如何建立一个正则表达式来捕获由单个空格分隔的单词？

Question

I want to build a regular expression that captures

Fee fie foe foo!

but when there are more than one space:

Fee fie  foe foo!

only captures "Fee fie"

My regex looks something like this:

words_re = re.compile(r"\w[-\w .,!]*")

which you can see captures any sequence starting with alphanumeric and followed by any combination of alphanumeric, spaces, and a few chosen punctuation marks. I just want to limit this to just one space at a time.

Alternatively, a variant of string.split() that returns the separating whitespace spans would do it for me too.

The closest I've gotten is this:

words_re = re.compile(r"\w[-\w.,!]*|\s+")
l = words_re.findall(s)

but I then need to search the returned list for sub-lists containing only single-space separators and then rebuild the strings from those.

One thought I had was to take the result from the above expression and then further split it with string.split(" ") to break it into sub-groups that were separated two spaces, but then what about the three-space case and so forth?

Answer 1

This will work

^(\w+(?:\s[-.!\w]+)*(?:[-.!\w]*$))

Regex Demo

If you want to match only upto one space string, you can use (This will match only from starting. You can remove the anchor if you want for capturing all possibility)

^(\w[-.!\w]*(?:\s[-.!\w]+)*)

Regex Demo

Answer 2

Try out

^((?:\w+(?: |[^ ]$))+)

You can see it live here

• We first match a word with \\w
• Then we allow it to be followed by one space, or anything but a space if that reach the end of the string (?: |[^ ]$)
• We repeat to match every words followed by one space or until end is reached +

Answer 3

Alternative solution without using a regex:

import itertools

def up_to_double_space(str):
    return " ".join(itertools.takewhile(lambda word: word, str.split(" ")))

up_to_double_space("Fee fie foe foo!")
# 'Fee fie foe foo!'
up_to_double_space("Fee fie  foe foo!")
# 'Fee fie'

Answer 4

This is more of a comment than a solution, but I lack the rep for that, but there is a split solution that might work for you. split takes a single argument and will split on that. If you use the space as the argument an empty sting is inserted in the list (from between the two spaces). The downside is that other whitespace (tab, etc) will not cause a split.

In [15]: x = 'fie fie  foo fum'

In [16]: x.split(' ')
Out[16]: ['fie', 'fie', '', 'foo', 'fum']

In [17]: x.split(' ')[:x.split(' ').index('')]
Out[17]: ['fie', 'fie']

It's also not selective about your punctuation, which might be an issue.

In general I think a regex is the correct answer but in case this handles all your needs, it's a lot simpler to use and maintain.