Ygtjki

I have the following data frame that continues indefinitely horizontally and vertically with negative numbers only in the odd columns:

-1 2 3 4 -5 9
 2 3 -4 5 -6 11

And I want the 2nd, 4th and 6th complete columns (or every even column) and the minus signs only from the 1st, 3rd, and 5th (or every odd column), so I get this:

- 2 4 - 9
 3 - 5 - 11

And eventually end up with this:

-2 4 -9
 3 -5 -11

So I need the values from the even columns unchanged and of the odd columns, if there's a negative value, keep the - only and if there's a positive value, discard it.

Is there a way to do this with awk/sed?

This is about a far as I get:

awk ' for (i=2;i<=NF;i+=2) $i="" 1' FILE.txt | sed 's/[0-9,.]*//g' 

Here's one way:

$ awk 'for(i=1;i<=NF;i+=2)if($i<0)$i="-"else$i=""; ;1' file |
 sed 's/- */-/g; s/ */ /g'
-2 4 -9
 3 -5 -11

The awk script goes over all odd columns and sets their value to - if they are negative and empty if not. Then, the sed removes any spaces following a - and then replaces multiple consecutive spaces with a single one. Note that this means that the alignment will be broken since some fields will have two characters or more and others will have one. That won't be an issue if you're working with fields, they just don't look pretty.

The sed way:

sed -E '
 s/^(([ t]*-?[ t]*[0-9.]+[ t]+[0-9.]+)*)[ t]+-?[ t]*[0-9.]+$/1/;
 s/[0-9.]+[ t]+([0-9.]+)/1/g'

Output:

-2 4 -9
 3 -5 -11

The first expression kills the trailing column if there are an odd number of columns. It does that by looking for 0 or more pairs <number> <number>, where the first number can be negative.

Edit: A shorter sed solution, inspired by @mikeserv:

sed -E '
 s/[0-9.]+[ t]*([0-9.]*)/1/g;
 s/[- t]*$//'

The same thing with perl:

perl -lpe 's/^((s*-?s*[d.]+s*[d.]+)*)s+-?s*[d.]+$/$1/o; s/[d.]+s+([d.]+)/$1/g'

Another way with perl (probably the cleanest one):

perl -lpe '$a = 1; s/([d.]+s*)/$a++ % 2 ? "" : $1/eg; s/[-s]*$//o'

A perl one:

$ perl -anle 'BEGIN$,=" "
 print map$_=$F[$_]=~/^-/?"-$F[$_+1]":" $F[$_+1]"grep!($_%2)0..$#F' file
-2 4 -9
 3 -5 -11

• 
  -an split input to @F array


• 
  BEGIN$,=" " set output field separator to a space


• 
  grep!($_%2)0..$#F get all even indexes in @F array, which are indexes of odd elements


• 
  map$_=$F[$_]=~/^-/?"-$F[$_+1]":" $F[$_+1]" check if odd element start with -, then append - to next even element, else append a space

As @terdon's answer but without the sed:

awk ' for(i=1;i<=NF;i+=2)
 if ($i<0) $(i+1)*=-1;
 $i = "";
 
 print
 '

A python solution

python -c 'from __future__ import print_function; 
import sys, math;
for line in sys.stdin:
 x = [int(y) for y in line.split()]
 print(*[int(math.copysign(b, a)) for a, b in zip(x[::2], x[1::2])], sep=" ")
' <file

A simple mathematics-based awk solution:

$ cat <<M | awk 'for(i=2;i<=NF;i+=2)printf "%4s",($(i-1)<0?-1:1)*$iprint ""'
-1 2 3 4 -5 9
2 3.2 -4 5 -6
M

 -2 4 -9
 3.2 -5

• Loop from the second (i=2) to the last field (i<=NF).


• Multiply the previous field ($(i-1)) with either -1 or 1.


• Format the output nicely (printf "%4s"), and print a trailing newline (print "").

The only caveat to this is that if you have an odd number of columns, the last field will not display anything at all. I hope this is what you expect. Apparently this is what you expect. :)

(edited to work with decimal values, and to make the loop conditions more aligned with the question while saving 2 characters.)