Replace the whole word with AWK

Question

In the following example:

echo "manoeuvre man track" | awk -v replace="man" '{ gsub(replace, ""); print }'

I get this result:

oeuvre  track

I would like it to replace text only when it finds the whole word, so as a result, I would get:

manoeuvre  track

(only the word man was removed)

--

• I tried this with word boundaries (\b and \y), but I cannot understand how to apply them here.

• I made it work with a for loop, but I thought maybe there is a more straightforward way using gsub?

• I only have AWK and don't have gAWK.

# awk --version
awk version 20200816

Answer 1

This may be what you're trying to do, using any awk, assuming you want the first full-field literal string match:

$ echo "manoeuvre man track" |
    awk -v replace="man" '
        s=index(" "$0" "," "replace" ") {
            $0 = substr($0,1,s-2) substr($0,s+length(replace))
        }
        { print }
    '
manoeuvre track

or this if you want to do a full-field regexp match for all fields:

$ echo "manoeuvre man track" |
    awk -v replace="man" '
        {
            $0=" "$0" "
            gsub(" "replace" "," ")
            gsub(/^ | $/,"",$0)
            print
        }
    '
manoeuvre track

There are lots of other possibilities...

Answer 2

Building on to the Ed's solution to address multiple search word appearing next to each other:

cat file
manoeuvre man man man mantrack man

awk -v replace="man" '
{
   $0 = " " $0 " "                # pad space on each side of the line
   gsub("( " replace ")+ ", " ")  # replace 1+ repetitions of man with a space
   gsub(/^ | $/, "")              # remove space from both sides of the line
}
1' file

manoeuvre mantrack

Answer 3

"Vanilla" awk is not a good fit for this task. I'd use perl:

$ echo "manoeuvre man track" |perl -pe 's/\bman\b//g'
manoeuvre  track

or if you want to pass "man" as a parameter:

$ echo "manoeuvre man track" |perl -spe 's/\b${word}\b//g' -- -word=man
manoeuvre  track

---

One last point: if you want the "word" treated as a literal string, use the \Q...\E regex markers:

$ echo "manoeuvre man m.n trackm.n" |perl -spe 's/\b${word}\b//g' -- -word=m.n
manoeuvre   trackm.n

$ echo "manoeuvre man m.n trackm.n" |perl -spe 's/\b\Q${word}\E\b//g' -- -word=m.n
manoeuvre man  trackm.n

Answer 4

Since you cannot use a GNU awk, you CAN'T use word boundaries. You would need a tool that does. For example, Perl.

If you can use Perl, you can use a solution like Glenn's one, but I'd use a but modified approach:

#!/bin/bash
text="manoeuvre man man trackman"
replace="man"
perl -spe 's/\s*(?!\B\w)\Q${replace}\E(?<!\w\B)//g' -- -replace="$replace" <<< "$text"
## => manoeuvre trackman

See this online demo.

Details:

• s - allows using custom switches
• p - reads the file line by line
• e - signals the end of options, and the next thing will be the command
• \s*(?!\B\w)\Q${replace}\E(?<!\w\B) - a regex that matches
  • \s* - one or more whitespaces
  • (?!\B\w) - a left-hand adaptive dynamic word boundary
  • \Q${replace}\E - a replace "word" (that can be any string with any special and non-special chars) where \Q and \E quote (escape) all special regex meta chars automatically (as they are treated as literal chars)
  • (?<!\w\B) - a right-hand adaptive dynamic word boundary.

Note that adaptive word boundaries will work best in cases when you do not know if the "word" one passes to the regex pattern starts or ends with special (i.e. "non-word") characters.

If you had a GNU awk, you could use something like:

awk -v replace="man" '{gsub("\\<"replace"\\>", "")}1'

See the online demo:

#!/bin/bash
s="manoeuvre man track"
awk -v replace="man" '{gsub("\\<"replace"\\>", "")}1' <<< "$s"
# => manoeuvre  track

NOTE:

• Word boundaries here are \< and \>, and to use a literal \ char in the awk command, it needs doubling
• replace is a variable containing just word chars, so it is OK to use \< and \> word boundaries to wrap this value to match as a whole word (it would be more complicated if it contained non-word chars)
• To create a pattern from a variable, you need to concatenate the values, here, it is done with "\\<"replace"\\>".
• Note that print is replaced with 1 in the code above.

Answer 5

Worst case, manually hack the likely occurrences -
(edit - Thanks, anubhava)

$: echo "man man man man man - manual man is really manners man" | 
    awk -v replace="man" '{
      while ($0~"[[:space:]]"replace"[[:space:]]") {
       gsub("[[:space:]]+"replace"[[:space:]]+"," ");
      }
      gsub("^"replace"[[:space:]]+","");
      gsub("[[:space:]]+"replace"$","");
   } 1'
- manual is really manners

As anubhava pointed out in the comments, the space-on-either-side check can be tricked by multiple consecutive occurrences of the pattern, so the scan needs to make multiple passes to be sure.
If you care to try sed it can do boundary markers and doesn't have that problem -

$:  echo "man man man man man - manual man is really manners man" | sed -E 's/\<man\>/ /g; s/  +/ /g;'
 - manual is really manners

Note that multiples at the front still leaves a leading space -you can always add one more check and clean to the pattern list.

If you have GNU awk, and maybe don't care about the double spaces -

$: echo "manual man has manners" | awk '{gsub("\\<man\\>","")}1'
manual  has manners

Be sure to double-slash the metachars. If you really want to supply the pattern as a variable on the command line, especially with GNU awk using border markers, watch out for double-parsing. You will still need double-backslash quoting if you use single-ticks, or four if you double-quote...

$: echo "manual man has manners" | awk -v replace='\\<man\\>' '{gsub(replace,"")}1'
manual has manners
$: echo "manual man has manners" | awk -v replace="\\\\<man\\\\>" '{gsub(replace,"")}1'
manual has manners