python regular expression

Question

given a file that looks like this:

mmm     55  v1235

mmm     111 v1241

mmm     22  v2453

mmm     1   v3464

mmm     555 v5353

I want the result to be ( replace all digits with white spaces):

mmm         v1235

mmm         v1241

mmm         v2453

mmm         v3464

mmm         v5353

Well I can find the length subtract and and turn to white space. can I do it in one line ?

something like that, but in sub second parameter is a string and not a regex:

new_line = re.sub(r'mmm \d+', r'mmm \s+', line)

ok in that case use: line = re.sub(r'\d(?=\d* )', ' ', line); — anubhava
– anubhava, Commented May 10, 2017 at 8:26

anubhava · Accepted Answer · 2017-05-10 08:28:24Z

1

Using a lookahead you can check if a matching digit is followed by 0 or more digits and a space as:

line = re.sub(r'\d(?=\d* )', ' ', line);

RegEx Demo

(?=\d* ) is positive lookahead that asserts we have 0 or more digits and a space next to a matching digit.

answered May 10, 2017 at 8:28

anubhava

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Sraw · Accepted Answer · 2017-05-10 08:20:25Z

0

new_line = re.sub(r'mmm\s\d+\s?v', 'mmm v', line)

Seems work?

answered May 10, 2017 at 8:20

Sraw

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Wiktor Stribiżew · Accepted Answer · 2017-05-10 08:20:48Z

0

You may put the mmm with 5 spaces into a lookbehind and process the match within a lambda expression:

import re
s = '''mmm     55  v

mmm     111 v

mmm     22  v

mmm     1   v

mmm     555 v'''
res = re.sub(r'(?<=mmm {5})[0-9]+', lambda x: " "*len(x.group()), s)
print(res)

See the Python demo.

The (?<=mmm {5})[0-9]+ pattern matches 1 or more digits that are preceded with mmm and 5 regular spaces. The lambda x: " "*len(x.group()) code replaces the digits with the same amount of spaces.

Or just wrap the two parts of the pattern with capturing groups and use .group(1) and .group(2):

res = re.sub(r'(mmm     )([0-9]+)', lambda x: "{}{}".format(x.group(1), " "*len(x.group(2))), s)

See another demo.

answered May 10, 2017 at 8:20

Wiktor Stribiżew

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