result.update(custom_dict_product([dict1,dict3]))
              result.update(custom_dict_product([dict1,dict2,dict3]))
              result
              #'A_B': 'a and b',
              # 'A_B_F': 'a and b and f',
              # 'A_B_G': 'a and b and g',
              # 'A_C': 'a and c',
              # 'A_C_F': 'a and c and f',
              # 'A_C_G': 'a and c and g',
              # 'A_D': 'a and d',
              # 'A_D_F': 'a and d and f',
              # 'A_D_G': 'a and d and g',
              # 'A_E': 'a and e',
              # 'A_E_F': 'a and e and f',
              # 'A_E_G': 'a and e and g',
              # 'A_F': 'a and f',
              # 'A_G': 'a and g'


              The function takes the given dictionaries and gets their keys and values, which is done by map(dict.keys, dictionaries))and map(dict.values, dictionaries)). The results of the first call



              list(it.product(*map(dict.keys, [dict1,dict2])))
              # [('A', 'C'), ('A', 'E'), ('A', 'B'), ('A', 'D')]


              The tuples insides this list are then forced to your desired structure with join(and again an map call to do this for every element):



              "_".join(('A', 'C'))
              # 'A_C'
              list(map("_".join, it.product(*map(dict.keys, [dict1,dict2]))))
              # ['A_C', 'A_E', 'A_B', 'A_D']


              Finally the two resulting lists are transformed to tuples of (keys, values) with the call of zip and handed to the dictionary creation.






              share|improve this answer










              New contributor




              Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.
























                0














                I created a (not so nice) function to do your task with arbitrary number of dictionaries.



                (Explanation below)



                import itertools as it

                dict1 =
                "A": "a"


                dict2 =
                "B": "b",
                "C": "c",
                "D": "d",
                "E": "e"


                dict3 =
                "F": "f",
                "G": "g"




                def custom_dict_product(dictionaries):
                return dict(zip(map("_".join, it.product(*map(dict.keys, dictionaries))),
                map(" and ".join, it.product(*map(dict.values, dictionaries)))))

                result = custom_dict_product([dict1,dict2])
                result.update(custom_dict_product([dict1,dict3]))
                result.update(custom_dict_product([dict1,dict2,dict3]))
                result
                #'A_B': 'a and b',
                # 'A_B_F': 'a and b and f',
                # 'A_B_G': 'a and b and g',
                # 'A_C': 'a and c',
                # 'A_C_F': 'a and c and f',
                # 'A_C_G': 'a and c and g',
                # 'A_D': 'a and d',
                # 'A_D_F': 'a and d and f',
                # 'A_D_G': 'a and d and g',
                # 'A_E': 'a and e',
                # 'A_E_F': 'a and e and f',
                # 'A_E_G': 'a and e and g',
                # 'A_F': 'a and f',
                # 'A_G': 'a and g'


                The function takes the given dictionaries and gets their keys and values, which is done by map(dict.keys, dictionaries))and map(dict.values, dictionaries)). The results of the first call



                list(it.product(*map(dict.keys, [dict1,dict2])))
                # [('A', 'C'), ('A', 'E'), ('A', 'B'), ('A', 'D')]


                The tuples insides this list are then forced to your desired structure with join(and again an map call to do this for every element):



                "_".join(('A', 'C'))
                # 'A_C'
                list(map("_".join, it.product(*map(dict.keys, [dict1,dict2]))))
                # ['A_C', 'A_E', 'A_B', 'A_D']


                Finally the two resulting lists are transformed to tuples of (keys, values) with the call of zip and handed to the dictionary creation.






                share|improve this answer










                New contributor




                Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






















                  0












                  0








                  0







                  I created a (not so nice) function to do your task with arbitrary number of dictionaries.



                  (Explanation below)



                  import itertools as it

                  dict1 =
                  "A": "a"


                  dict2 =
                  "B": "b",
                  "C": "c",
                  "D": "d",
                  "E": "e"


                  dict3 =
                  "F": "f",
                  "G": "g"




                  def custom_dict_product(dictionaries):
                  return dict(zip(map("_".join, it.product(*map(dict.keys, dictionaries))),
                  map(" and ".join, it.product(*map(dict.values, dictionaries)))))

                  result = custom_dict_product([dict1,dict2])
                  result.update(custom_dict_product([dict1,dict3]))
                  result.update(custom_dict_product([dict1,dict2,dict3]))
                  result
                  #'A_B': 'a and b',
                  # 'A_B_F': 'a and b and f',
                  # 'A_B_G': 'a and b and g',
                  # 'A_C': 'a and c',
                  # 'A_C_F': 'a and c and f',
                  # 'A_C_G': 'a and c and g',
                  # 'A_D': 'a and d',
                  # 'A_D_F': 'a and d and f',
                  # 'A_D_G': 'a and d and g',
                  # 'A_E': 'a and e',
                  # 'A_E_F': 'a and e and f',
                  # 'A_E_G': 'a and e and g',
                  # 'A_F': 'a and f',
                  # 'A_G': 'a and g'


