Python regex for matching odd number of consecutive backslashes

Question

I want to find all odd number of continuous backslashes.

I tried the following expression:

(?<!\\)(\\)(?!\\)

However, it matches a single backslash if present.

Answer 1

You can try this too:

(?<!\\)(?:\\{2})*\\(?!\\)

Explanation

1. (?<!\\\\) No backslash in the back from the current location
2. (?:\\\\{2}) matches each next consecutive pairs of backslash , zero or more occasion
3. \\\\ matches a single backslash
4. (?!\\\\) then checks if there is no more backslash immediately

Answer 2

You may use

r'(?<!\\)\\(?:\\{2})*(?!\\)'

See the regex demo .

Details :

• (?<!\\\\) - no backslash can appear right before the current location
• \\\\ - a single backslash
• (?:\\\\{2})* - 0 or more occurrences of 2 backslashes
• (?!\\\\) - no \\ can appear immediately to the right at the current location.

Note that in Python it is a good idea to use raw string literal when using literal backslashes in a regex pattern (to reduce the number of escapes and make the pattern look tidier, and avoid any issues related to misinterpreted backslashes.)

Answer 3

If you're looking for speed, this benchmarks the best:
\\\\(?<!\\\\\\\\)(?:\\\\\\\\)*(?!\\\\)

Regex1:   \\(?<!\\\\)(?:\\\\)*(?!\\)
Options:  < none >
Completed iterations:   50  /  50     ( x 1000 )
Matches found per iteration:   3
Elapsed Time:    0.23 s,   229.23 ms,   229231 µs

Regex2:   (?<!\\)\\(?:\\{2})*(?!\\)
Options:  < none >
Completed iterations:   50  /  50     ( x 1000 )
Matches found per iteration:   3
Elapsed Time:    0.83 s,   829.68 ms,   829684 µs

Regex3:   (?<!\\)(?:\\{2})*\\(?!\\)
Options:  < none >
Completed iterations:   50  /  50     ( x 1000 )
Matches found per iteration:   3
Elapsed Time:    1.09 s,   1086.34 ms,   1086340 µs