Python正则表达式查找文本中的所有单词

Question

This sounds very simple, I know, but for some reason I can't get all the results I need

Word in this case is any char but white-space that is separetaed with white-space for example in the following string: "Hello there stackoverflow." the result should be: ['Hello','there','stackoverflow.']

My code:

import re

word_pattern = "^\S*\s|\s\S*\s|\s\S*$"
result = re.findall(word_pattern,text)
print result

but after using this pattern on a string like I've shown it only puts the first and the last words in the list and not the words separeted with two spaces

What is the problem with this pattern?

Answer 1

Use the \\b boundary test instead:

r'\b\S+\b'

Result:

>>> import re
>>> re.findall(r'\b\S+\b', 'Hello there StackOverflow.')
['Hello', 'there', 'StackOverflow']

or not use a regular expression at all and just use .split() ; the latter would include the punctiation in a sentence (the regex above did not match the . in the sentence).

Answer 2

to find all words in a string best use split

>>> "Hello there stackoverflow.".split()
['Hello', 'there', 'stackoverflow.']

but if you must use regular expressions, then you should change your regex to something simpler and faster: r'\\b\\S+\\b' .

• r turns the string to a 'raw' string. meaning it will not escape your characters.
• \\b means a boundary, which is a space, newline, or punctuation.
• \\S you should know, is any non-whitespace character.
• + means one or more of the previous.

so together it means find all visible sets of characters (words/numbers).

Answer 3

How about simply using -

>>> s = "Hello there stackoverflow."
>>> s.split()
['Hello', 'there', 'stackoverflow.']

Answer 4

The other answers are good. Depending on what you want (eg. include/exclude punctuation or other non-word characters) an alternative could be to use a regex to split by one or more whitespace characters:

re.split(r'\s+', 'Hello there   StackOverflow.')
['Hello', 'There', 'StackOverflow.']