                  The function takes the given dictionaries and gets their keys and values, which is done by map(dict.keys, dictionaries))and map(dict.values, dictionaries)). The results of the first call



                  list(it.product(*map(dict.keys, [dict1,dict2])))
                  # [('A', 'C'), ('A', 'E'), ('A', 'B'), ('A', 'D')]


                  The tuples insides this list are then forced to your desired structure with join(and again an map call to do this for every element):



                  "_".join(('A', 'C'))
                  # 'A_C'
                  list(map("_".join, it.product(*map(dict.keys, [dict1,dict2]))))
                  # ['A_C', 'A_E', 'A_B', 'A_D']


                  Finally the two resulting lists are transformed to tuples of (keys, values) with the call of zip and handed to the dictionary creation.






                  share|improve this answer










                  New contributor




                  Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.










                  I created a (not so nice) function to do your task with arbitrary number of dictionaries.



                  (Explanation below)



                  import itertools as it

                  dict1 =
                  "A": "a"


                  dict2 =
                  "B": "b",
                  "C": "c",
                  "D": "d",
                  "E": "e"


                  dict3 =
                  "F": "f",
                  "G": "g"




                  def custom_dict_product(dictionaries):
                  return dict(zip(map("_".join, it.product(*map(dict.keys, dictionaries))),
                  map(" and ".join, it.product(*map(dict.values, dictionaries)))))

                  result = custom_dict_product([dict1,dict2])
                  result.update(custom_dict_product([dict1,dict3]))
                  result.update(custom_dict_product([dict1,dict2,dict3]))
                  result
                  #'A_B': 'a and b',
                  # 'A_B_F': 'a and b and f',
                  # 'A_B_G': 'a and b and g',
                  # 'A_C': 'a and c',
                  # 'A_C_F': 'a and c and f',
                  # 'A_C_G': 'a and c and g',
                  # 'A_D': 'a and d',
                  # 'A_D_F': 'a and d and f',
                  # 'A_D_G': 'a and d and g',
                  # 'A_E': 'a and e',
                  # 'A_E_F': 'a and e and f',
                  # 'A_E_G': 'a and e and g',
                  # 'A_F': 'a and f',
                  # 'A_G': 'a and g'


                  The function takes the given dictionaries and gets their keys and values, which is done by map(dict.keys, dictionaries))and map(dict.values, dictionaries)). The results of the first call



                  list(it.product(*map(dict.keys, [dict1,dict2])))
                  # [('A', 'C'), ('A', 'E'), ('A', 'B'), ('A', 'D')]


                  The tuples insides this list are then forced to your desired structure with join(and again an map call to do this for every element):



                  "_".join(('A', 'C'))
                  # 'A_C'
                  list(map("_".join, it.product(*map(dict.keys, [dict1,dict2]))))
                  # ['A_C', 'A_E', 'A_B', 'A_D']


                  Finally the two resulting lists are transformed to tuples of (keys, values) with the call of zip and handed to the dictionary creation.







                  share|improve this answer










                  New contributor




                  Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|improve this answer



                  share|improve this answer








                  edited Apr 11 at 16:45





















                  New contributor




                  Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered Apr 11 at 16:38









                  Sparky05Sparky05

                  2056




                  2056




                  New contributor




                  Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  Sparky05 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





















                      0














                      Here a dirty, but working, solution that makes use of itertools



                      from itertools import product, combinations


                      # create a list and sum dict to be used later
                      t = [dict1, dict2, dict3]
                      k =
                      for d in t:
                      k.update(d)


                      # iterate over "i" order of combinations ("dict1_X" or "dict1_X_Y") and
                      # the cartesian product of keys for each combination

                      results =
                      for i in range(2, 4):
                      a = [
                      [
                      results.update("_".join(y): " and ".join([k[j] for j in y]))
                      for y in product(*x)
                      ]
                      for x in combinations(t, i)
                      if dict1 in x
                      ]

                      results


                      Output:



                      'A_B': 'a and b',
                      'A_B_F': 'a and b and f',
                      'A_B_G': 'a and b and g',
                      'A_C': 'a and c',
                      'A_C_F': 'a and c and f',
                      'A_C_G': 'a and c and g',
                      'A_D': 'a and d',
                      'A_D_F': 'a and d and f',
                      'A_D_G': 'a and d and g',
                      'A_E': 'a and e',
                      'A_E_F': 'a and e and f',
                      'A_E_G': 'a and e and g',
                      'A_F': 'a and f',
                      'A_G': 'a and g'





                      share|improve this answer










                      New contributor




                      Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.
























