Remove 4 chars before the last char with sed (or awk)

Question

I would like to remove the 4 chars before the last char.

Input:

abc2a982
e1kei9e5bc5
e1edeaww58476

Expected output:

abc2
e1kei95
e1edeaww6

So far I have tried:

cat file | while read line; do echo $line | sed 's/.\{4}$\1/';done

I guess there should be something else instead of \1.

Possible duplicate of [How can I strip first X characters from string using sed?](https://stackoverflow.com/questions/11469989/how-can-i-strip-first-x-characters-from-string-using-sed) — l'L'l, Dec 21 '17 at 18:31

score 4 · Accepted Answer · edited Dec 21 '17 at 18:43

4

% cat input | sed 's/....\(.\)$/\1/'
abc2
e1kei95
e1edeaww6

edited Dec 21 '17 at 18:43

Inian

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answered Dec 21 '17 at 18:28

Gambit Support

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LOL, unless "cat input" is a placeholder for a much more complicated pipeline. – Gambit Support Dec 21 '17 at 18:33
Actually I tried this solution, but my error was that I put '\$.\$' for all '.' but the first. – achille Dec 21 '17 at 18:43
@achille - that wasn't included in your question, but it should still work. You could just count the parenthesized expressions, and select the one you want to place in your replacement string. `'s/....$.$$/\1/'` is functionally equivalent to `'s/$.$$.$$.$$.$$.$$/\5/'`. – ghoti Dec 21 '17 at 18:47

score 0 · Answer 2 · edited Dec 21 '17 at 19:52

If you had the urge to do this with bash alone, and avoid using sed, you could use parameter expansion to manipulate your strings.

while read -r line; do
  allbutlast1="${line%?}"          # strip the last character from $line
  lastchar="${line#$allbutlast1}"  # strip what we just captured from start of line
  allbutlast5="${line%?????}"      # strip the last 5 characters
  printf '%s%s\n' "$allbutlast5" "$lastchar"
done

Or if you're using bash as your shell, you have additional options:

while read -r line; do
  printf '%s%s\n' "${line:0:$(( ${#line} - 5))}" "${line:$(( ${#line} - 1 ))}"
done

^{(bash code compacted to save on throw-away variables.)}

The POSIX code (first example) uses parameter expansions ${var%...} and ${var#...} to construct the output. The Bash code uses ${var:start:length} notation, with arithmetic expansion, $(( ... )).

This answer is included mostly for academic purposes. You'll get better performance using awk or sed than processing input line by line with a shell script.

Speaking of which, an awk solution might mirror the bash solution:

awk '{print substr($0,1,length($0)-5) substr($0,length($0))}'

Note that while bash's ${var:start:length} notation starts numbering characters at zero, awk's substr() function starts at one.

Bash 4.3 or newer can do `printf '%s%s\n' "${line:0:-5}" "${line:(-1)}"`. — Benjamin W., Dec 21 '17 at 19:50

Remove 4 chars before the last char with sed (or awk)

2 Answers2