By using (,), you are capturing the group, if you simply remove them you will not have this problem. >>> str1 = “a b c d” >>> re.split(” +”, str1) [‘a’, ‘b’, ‘c’, ‘d’] However there is no need for regex, str.split without any delimiter specified will split this by whitespace for you. This would … Read more
This is not a bug, but a bug fix in Python 3.7 from the commit fbb490fd2f38bd817d99c20c05121ad0168a38ee. In regex, a non-zero-width match moves the pointer position to the end of the match, so that the next assertion, zero-width or not, can continue to match from the position following the match. So in your example, after .* … Read more
Use regular expressions: import re for line in open(‘m.txt’): match = re.search(‘uniprotkb:P(\d+)’, line) if match: print match.group(1)
Simply use \1 instead of $1: In [1]: import re In [2]: method = ‘images/:id/huge’ In [3]: re.sub(r'(:[a-z]+)’, r'<span>\1</span>’, method) Out[3]: ‘images/<span>:id</span>/huge’ Also note the use of raw strings (r’…’) for regular expressions. It is not mandatory but removes the need to escape backslashes, arguably making the code slightly more readable.
The problem you have is that if the regex that re.findall tries to match captures groups (i.e. the portions of the regex that are enclosed in parentheses), then it is the groups that are returned, rather than the matched string. One way to solve this issue is to use non-capturing groups (prefixed with ?:). >>> … Read more
No need for regex here: result = input_str.split(‘@’)[-1]