How to construct and plot the following accumulated list? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How to construct pairs in a list?Select rows of matrix that are higher than a given rowHow do I select only those lists whose elements are all above a certain number?Selecting every n-th element from a listFinding pairs where the intersection of them is empty set from a nested listHow to compare and remove x-coordinates (and the associated y-coordinates) that are less than previous x-coordinates?ListPlot with colorsModifying a function so that it will take multiple conditionsHow to construct a list of lengths efficientlySelecting cases from a list based on two conditions
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How to construct and plot the following accumulated list?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How to construct pairs in a list?Select rows of matrix that are higher than a given rowHow do I select only those lists whose elements are all above a certain number?Selecting every n-th element from a listFinding pairs where the intersection of them is empty set from a nested listHow to compare and remove x-coordinates (and the associated y-coordinates) that are less than previous x-coordinates?ListPlot with colorsModifying a function so that it will take multiple conditionsHow to construct a list of lengths efficientlySelecting cases from a list based on two conditions
$begingroup$
I have a list, say it is
a = 1,3,3.2,3.9,4,4.4,4.9,5,7,8
To get the accumulate:
acc = Accumlate[a]
I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.
Then I want to take the derivative of the plot to plot it as well.
plotting list-manipulation
edited Apr 13 at 16:26
Roman
5,56111131
asked Apr 13 at 11:22
HamzaHamza
225
$endgroup$
add a comment |
$begingroup$
I have a list, say it is
a = 1,3,3.2,3.9,4,4.4,4.9,5,7,8
To get the accumulate:
acc = Accumlate[a]
I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.
Then I want to take the derivative of the plot to plot it as well.
plotting list-manipulation
edited Apr 13 at 16:26
Roman
5,56111131
asked Apr 13 at 11:22
HamzaHamza
225
$endgroup$
1
$begingroup$
The plotting is not a problem seeListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
$endgroup$
– Hugh
Apr 13 at 11:26
$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[Do[Print[acc[[i]], Count[a[[i]]], i, 10]] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04
$begingroup$
Your combined use ofDoandPrintdoesn't do what you expect.Printmerely prints to the screen but does not build a list; useTableinstead. Also,Countis something else.
$endgroup$
– Roman
Apr 13 at 16:29
add a comment |
$begingroup$
I have a list, say it is
a = 1,3,3.2,3.9,4,4.4,4.9,5,7,8
To get the accumulate:
acc = Accumlate[a]
I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.
Then I want to take the derivative of the plot to plot it as well.
plotting list-manipulation
edited Apr 13 at 16:26
Roman
5,56111131
asked Apr 13 at 11:22
HamzaHamza
225
$endgroup$
I have a list, say it is
a = 1,3,3.2,3.9,4,4.4,4.9,5,7,8
To get the accumulate:
acc = Accumlate[a]
I want to plot the number of elements less than or equal to a certain element against the accumulated value of the list elements.
The pairs should look like (acc[[i]], number of elements less than a[[i]]). For instance, (1, 1), (4, 2), (7.2,3), etc.
Then I want to take the derivative of the plot to plot it as well.
plotting list-manipulation
plotting list-manipulation
edited Apr 13 at 16:26
Roman
5,56111131
asked Apr 13 at 11:22
HamzaHamza
225
edited Apr 13 at 16:26
Roman
5,56111131
asked Apr 13 at 11:22
HamzaHamza
225
edited Apr 13 at 16:26
Roman
5,56111131
edited Apr 13 at 16:26
Roman
5,56111131
edited Apr 13 at 16:26
Roman
5,56111131
5,56111131
asked Apr 13 at 11:22
HamzaHamza
225
asked Apr 13 at 11:22
HamzaHamza
225
asked Apr 13 at 11:22
HamzaHamza
225
225
1
$begingroup$
The plotting is not a problem seeListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
$endgroup$
– Hugh
Apr 13 at 11:26
$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[Do[Print[acc[[i]], Count[a[[i]]], i, 10]] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04
$begingroup$
Your combined use ofDoandPrintdoesn't do what you expect.Printmerely prints to the screen but does not build a list; useTableinstead. Also,Countis something else.
