Python regex “OR” gives empty string when using findall

问题

I'm using a simple regex (.*?)(\d+[.]\d+)|(.*?)(\d+) to match int/float/double value in a string. When doing findall the regex shows empty strings in the output. The empty strings gets removed when I remove the | operator and do an individual match. I had also tried this on regex101 it doesn't show any empty string. How can I remove this empty strings ? Here's my code:

>>>import re
>>>match_float = re.compile('(.*?)(\d+[.]\d+)|(.*?)(\d+)')
>>>match_float.findall("CA$1.90")  
>>>match_float.findall("RM1")

Output:

>>>[('CA$', '1.90', '', '')]
>>>[('', '', 'RM', '1')]

回答1:

Since you defined 4 capturing groups in the pattern, they will always be part of the re.findall output unless you remove them (say, by using filter(None, ...).

However, in the current situation, you may "shrink" your pattern to

r'(.*?)(\d+(?:\.\d+)?)'

See the regex demo

Now, it will only have 2 capturing groups, and thus, findall will only output 2 items per tuple in the resulting list.

Details:

• (.*?) - Capturing group 1 matching any zero or more chars other than line break chars, as few as possible up to the first occurrence of ...
• (\d+(?:\.\d+)?) - Capturing group 2:
  • \d+ - one of more digits
  • (?:\.\d+)? - an optional *non-*capturing group that matches 1 or 0 occurrences of a . and 1+ digits.

See the Python demo:

import re
rx = r"(.*?)(\d+(?:[.]\d+)?)"
ss = ["CA$1.90", "RM1"]
for s in ss:
    print(re.findall(rx, s))
# => [('CA$', '1.90')]  [('RM', '1')]

来源：https://stackoverflow.com/questions/45876194/python-regex-or-gives-empty-string-when-using-findall