Bash How can I wrap text at a certain number of space separated inputs

Question

I have a long string like: "1 2 6 9 18 19 25 67 89 102 140 187" and I want to fold or wrap it after a specified number of inputs or spaces using (e.g. fold on the third input) rather an the number of characters, which varies.

Answer 1

The simplest answer probably is:

echo "1 2 6 9 18 19 25 67 89 102 140 187 99 12" | xargs -n 3

Answer 2

Here's a perl script that folds stdin on words (i.e. strings separated by spaces). You specify the number of "words" on the command line.

Save it as, e.g., fold-words.pl and make it executable with chmod +x fold-words.pl

#! /usr/bin/perl 

use strict;

my $max = shift ;

while (<>) { 
    my $count = 0;
    foreach my $word (split) {
        print "$word " ;
        $count++ ;
        print "\n" if ($count % $max == 0)
    }
    print "\n" if ($count % $max != 0);
    $count=0;
}

Example output:

$ echo "1 2 6 9 18 19 25 67 89 102 140 187 99 12"  | ./fold-words.pl 3
1 2 6 
9 18 19 
25 67 89 
102 140 187 
99 12 

Note that it's possible to have a more stringent (or even bizarre) definition of "words" by changing the (split) to use any arbitrary regular expression. e.g. (split /\t/) will split only on single tabs, rather than the default of "one or more white space characters".

Answer 3

Here's a simple awk version; change the 3 in %3 if you want a different number of fields per line:

awk '{ for(i=1; i<NF; i++) { printf $i OFS; if (i%3 == 0) { print "" }} printf $i}'

Sample run:

$ str="one two three four five"
$ echo $str | awk '{ for(i=1; i<NF; i++) { printf $i OFS; if (i%3 == 0) { print "" }} printf $i}'
one two three
four five