Split a string on every third space

Question

How can I split a string on every third space? I'm wondering if python has built-in syntax for this or if it can be done using a list comprehension.

"a bb c dd ee f" -> ["a bb c", "dd ee f"]

you could do matching re.findall(r'\S+\s\S+\s\S+', "a bb c dd ee f") — Avinash Raj
– Avinash Raj, Commented Apr 13, 2015 at 19:43

Nir Alfasi · Accepted Answer · 2015-04-13 19:25:33Z

5

re.split(r'(.*?\s.*?\s.*?)\s', "a bb c dd ee f")

and in order to remove empty strings from the result:

[x for x in re.split(r'(.*?\s.*?\s.*?)\s', "a bb c dd ee f") if x]

answered Apr 13, 2015 at 19:25

Nir Alfasi

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2 Comments

MattDMo Over a year ago

You could also do re.split(r'(.*?\s.*?\s.*?)\s', "a bb c dd ee f")[1:]

Nir Alfasi Over a year ago

@MattDMo true, but since this string is only an example, and in a real case there might be additional empty strings, I chose a more general approach (even if more verbose).

Kasravnd · Accepted Answer · 2015-04-13 19:36:45Z

1

As a more general way you can use a function :

>>> def spliter(s,spl,ind):
...    indx=[i for i,j in enumerate(s) if j==spl][ind-1]
...    return [s[:indx],s[indx+1:]]
... 
>>> s="a bb c dd ee f"
>>> spliter(s,' ',3)
['a bb c', 'dd ee f']

edited Apr 13, 2015 at 19:36

answered Apr 13, 2015 at 19:28

Kasravnd

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