                        0














                        Here a dirty, but working, solution that makes use of itertools



                        from itertools import product, combinations


                        # create a list and sum dict to be used later
                        t = [dict1, dict2, dict3]
                        k =
                        for d in t:
                        k.update(d)


                        # iterate over "i" order of combinations ("dict1_X" or "dict1_X_Y") and
                        # the cartesian product of keys for each combination

                        results =
                        for i in range(2, 4):
                        a = [
                        [
                        results.update("_".join(y): " and ".join([k[j] for j in y]))
                        for y in product(*x)
                        ]
                        for x in combinations(t, i)
                        if dict1 in x
                        ]

                        results


                        Output:



                        'A_B': 'a and b',
                        'A_B_F': 'a and b and f',
                        'A_B_G': 'a and b and g',
                        'A_C': 'a and c',
                        'A_C_F': 'a and c and f',
                        'A_C_G': 'a and c and g',
                        'A_D': 'a and d',
                        'A_D_F': 'a and d and f',
                        'A_D_G': 'a and d and g',
                        'A_E': 'a and e',
                        'A_E_F': 'a and e and f',
                        'A_E_G': 'a and e and g',
                        'A_F': 'a and f',
                        'A_G': 'a and g'





                        share|improve this answer










                        New contributor




                        Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






















                          0












                          0








                          0







                          Here a dirty, but working, solution that makes use of itertools



                          from itertools import product, combinations


                          # create a list and sum dict to be used later
                          t = [dict1, dict2, dict3]
                          k =
                          for d in t:
                          k.update(d)


                          # iterate over "i" order of combinations ("dict1_X" or "dict1_X_Y") and
                          # the cartesian product of keys for each combination

                          results =
                          for i in range(2, 4):
                          a = [
                          [
                          results.update("_".join(y): " and ".join([k[j] for j in y]))
                          for y in product(*x)
                          ]
                          for x in combinations(t, i)
                          if dict1 in x
                          ]

                          results


                          Output:



                          'A_B': 'a and b',
                          'A_B_F': 'a and b and f',
                          'A_B_G': 'a and b and g',
                          'A_C': 'a and c',
                          'A_C_F': 'a and c and f',
                          'A_C_G': 'a and c and g',
                          'A_D': 'a and d',
                          'A_D_F': 'a and d and f',
                          'A_D_G': 'a and d and g',
                          'A_E': 'a and e',
                          'A_E_F': 'a and e and f',
                          'A_E_G': 'a and e and g',
                          'A_F': 'a and f',
                          'A_G': 'a and g'





                          share|improve this answer










                          New contributor




                          Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.










                          Here a dirty, but working, solution that makes use of itertools



                          from itertools import product, combinations


                          # create a list and sum dict to be used later
                          t = [dict1, dict2, dict3]
                          k =
                          for d in t:
                          k.update(d)


                          # iterate over "i" order of combinations ("dict1_X" or "dict1_X_Y") and
                          # the cartesian product of keys for each combination

                          results =
                          for i in range(2, 4):
                          a = [
                          [
                          results.update("_".join(y): " and ".join([k[j] for j in y]))
                          for y in product(*x)
                          ]
                          for x in combinations(t, i)
                          if dict1 in x
                          ]

                          results


                          Output:



                          'A_B': 'a and b',
                          'A_B_F': 'a and b and f',
                          'A_B_G': 'a and b and g',
                          'A_C': 'a and c',
                          'A_C_F': 'a and c and f',
                          'A_C_G': 'a and c and g',
                          'A_D': 'a and d',
                          'A_D_F': 'a and d and f',
                          'A_D_G': 'a and d and g',
                          'A_E': 'a and e',
                          'A_E_F': 'a and e and f',
                          'A_E_G': 'a and e and g',
                          'A_F': 'a and f',
                          'A_G': 'a and g'






                          share|improve this answer










                          New contributor




                          Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          share|improve this answer



                          share|improve this answer








                          edited Apr 11 at 18:21





















                          New contributor




                          Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.









