replace word using regex with specific conditions

Question

How would I use the python "re" module to replace a word, i.e., 'co' with an empty string, i.e., '' in a given text, only if:

the word is at the end of the text and there's a space before the word
the word is not the final word in the text but there's a space at beginning, and then another space at the end of the word

I.e.,

# word is not the final word in the text but there's a space at beginning, and then another space at the end of the word
txt = 'A co is mine'
txt_after_replace = 'A is mine'
txt = 'A column is mine'
txt_ater_replace = 'A column is mine'
# word is the end of the text and there's a space before the word
txt = 'my co'
txt_after_replace = 'my'
txt = 'my column'
txt_after_replace = 'my column'

If I do: txt.replace(' co', '') these two cases will fail: txt = 'my column', txt_ater_replace = 'A column is mine'. Since it won't check for the end of text right after the word or for a space in the text right after the word.

I think the re.sub module would come to the rescue here but I'm unsure how.

This should work for any general word, i.e., 'co' in this case.

I don't think you can match both criteria at the same time. A regex for the first would be \b\w+$ (which would slightly over-match if there's punctuation before the word) and the second one could use capture groups " (\w+) .*$" — Beefster
– Beefster, Commented Jan 11, 2018 at 0:07
@Beefster thanks for the help, it turns out to be possible based on answer I accepted. — Dnaiel
– Dnaiel, Commented Jan 11, 2018 at 0:15
I realized after I said that that you meant "either" of the conditions. — Beefster
– Beefster, Commented Jan 11, 2018 at 0:17
@Beefster get it, thanks, not clearer Q, I know, apologies for the english... ;-) — Dnaiel
– Dnaiel, Commented Jan 11, 2018 at 0:18

Rahul · Accepted Answer · 2018-01-11 00:33:53Z

1

You can use alternation to match both criteria using following regex.

Regex: (?:\sco\s|\sco$)

Explanation:

\sco\s matches co preceded and succeed by a space.
\sco$ matches co at end preceded by a space.

Regex101 Demo

In python:

import re
str = "coworker in my company are not so co operative. silly co"
res = re.sub(r'(?:\sco\s|\sco$)', ' ', str)
print(res)

Result: coworker in my company are not so operative. silly

Ideone Demo

edited Jan 11, 2018 at 0:33

answered Jan 11, 2018 at 0:11

Rahul

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6 Comments

sKwa Over a year ago

or even better (\s?\bco\b\s?)

Dnaiel Over a year ago

@sKwa that one seems to fail in this case: re.sub('(\s?\bco\b\s?)', ' ', 'my co sa')

Rahul Over a year ago

@sKwa: Optional space will skip even if it's present.

Dnaiel Over a year ago

@Rahul lol "coworker in my company are not so co operative. silly co"

sKwa Over a year ago

@Dnaiel, I used in Rahul's feedle and it works in feedle O_o

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Skycc · Accepted Answer · 2018-01-11 02:26:52Z

0

you can use lookahead regex

\sco(?=$|\s)

explanation:

space follow by co, then assert what follow by co must be either space or end of text

python code

import re
txt = 'A co is mine, A column is mine, my column, my co'
new_txt = re.sub('\sco(?=$|\s)', '', txt)
# 'A is mine, A column is mine, my column, my'

answered Jan 11, 2018 at 2:26

Skycc

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