python, re.sub add a white space every 'x' characters

Question

string = 'ABCDEFGHIJLM'

I'm trying to achieve the following result using

re.sub(SOME CODE HERE):
'ABC DEF GHI JLM'

Is it possible to do that?

To format the table/code/error correctly, please add **4 leading** space to each line of the table. Thanks! — , Nov 12 '17 at 03:03
Possible duplicate of [Regex insert space every third character, except at end of line](https://stackoverflow.com/questions/32981050/regex-insert-space-every-third-character-except-at-end-of-line) — Van Peer, Nov 12 '17 at 03:11

Sven Zwei · Accepted Answer · 2017-11-12T04:19:13.383

Just match three letters "(\w\w\w)" and replace them with themselves plus a space ("\1 "):

print re.sub(r'(\w\w\w)', r'\1 ', 'ABCDEFGHIJLM')

Prints:

'ABC DEF GHI JLM '

To get rid of the trailing space, you can put a negative look-ahead of "Not end of string (?!$)":

print re.sub(r'(\w\w\w)(?!$)', r'\1 ', 'ABCDEFGHIJLM')

Prints:

'ABC DEF GHI JLM'

If you want to make the group size a parameter, you can specify the letter count as a quantifier ({n}) after the \w to define the size. E.g.:

group_size = 2
print re.sub(r'(\w{%d})(?!$)' % group_size, r'\1 ', 'ABCDEFGHIJLM')

Prints:

'AB CD EF GH IJ LM'

You need to defined it for a general case of n characters – skrubber Nov 12 '17 at 03:25 — skrubber, Nov 12 '17 at 03:25

score 0 · Answer 2 · answered Nov 12 '17 at 03:13

0

Use re findall and join this way:

 s='ABCDEFGHIJLM'
 n=3
 pattern= '.'*n 
 ' '.join(re.findall(pattern, s))
'ABC DEF GHI JLM'

answered Nov 12 '17 at 03:13

skrubber

2 Answers2