Regex: Remove spaces in some punctuation

Question

I have some strings such as abc pre - school unit or abc pre / school district that I would need to delete additional spaces before and after hyphen and slash. These examples will become abc pre-school unit and abc pre/school district.

I attempted this solution, but this works just replacing either slash or hyphen with hyphen. How can I delete the spaces to get these strings?

abc pre-school unit abc pre/school district

import re

text= ['abc pre - school unit', 'abc pre / school district']

for name in text:
    tmp= re.sub("\s+[-/]\s+" , "-", name)

    print(tmp)

score 2 · Accepted Answer · answered May 26 '20 at 00:39

You could capture the symbol and then replace with that:

text = ['abc pre - school unit', 'abc pre / school district']

for name in text:
    tmp = re.sub("\s+([/-])\s+" , "\\1", name)
    print(tmp)

This prints:

abc pre-school unit
abc pre/school district

BENY · Answer 2 · 2020-05-26T00:43:31.747

1

In your case you need assign it back as well

text= ['abc pre - school unit', 'abc pre / school district']
tmp=[]
for name in text:
    tmp.append(re.sub("\s+([-/])\s+" , r'\1', name))

tmp
['abc pre-school unit', 'abc pre/school district']

Or

newlist=list(map(lambda x : re.sub("\s+([-/])\s+" , r'\1', x),text))

edited May 26 '20 at 00:43

answered May 26 '20 at 00:38

BENY

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The `Or` part of your answer is identical to mine, and was added substantially after I posted. – Tim Biegeleisen May 26 '20 at 02:11

score 1 · Answer 3 · answered May 26 '20 at 00:39

1

in re.sub, you can capture a pattern by putting it in braces. You can refer to it in the replacement by using positional arguments such as \1, \2, \3

So the solution would be: for name in text: tmp.append(re.sub("\s+([-/])\s+" , "\1", name))

answered May 26 '20 at 00:39

Catch22

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Regex: Remove spaces in some punctuation

3 Answers3