Replacing a character at a random position using sed?

Question

I'm trying to replace a character in a file at a random position. My file looks something like:

aab  
babab  
abab  

I'm trying to replace a random character for 'c'. So the output might look like:

aab  
bcbab  
abab 

I have tried removing all line breaks and saving in a file new_string.txt and then using sed but it isn't working.

This is the code I have tried:

rand1="$(shuf -i 0-$tot_len -n 1)"
sed "s/^\(.\{"${rand1}"\}\)./\1G/" new_string.txt

I keep getting the error:

sed: -e expression #1, char 25: Invalid content of \{\}

Answer 1

No need for the curly brackets in your variable, and the variable should be quoted as well. Use:

sed "s/^\(.\{$rand1\}\)./\1G/" new_string.txt

---

UPDATE: as stated below in comments:

The original code is fine, however the integer for $rand1 is too large for sed. I found that the maximum value can be 32767 for GNU sed, i.e. sed still takes 16bit integers only.

You can obtain that limit for the system's regular expression library (though GNU sed generally uses a builtin version) with:

$ getconf RE_DUP_MAX
32767

POSIX requires that limit to be at least _POSIX_RE_DUP_MAX (255), and that's the maximum you can expect portably (some systems like Solaris or OS/X have it as low as that).

Answer 2

On a GNU system, to substitute one character (other than newline) at random, you could do:

file=myfile.txt
offset=$(grep -bo . < "$file" | cut -d: -f1 | shuf -n1)
[ -z "$offset" ] || # file doesn't have non-newline characters
  printf c | dd bs=1 seek="$offset" of="$file" conv=notrunc status=none

(with old versions of GNU dd (prior to 8.20), replace status=none with 2> /dev/null).

grep -bo . < "$file" would give you the offset in number of bytes in the file of each non-newline character. For instance, with a file encoded in UTF-8 that contains:

$3
£1
€2

That gives us:

$ grep -bo . < "$file"
0:$
1:3
3:£
5:1
7:€
10:2

With cut -d: -f1, we retain the part before the first colon. Then, we pick one of those offsets at random with shuf -n1.

That assumes the replacement character has the same size as the replaced one. For instance, replacing that £ above (2 bytes) with c (1 byte) would leave the file with c followed by an invalid character.

To work around that, we can't overwrite the file in-place anymore as we'd need to shift data around.

We'd need something like:

perl -C -0777 -pi -e "substr \$_, $offset, 1, 'c'" -- "$file"

instead. With -C, perl honours the locale for what constitutes a character. -0777 -p turns on the slurp mode where the content of $file is slurped into $_ (see Security implications of running perl -ne '…' * though for security considerations with that construct). -pi gives you in-place editing, $_ is written back to the file after the code is run. Then we call substr to substitute the 1 character at the given offset with c.

Answer 3

Try this:

sed 's/^\(.\{'"${rand1}"'\}\)./\1G/'  new_string.txt

Answer 4

With new GNU sed you can do it even without \newline remove

sed -z "s/./@/$(($RANDOM%$(wc -m < file.txt)))" file.txt