Find and replace multiple comma/space instances in a string, Python

Question

I have a string with multiple successive instances of , (comma+space) that I want to replace with a single instance. Is there a clean way to do so? I suppose RegEx can be helpful.

A naive example:

s = 'a, b, , c, , , d, , , e, , , , , , , f

The desired output:

'a, b, c, d, e, f

Naturally, the text can change, so the search should be for successive instances of ,.

Not sure I understand the question. This is a simplified example of something else :) — Oleg Melnikov, Nov 28 '15 at 20:20

Dušan Maďar · Accepted Answer · 2015-11-28T20:07:42.190

9

So the regular expression searches for two or more instances of , (comma + space) and then in sub function you replace it with only a single ,.

import re
pattern = re.compile(r'(,\s){2,}')

test_string = 'a, b, , c, , , d, , , e, , , , , , , f'
print re.sub(pattern, ', ', test_string)
>>> a, b, c, d, e, f

and without a regular expression (as @Casimir et Hippolyte suggested in comment)

test_string = 'a, b, , c, , , d, , , e, , , , , , , f'
test_string_parts = test_string.split(',')
test_string_parts = [part.strip() for part in test_string_parts if part != ' ']
print ', '.join(test_string_parts)
>>> a, b, c, d, e, f

edited Nov 28 '15 at 20:07

answered Nov 28 '15 at 19:51

Dušan Maďar

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Thanks. Would not this work the same: `re.sub(r'(,\s){2,}', ', ', test_string)` ? Or, is there some notable difference or corner case? – Oleg Melnikov Nov 28 '15 at 20:01
1

It will work as well. More info about `compile` here: http://stackoverflow.com/questions/452104/is-it-worth-using-pythons-re-compile – Dušan Maďar Nov 28 '15 at 20:05
If instead of `', '` I have `'\n'` how would I adapt this find & replace? `(\\n){2,}` seems to work in https://regex101.com, but not in Python. Any suggestions? – Oleg Melnikov Nov 28 '15 at 20:16
Why not only `r'\n{2,}'`? – Dušan Maďar Nov 28 '15 at 20:23
tried `r'\n{2,}'`. It fails to catch `'\n '`in Python and in https://regex101.com – Oleg Melnikov Nov 28 '15 at 20:30
Thanks to @Casimir. This seems to work nicely: `functools.reduce( (lambda x, y: x + '\n ' + y if y else x), s.split('\n '))` – Oleg Melnikov Nov 28 '15 at 20:31
in that case `r'\n\s{2,}'` – Dušan Maďar Nov 28 '15 at 20:34
:D well, that works for me, but perhaps I am testing with a different string – Dušan Maďar Nov 28 '15 at 20:38

Casimir et Hippolyte · Answer 2 · 2015-11-28T20:15:39.847

2

You can use reduce:

>>> from functools import reduce
>>> reduce( (lambda x, y: x+', '+y if y else x), s.split(', '))

(Where x is the carry and y the item)

edited Nov 28 '15 at 20:15

answered Nov 28 '15 at 20:09

Casimir et Hippolyte

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AkaKaras · Answer 3 · 2015-11-29T19:15:19.820

1

the simplest way for your problem would be:

>>> s = 'a, b, , c, , , d, , , e, , , , , , , f'
>>> s = [x for x in s if x.isalpha()]
>>> print(s)
['a', 'b', 'c', 'd', 'e', 'f']

then, use join()

>>> ', '.join(s)
'a, b, c, d, e, f'

do it in one line:

>>> s = ', '.join([x for x in s if x.isalpha()])
>>> s
'a, b, c, d, e, f'

Just figure other way:

>>> s = 'a, b, , c, , , d, , , e, , , , , , , f'
>>> s = s.split()  #split all ' '(<- space)
>>> s
['a,', 'b,', ',', 'c,', ',', ',', 'd,', ',', ',', 'e,', ',', ',', ',', ',', ',', ',', 'f']
>>> while ',' in s:
...     s.remove(',')
>>> s
['a,', 'b,', 'c,', 'd,', 'e,', 'f']
>>> ''.join(s)
'a,b,c,d,e,f'

edited Nov 29 '15 at 19:15

answered Nov 29 '15 at 05:02

AkaKaras

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However... this only works for single-letter variables... Please read the question :) – Oleg Melnikov Nov 29 '15 at 06:08
just add a new method – AkaKaras Nov 29 '15 at 19:15

score 1 · Answer 4 · edited Dec 19 '18 at 08:31

1

s = ", , a, b, , c, , , d, , , e, , , ,  , , , f,,,,"
s = [o for o in s.replace(' ', '').split(',') if len(o)]
print (s)

edited Dec 19 '18 at 08:31

Tobias Wilfert

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answered Dec 19 '18 at 07:57

Gold

26
2

This answer came in low quality post. Add some explanation even though code is self explanatory – Harsha Biyani Dec 19 '18 at 08:40

score 0 · Answer 5 · answered May 04 '21 at 14:49

One more solution: go through the combination of the list and the same list shifted by one (in other words, by the pairs of consecutive items) and select the second item from each pair where the first (previous) item differs from the second (next) item:

s = 'a. b. . c. . . d. . . e. . . . . . . f'
test = []
for i in s:
    if i != ' ':
        test.append(i)


res = [test[0]] + [y for x,y in zip(test, test[1:]) if x!=y]

for x in res:
    print(x, end='')

yields

a.b.c.d.e.f
[Program finished]

Find and replace multiple comma/space instances in a string, Python

5 Answers5