$endgroup$
– Roman
Apr 13 at 16:29
add a comment |
1
$begingroup$
The plotting is not a problem seeListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?
$endgroup$
– Hugh
Apr 13 at 11:26
$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[Do[Print[acc[[i]], Count[a[[i]]], i, 10]] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04
$begingroup$
Your combined use ofDoandPrintdoesn't do what you expect.Printmerely prints to the screen but does not build a list; useTableinstead. Also,Countis something else.
$endgroup$
– Roman
Apr 13 at 16:29
1
1
$begingroup$
The plotting is not a problem see
ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?$endgroup$
– Hugh
Apr 13 at 11:26
$begingroup$
The plotting is not a problem see
ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?$endgroup$
– Hugh
Apr 13 at 11:26
$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[Do[Print[acc[[i]], Count[a[[i]]], i, 10]] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04
$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[Do[Print[acc[[i]], Count[a[[i]]], i, 10]] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04
$begingroup$
Your combined use of
Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.$endgroup$
– Roman
Apr 13 at 16:29
$begingroup$
Your combined use of
Do and Print doesn't do what you expect. Print merely prints to the screen but does not build a list; use Table instead. Also, Count is something else.$endgroup$
– Roman
Apr 13 at 16:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:
ListLinePlot[Transpose[Accumulate[a], Range[Length[a]]]]
If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].
If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], x, 1, Total[a]] and the first-derivative plot with
b = Interpolation[Transpose[Accumulate[Sort[a]], Range[Length[a]]],
InterpolationOrder -> 1];
Plot[b'[x], x, 1, Total[a]]
answered Apr 13 at 16:18
RomanRoman
5,56111131
$endgroup$
add a comment |
$begingroup$
try this
a = 1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8;
acc = Accumulate@a;
aa[x_] := Length@Select[a, # <= a[[x]] &];
list = Transpose[acc, aa /@ Range@Length@a]
ListLinePlot@list
1,1,4,2,7.2,3,11.1,4,15.1,5,19.5,6,24.4,7,29.4,8,36.4,9,44.4,10
answered Apr 13 at 12:45
J42161217J42161217
4,573324
$endgroup$
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:
ListLinePlot[Transpose[Accumulate[a], Range[Length[a]]]]
If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].
If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], x, 1, Total[a]] and the first-derivative plot with
b = Interpolation[Transpose[Accumulate[Sort[a]], Range[Length[a]]],
InterpolationOrder -> 1];
Plot[b'[x], x, 1, Total[a]]
answered Apr 13 at 16:18
RomanRoman
5,56111131
$endgroup$
add a comment |
$begingroup$
Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:
ListLinePlot[Transpose[Accumulate[a], Range[Length[a]]]]
If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].
If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], x, 1, Total[a]] and the first-derivative plot with
b = Interpolation[Transpose[Accumulate[Sort[a]], Range[Length[a]]],
InterpolationOrder -> 1];
Plot[b'[x], x, 1, Total[a]]
answered Apr 13 at 16:18
RomanRoman
5,56111131
$endgroup$
add a comment |
$begingroup$
Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:
ListLinePlot[Transpose[Accumulate[a], Range[Length[a]]]]
If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].
If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], x, 1, Total[a]] and the first-derivative plot with
b = Interpolation[Transpose[Accumulate[Sort[a]], Range[Length[a]]],
InterpolationOrder -> 1];
Plot[b'[x], x, 1, Total[a]]
answered Apr 13 at 16:18
RomanRoman
5,56111131
$endgroup$
Assuming that there are no duplicate elements in the list, and that the list is sorted in ascending order:
ListLinePlot[Transpose[Accumulate[a], Range[Length[a]]]]
If the list isn't sorted, you should replace Accumulate[a] with Accumulate[Sort[a]].