                          answered Apr 11 at 17:41









                          Lante DellarovereLante Dellarovere

                          26816




                          26816




                          New contributor




                          Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.





                          New contributor





                          Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          Lante Dellarovere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.



























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                              Cannot Extend partition with GParted The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Community Moderator Election ResultsCan't increase partition size with GParted?GParted doesn't recognize the unallocated space after my current partitionWhat is the best way to add unallocated space located before to Ubuntu 12.04 partition with GParted live?I can't figure out how to extend my Arch home partition into free spaceGparted Linux Mint 18.1 issueTrying to extend but swap partition is showing as Unknown in Gparted, shows proper from fdiskRearrange partitions in gparted to extend a partitionUnable to extend partition even though unallocated space is next to it using GPartedAllocate free space to root partitiongparted: how to merge unallocated space with a partition

                              대한민국 목차 국명 지리 역사 정치 국방 경제 사회 문화 국제 순위 관련 항목 각주 외부 링크 둘러보기 메뉴북위 37° 34′ 08″ 동경 126° 58′ 36″ / 북위 37.568889° 동경 126.976667°  / 37.568889; 126.976667ehThe Korean Repository문단을 편집문단을 편집추가해Clarkson PLC 사Report for Selected Countries and Subjects-Korea“Human Development Index and its components: P.198”“http://www.law.go.kr/%EB%B2%95%EB%A0%B9/%EB%8C%80%ED%95%9C%EB%AF%BC%EA%B5%AD%EA%B5%AD%EA%B8%B0%EB%B2%95”"한국은 국제법상 한반도 유일 합법정부 아니다" - 오마이뉴스 모바일Report for Selected Countries and Subjects: South Korea격동의 역사와 함께한 조선일보 90년 : 조선일보 인수해 혁신시킨 신석우, 임시정부 때는 '대한민국' 국호(國號) 정해《우리가 몰랐던 우리 역사: 나라 이름의 비밀을 찾아가는 역사 여행》“남북 공식호칭 ‘남한’‘북한’으로 쓴다”“Corea 대 Korea, 누가 이긴 거야?”국내기후자료 - 한국[김대중 前 대통령 서거] 과감한 구조개혁 'DJ노믹스'로 최단기간 환란극복 :: 네이버 뉴스“이라크 "韓-쿠르드 유전개발 MOU 승인 안해"(종합)”“해외 우리국민 추방사례 43%가 일본”차기전차 K2'흑표'의 세계 최고 전력 분석, 쿠키뉴스 엄기영, 2007-03-02두산인프라, 헬기잡는 장갑차 'K21'...내년부터 공급, 고뉴스 이대준, 2008-10-30과거 내용 찾기mk 뉴스 - 구매력 기준으로 보면 한국 1인당 소득 3만弗과거 내용 찾기"The N-11: More Than an Acronym"Archived조선일보 최우석, 2008-11-01Global 500 2008: Countries - South Korea“몇년째 '시한폭탄'... 가계부채, 올해는 터질까”가구당 부채 5000만원 처음 넘어서“‘빚’으로 내몰리는 사회.. 위기의 가계대출”“[경제365] 공공부문 부채 급증…800조 육박”“"소득 양극화 다소 완화...