In AWK, is it possible to specify "ranges" of fields?

Question

In AWK, is it possible to specify "ranges" of fields?

Example. Given a tab-separated file "foo" with 100 fields per line, I want to print only the fields 32 to 57 for each line, and save the result in a file "bar". What I do now:

awk 'BEGIN{OFS="\t"}{print $32, $33, $34, $35, $36, $37, $38, $39, $40, $41, $42, $43, $44, $45, $46, $47, $48, $49, $50, $51, $52, $53, $54, $55, $56, $57}' foo > bar

The problem with this is that it is tedious to type and prone to errors.

Is there some syntactic form which allows me to say the same in a more concise and less error prone fashion (like "$32..$57") ?

Answer 1

Besides the awk answer by @Jerry, there are other alternatives:

Using cut (assumes tab delimiter by default):

cut -f32-58 foo >bar

Using perl:

perl -nle '@a=split;print join "\t", @a[31..57]' foo >bar

Answer 2

Mildly revised version:

BEGIN { s = 32; e = 57; }

      { for (i=s; i<=e; i++) printf("%s%s", $(i), i<e ? OFS : "\n"); }

Answer 3

You can do it in awk by using RE intervals. For example, to print fields 3-6 of the records in this file:

$ cat file
1 2 3 4 5 6 7 8 9
a b c d e f g h i

would be:

$ gawk 'BEGIN{f="([^ ]+ )"} {print gensub("("f"{2})("f"{4}).*","\\3","")}' file
3 4 5 6
c d e f

I'm creating an RE segment f to represent every field plus it's succeeding field separator (for convenience), then I'm using that in the gensub to delete 2 of those (i.e the first 2 fields), remember the next 4 for reference later using \3, and then delete what comes after them. For your tab-separated file where you want to print fields 32-57 (i.e. the 26 fields after the first 31) you'd use:

gawk 'BEGIN{f="([^\t]+\t)"} {print gensub("("f"{31})("f"{26}).*","\\3","")}' file

The above uses GNU awk for it's gensub() function. With other awks you'd use sub() or match() and substr().

EDIT: Here's how to write a function to do the job:

gawk '
function subflds(s,e,   f) {
   f="([^" FS "]+" FS ")"
   return gensub( "(" f "{" s-1 "})(" f "{" e-s+1 "}).*","\\3","")
}
{ print subflds(3,6) }
' file
3 4 5 6
c d e f

Just set FS as appropriate. Note that this will need a tweak for the default FS if your input file can start with spaces and/or have multiple spaces between fields and will only work if your FS is a single character.

Answer 4

I'm late but this is quick at to the point so I'll leave it here. In cases like this I normally just remove the fields I don't need with gsub and print. Quick and dirty example, since you know your file is delimited by tabs you can remove the first 31 fields:

awk '{gsub(/^(\w\t){31}/,"");print}'

example of removing 4 fields because lazy:

printf "a\tb\tc\td\te\tf\n" | awk '{gsub(/^(\w\t){4}/,"");print}'

Output:

e   f

This is shorter to write, easier to remember and uses less CPU cycles than horrendous loops.

Answer 5

You can use a combination of loops and printf for that in awk:

#!/bin/bash

start_field=32
end_field=58

awk -v start=$start_field -v end=$end_field 'BEGIN{OFS="\t"}
{for (i=start; i<=end; i++) {
    printf "%s" $i;
    if (i < end) {
        printf "%s", OFS;
    } else {
        printf "\n";
    }
}}'

This looks a bit hacky, however:

• it properly delimits your output based on the specified OFS, and
• it makes sure to print a new line at the end for each input line in the file.

Answer 6

I do not know a way to do field range selection in awk. I know how to drop fields at the end of the input (see bellow), but not easily at the beginning. Bellow, the hard way to drop fields at the beginning.

If you know a character c that is not included in your input, you could use the following awk script:

BEGIN { s = 32; e = 57; c = "#"; }
{ NF = e            # Drop the fields after e.
  $s = c $s         # Put a c in front of the s field.
  sub(".*"c, "")    # Drop the chars before c.
  print             # Print the edited line.
}

EDIT:

And I just thought that you can always find a character that is not in the input: use \n.

Answer 7

Unofrtunately don't seem to have access to my account anymore, but also don't have 50 rep to add a comment anyway.

Bob's answer can be simplified a lot using 'seq':

echo $(seq -s ,\$ 5 9| cut -d, -f2-)
$6,$7,$8,$9

The minor disadvantage is you have to specify your first field number as one lower. So to get fields 3 through 7, I specify 2 as the first argument.

seq -s ,\$ 2 7 sets field seperator for seq at ',$' and yields 2,$3,$4,$5,$6,$7

cut -d, -f2- sets field delimiter at ',' and basically cuts of everything before the first comma, by showing everything from the second field on. Thus resulting in $3,$4,$5,$6,$7

When combined with Bob's answer, we get:

    $ cat awk.txt

    1 2 3 4 5 6 7 8 9

    a b c d e f g h i

    $ awk "{print $(seq -s ,\$ 2 7| cut -d, -f2-)}" awk.txt

    3 4 5 6 7

    c d e f g

    $

Answer 8

I use this simple function, which does not check that the field range exists in the line.

function subby(f,l, s) {
  s = $f
  for(i=f+1;i<=l;i++)
    s = sprintf("%s %s",s,$i)

  return s
}

Answer 9

(I know OP requested "in AWK" but ... )

Using bash expansion on the command line to generate arguments list;

$ cat awk.txt

1 2 3 4 5 6 7 8 9

a b c d e f g h i

$ awk "{print $(c="" ;for i in {3..7}; do c=$c\$$i, ; done ; c=${c%%,} ; echo $c ;)}" awk.txt

3 4 5 6 7
c d e f g

explanation ;

c="" # var to hold args list
for i in {3..7} # the required variable range 3 - 7
do 
   # replace c's value with concatenation of existing value, literal $, i value and a comma
   c=$c\$$i, 
done 
c=${c%%,} # remove trailing/final comma
echo $c #return the list string

placed on single line using semi-colons, inside $() to evaluate/expand in place.