If you need the derivative, it may be easier to first construct an interpolating function. Here I make the interpolation linear (InterpolationOrder -> 1) but you can change this. You can get the above plot with Plot[b[x], x, 1, Total[a]] and the first-derivative plot with
b = Interpolation[Transpose[Accumulate[Sort[a]], Range[Length[a]]],
InterpolationOrder -> 1];
Plot[b'[x], x, 1, Total[a]]
answered Apr 13 at 16:18
RomanRoman
5,56111131
edited Apr 13 at 16:30
answered Apr 13 at 16:18
RomanRoman
5,56111131
answered Apr 13 at 16:18
RomanRoman
5,56111131
answered Apr 13 at 16:18
RomanRoman
5,56111131
5,56111131
add a comment |
add a comment |
$begingroup$
try this
a = 1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8;
acc = Accumulate@a;
aa[x_] := Length@Select[a, # <= a[[x]] &];
list = Transpose[acc, aa /@ Range@Length@a]
ListLinePlot@list
1,1,4,2,7.2,3,11.1,4,15.1,5,19.5,6,24.4,7,29.4,8,36.4,9,44.4,10
answered Apr 13 at 12:45
J42161217J42161217
4,573324
$endgroup$
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
add a comment |
$begingroup$
try this
a = 1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8;
acc = Accumulate@a;
aa[x_] := Length@Select[a, # <= a[[x]] &];
list = Transpose[acc, aa /@ Range@Length@a]
ListLinePlot@list
1,1,4,2,7.2,3,11.1,4,15.1,5,19.5,6,24.4,7,29.4,8,36.4,9,44.4,10
answered Apr 13 at 12:45
J42161217J42161217
4,573324
$endgroup$
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
add a comment |
$begingroup$
try this
a = 1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8;
acc = Accumulate@a;
aa[x_] := Length@Select[a, # <= a[[x]] &];
list = Transpose[acc, aa /@ Range@Length@a]
ListLinePlot@list
1,1,4,2,7.2,3,11.1,4,15.1,5,19.5,6,24.4,7,29.4,8,36.4,9,44.4,10
answered Apr 13 at 12:45
J42161217J42161217
4,573324
$endgroup$
try this
a = 1, 3, 3.2, 3.9, 4, 4.4, 4.9, 5, 7, 8;
acc = Accumulate@a;
aa[x_] := Length@Select[a, # <= a[[x]] &];
list = Transpose[acc, aa /@ Range@Length@a]
ListLinePlot@list
1,1,4,2,7.2,3,11.1,4,15.1,5,19.5,6,24.4,7,29.4,8,36.4,9,44.4,10
answered Apr 13 at 12:45
J42161217J42161217
4,573324
edited Apr 13 at 14:12
answered Apr 13 at 12:45
J42161217J42161217
4,573324
answered Apr 13 at 12:45
J42161217J42161217
4,573324
answered Apr 13 at 12:45
J42161217J42161217
4,573324
4,573324
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
add a comment |
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
The y-axis should represent the number of elements less than or equal a[[i]] not the element itself.
$endgroup$
– Hamza
Apr 13 at 13:57
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
$begingroup$
@Hamza fixed....
$endgroup$
– J42161217
Apr 13 at 14:12
add a comment |
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1
$begingroup$
The plotting is not a problem see
ListPlot. Also, if you want a derivative you will have to define this as you don't have a continuous function. What have you tried so far?$endgroup$
– Hugh
Apr 13 at 11:26
$begingroup$
The plot is an issue when I have very large number of pairs like 1000 pairs. I am trying to do it using ListPlot and a Do loop but it does not work. I am doing: ListLinePlot[Do[Print[acc[[i]], Count[a[[i]]], i, 10]] but it only prints the pairs for me with an empty plot.
$endgroup$
– Hamza
Apr 13 at 12:04
$begingroup$
Your combined use of
DoandPrintdoesn't do what you expect.Printmerely prints to the screen but does not build a list; useTableinstead. Also,Countis something else.$endgroup$
– Roman
Apr 13 at 16:29