불평등은 여전"”“공정사회·공생발전 한참 멀었네”iSuppli,08年2QのDRAMシェア・ランキングを発表(08/8/11)South Korea dominates shipbuilding industry | Stock Market News & Stocks to Watch from StraightStocks한국 자동차 생산, 3년 연속 세계 5위자동차수출 '현대-삼성 웃고 기아-대우-쌍용은 울고' 과거 내용 찾기동반성장위 창립 1주년 맞아Archived"중기적합 3개업종 합의 무시한 채 선정"李대통령, 사업 무분별 확장 소상공인 생계 위협 질타삼성-LG, 서민업종인 빵·분식사업 잇따라 철수상생은 뒷전…SSM ‘몸집 불리기’ 혈안Archived“경부고속도에 '아시안하이웨이' 표지판”'철의 실크로드' 앞서 '말(言)의 실크로드'부터, 프레시안 정창현, 2008-10-01“'서울 지하철은 안전한가?'”“서울시 “올해 안에 모든 지하철역 스크린도어 설치””“부산지하철 1,2호선 승강장 안전펜스 설치 완료”“전교조, 정부 노조 통계서 처음 빠져”“[Weekly BIZ] 도요타 '제로 이사회'가 리콜 사태 불러들였다”“S Korea slams high tuition costs”““정치가 여론 양극화 부채질… 합리주의 절실””“〈"`촛불집회'는 민주주의의 질적 변화 상징"〉”““촛불집회가 민주주의 왜곡 초래””“국민 65%, "한국 노사관계 대립적"”“한국 국가경쟁력 27위‥노사관계 '꼴찌'”“제대로 형성되지 않은 대한민국 이념지형”“[신년기획-갈등의 시대] 갈등지수 OECD 4위…사회적 손실 GDP 27% 무려 300조”“2012 총선-대선의 키워드는 '국민과 소통'”“한국 삶의 질 27위, 2000년과 2008년 연속 하위권 머물러”“[해피 코리아] 행복점수 68점…해외 평가선 '낙제점'”“한국 어린이·청소년 행복지수 3년 연속 OECD ‘꼴찌’”“한국 이혼율 OECD중 8위”“[통계청] 한국 이혼율 OECD 4위”“오피니언 [이렇게 생각한다] `부부의 날` 에 돌아본 이혼율 1위 한국”“Suicide Rates by Country, Global Health Observatory Data Repository.”“1. 또 다른 차별”“오피니언 [편집자에게] '왕따'와 '패거리 정치' 심리는 닮은꼴”“[미래한국리포트] 무한경쟁에 빠진 대한민국”“대학생 98% "외모가 경쟁력이라는 말 동의"”“특급호텔 웨딩·200만원대 유모차… "남보다 더…" 호화病, 고질병 됐다”“[스트레스 공화국] ① 경쟁사회, 스트레스 쌓인다”““매일 30여명 자살 한국, 의사보다 무속인에…””“"자살 부르는 '우울증', 환자 중 85% 치료 안 받아"”“정신병원을 가다”“대한민국도 ‘묻지마 범죄’,안전지대 아니다”“유엔 "학생 '성적 지향'에 따른 차별 금지하라"”“유엔아동권리위원회 보고서 및 번역본 원문”“고졸 성공스토리 담은 '제빵왕 김탁구' 드라마 나온다”“‘빛 좋은 개살구’ 고졸 취업…실습 대신 착취”원본 문서“정신건강, 사회적 편견부터 고쳐드립니다”‘소통’과 ‘행복’에 목 마른 사회가 잠들어 있던 ‘심리학’ 깨웠다“[포토] 사유리-곽금주 교수의 유쾌한 심리상담”“"올해 한국인 평균 영화관람횟수 세계 1위"(종합)”“[게임연중기획] 게임은 문화다-여가활동 1순위 게임”“영화속 ‘영어 지상주의’ …“왠지 씁쓸한데””“2월 `신문 부수 인증기관` 지정..방송법 후속작업”“무료신문 성장동력 ‘차별성’과 ‘갈등해소’”대한민국 국회 법률지식정보시스템"Pew Research Center's Religion & Public Life Project: South Korea"“amp;vwcd=MT_ZTITLE&path=인구·가구%20>%20인구총조사%20>%20인구부문%20>%20 총조사인구(2005)%20>%20전수부문&oper_YN=Y&item=&keyword=종교별%20인구& amp;lang_mode=kor&list_id= 2005년 통계청 인구 총조사”원본 문서“한국인이 좋아하는 취미와 운동 (2004-2009)”“한국인이 좋아하는 취미와 운동 (2004-2014)”Archived“한국, `부분적 언론자유국' 강등〈프리덤하우스〉”“국경없는기자회 "한국, 인터넷감시 대상국"”“한국, 조선산업 1위 유지(S. Korea Stays Top Shipbuilding Nation) RZD-Partner Portal”원본 문서“한국, 4년 만에 ‘선박건조 1위’”“옛 마산시,인터넷속도 세계 1위”“"한국 초고속 인터넷망 세계1위"”“인터넷·휴대폰 요금, 외국보다 훨씬 비싸”“한국 관세행정 6년 연속 세계 '1위'”“한국 교통사고 사망자 수 OECD 회원국 중 2위”“결핵 후진국' 한국, 환자가 급증한 이유는”“수술은 신중해야… 자칫하면 생명 위협”대한민국분류대한민국의 지도대한민국 정부대표 다국어포털대한민국 전자정부대한민국 국회한국방송공사about korea and information korea브리태니커 백과사전(한국편)론리플래닛의 정보(한국편)CIA의 세계 정보(한국편)마리암 부디아 (Mariam Budia),『한국: 하늘이 내린 한 폭의 그림』, 서울: 트랜스라틴 19호 (2012년 3월)대한민국ehehehehehehehehehehehehehehWorldCat132441370n791268020000 0001 2308 81034078029-6026373548cb11863345f(데이터)00573706ge128495