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
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
In a regular expression, you need to specify that you’re in multiline mode: >>> import re >>> s = “””cat … dog””” >>> >>> re.match(r’cat\ndog’,s,re.M) <_sre.SRE_Match object at 0xcb7c8> Notice that re translates the \n (raw string) into newline. As you indicated in your comments, you don’t actually need re.M for it to match, but … Read more
Yes, but you have to OR them together: x = re.findall(pattern=r’CAT.+?END’, string=’Cat \n eND’, flags=re.I | re.DOTALL)
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.