python regex negative lookahead

Question

when I use negative lookahead on this string

1pt 22px 3em 4px

like this

/\d+(?!px)/g

i get this result

(1, 2, 3)

and I want all of the 22px to be discarded but I don't know how should I do that

If the letters are there all the time, just use \d+(?!px)[a-z]+ — user557597
– user557597, Commented Apr 3, 2017 at 20:30

Wiktor Stribiżew · Accepted Answer · 2017-04-03 20:17:35Z

5

Add a digit pattern to the lookahead:

\d+(?!\d|px)

See the regex demo

This way, you will not allow a digit to match after 1 or more digits are already matched.

Another way is to use an atomic group work around like

(?=(\d+))\1(?!px)

See the regex demo. Here, (?=(\d+)) captures one or more digits into Group 1 and the \1 backreference will consume these digits, thus preventing backtracking into the \d+ pattern. The (?!px) will fail the match if the digits are followed with px and won't be able to backtrack to fetch 2.

Both solutions will work with re.findall.

answered Apr 3, 2017 at 20:17

Wiktor Stribiżew

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1 Comment

Wiktor Stribiżew Over a year ago

See the Python demo. Remember to use raw string literals when (r'pattern') when declaring regex patterns.

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