Replace each character of white space at the end of each line with '_'

Question

I'm wanting to replace each character of white space at the end of each line with '_'. I found a similar question and answer for leading whitespace. But have failed to reconstruct it for trailing whitespace. Here's the link: https://stackoverflow.com/questions/9222281/replace-leading-whitespace-with-sed-or-similar

If anyone can think of a quicker or better way, that would also be great. I also appreciate good explanations, as that way I learn quicker :)

Input:
foo bar
 foo bar oof
  line 3a  
  line fo a

Output:
foo bar_____
 foo bar oof
  line 3a___
  line fo a_

Answer 1

With GNU sed, replacing all spaces at eol by underscore:

sed ':x;s/ \( *\)$/_\1/;tx'  

Answer 2

More efficient to use perl:

perl -lpe 's/(\s+)$/"_" x length($1)/e' input.txt

which only has to do one substitution per line with trailing whitespace, instead of looping.

Answer 3

With GNU awk for the 3rd arg to match() and gensub():

$ awk 'match($0,/(.*[^ ])(.*)/,a){$0=a[1] gensub(/ /,"_","g",a[2])} 1' file
foo bar_____
 foo bar oof
  line 3a___
  line fo a_

With any awk:

$ awk 'match($0,/ +$/){tail=substr($0,RSTART,RLENGTH); gsub(/ /,"_",tail); $0=substr($0,1,RSTART-1) tail} 1' file
foo bar_____
 foo bar oof
  line 3a___
  line fo a_

To replace leading blanks too by tweaking the above gawk solution:

$ awk 'match($0,/^( *)(.*[^ ])(.*)/,a){$0=gensub(/ /,"_","g",a[1]) a[2] gensub(/ /,"_","g",a[3])} 1' file
foo bar_____
_foo bar oof
__line 3a___
__line fo a_

Answer 4

With awk

awk -F '[ \t]+$' 'NF>1{t=substr($0,length($1)+1);gsub(/./,"_",t); $0=$1 t} 1'

This also handles a mixture of trailing tabs and spaces. The field separator (-F, FS) can be easily adjusted to only match spaces or also match other kind of invisible characters, provided that it's kept anchored at the end with $.

To make that work for leading blanks, everything should be mirrored not just $ to ^:

awk -F '^[ \t]+' 'NF>1{h=substr($0,1,length()-length($2));gsub(/./,"_",h); $0=h $2} 1'

To make it work for both leading and trailing blanks, the logic should be inverted; set the field separator to a pattern not matching leading and trailing blanks:

awk -F '[^ \t](.*[^ \t]|$)' '{s=$0; h=gsub(/./,"_",$1); t=gsub(/./,"_",$2); print $1 substr(s,h+1, length(s)-h-t) $2}'

Or the same with adjustable pattern:

awk -v ns='[^ \t]' 'BEGIN{FS=ns"(.*"ns"|$)"}{s=$0; h=gsub(/./,"_",$1); t=gsub(/./,"_",$2); print $1 substr(s,h+1, length(s)-h-t) $2}'

Different from @EdMorton's solutions, these handle correctly lines which contain only spaces and will work with any implementation of awk, not just GNU awk (gawk): eg. with mawk or bwk ("original-awk"), which are both much faster than gawk. But even when used with gawk, the last solution will be almost twice as fast as @EdMorton's.

With sed

With sed, the only solution I can think of is to substitute repeatedly in a loop; if there are many trailing spaces and long lines, this can get slow fast:

sed -e :x -e 's/ \( *\)$/_\1/;tx'

Notice that sed ':x;s/ \( *\)$/_\1/;tx' is not standard sed; :label is not one of the commands which can be terminated by a ;:

Editing commands other than {...}, a, b, c, i, r, t, w, :, and # can be followed by a <semicolon>, optional <blank> characters, and another editing command.

With perl

Here is an alternate perl solution, which is NOT really an improvement upon the existing perl answer, but which, since it doesn't use the e flag of s///, could be theoretically adapted to some other tool providing a sed-like s/// and perl/pcre-like zero-width assertions in its regexes:

perl -ple 's/\s(?=\s*$)/_/g'

Answer 5

If you're trying to add whitespace to even out the lines:

$ cat -A file
foo bar$
 foo bar oof$
  line 3a  $
  line fo a$

Lines are not even

perl -MList::Util=max -lne '
    push @lines, $_
}
END {
    $wid = max map {length} @lines;
    for $line (@lines) {
        $padded = sprintf "%-*s", $wid, $line;
        $padded =~ s/(\s+)$/"_" x length($1)/e;
        print $padded
    }
' file

foo bar_____
 foo bar oof
  line 3a___
  line fo a_

Answer 6

You can do this via the following simple methods also :

Perl splits on nonwhitespace and perform the transliteration of space to underscore on the last field. Then print all fields.

$ perl -F'(\S+)'  -lane '
    $F[-1] =~ tr/ /_/ if @F;
    print @F;
' file 

Using GNU sed we perform the equivalent of lookahead:

$ sed -re '
    y/ /\n/
    :loop
      s/\n(\S| )/ \1/
    tloop
    y/\n/_/
' file 

This will work posixly (change \S to [^[:space:]])

$ sed -e '
    h;s/\s*$//;x;s/.*\S//
    y/ /_/;x;G;s/\n//
' file 

This method will work only in gnu sed as it makes use of s///e extended flag.

$ sed -re '
    s/(.*\S)(.*)/echo "\1""$(echo "\2" | tr " " _)"/e
' file