python - regex why does `findall` find nothing, but `search` works?

Question

>>> reg = re.compile(r'^\d{1,3}(,\d{3})*$')
>>> str = '42'
>>> reg.search(str).group()
'42'
>>> reg.findall(str)
['']
>>> 

python regex Why does reg.findall find nothing, but reg.search works in this piece of code above?

Answer 1

When you have capture groups (wrapped with parenthesis) in the regex, findall will return the match of the captured group; And in your case the captured group matches an empty string; You can make it non capture with ?: if you want to return the whole match; re.search ignores capture groups on the other hand. These are reflected in the documentation:

re.findall:

Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.

re.search:

Scan through string looking for the first location where the regular expression pattern produces a match, and return a corresponding MatchObject instance. Return None if no position in the string matches the pattern; note that this is different from finding a zero-length match at some point in the string.

import re
reg = re.compile(r'^\d{1,3}(?:,\d{3})*$')
s = '42'
reg.search(s).group()
​# '42'

reg.findall(s)
# ['42']