Python split string exactly on one space. if double space make " word" not "word"

Question

I have the following string.

words = "this is a book and i like it"

What i want is that when i split it by one space i get the following. wordList = words.split(" ") print wordList << ['this','is','a',' book','and','i',' like','it']

Simple words.split(" ") function splits the string but incase of double space it remove both spaces which gives 'book' and 'like'. and what i need is ' book' and ' like' keeping extra spaces intact in the split output in case of double, triple... n spaces

Answer 1

You can split on whitespace that is not preceded by white space using look behind (?<=) syntax:

import re

re.split("(?<=\\S) ", words)
# ['this', 'is', 'a', ' book', 'and', 'i', ' like', 'it']

Or similarly, use negative look behind:

re.split("(?<!\\s) ", words)
# ['this', 'is', 'a', ' book', 'and', 'i', ' like', 'it']

Answer 2

Just another regex solution: if you need to split with a single left-most whitespace char, use \s? to match one or zero whitespaces, and then capture 0+ remaining whitespaces and the subsequent non-whitespace chars.

One very important step: run rstrip on the input string before running the regex to remove all the trailing whitespace, since otherwise, its performance will decrease greatly.

import re
words = "this is a  book and i  like it"
print(re.findall(r'\s?(\s*\S+)', words.rstrip()))
# => ['this', 'is', 'a', ' book', 'and', 'i', ' like', 'it']

See a Python demo. The re.findall returns just the captured substrings and since we only have one capturing group, the result is a list of those captures.

Also, here is a regex demo. Details:

• \s? - 1 or 0 (due to ? quantifier) whitespaces
• (\s*\S+) - Capturing group #1 matching
  • \s* - zero or more (due to the * quantifier) whitespace
  • \S+ - 1 or more (due to + quantifier) non-whitespace symbols.