I'm new to python regular expressions and was wondering if someone could help me out by walking me through what this means (I'll state what I think each bit means here as well).
Thanks!
RegExp:
r'(^.*def\W*)(\w+)\W*\((.*)\):'
r'...' = python definition of regular expression within the ''
(...) = a regex term
(^. = match the beginning of any character
*def\W* = ???
(\w+) = match any of [a, z] 1 or more times
\W*\ = ? i think its the same as the line above this but from 0+ more times instead of 1 but since it matches the def\W line above (which i dont really know the meaning of) i'm not sure.
((.*)\): = match any additional character within brackets ()
thanks!
It seems like a failed attempt to match a Python function signature:
import re
regex = re.compile(r""" # r'' means that \n and the like is two chars
# '\\','n' and not a single newline character
( # begin capturing group #1; you can get it: regex.match(text).group(1)
^ # match begining of the string or a new line if re.MULTILINE is set
.* # match zero or more characters except newline (unless
# re.DOTALL is set)
def # match string 'def'
\W* # match zero or more non-\w chars i.e., [^a-zA-Z0-9_] if no
# re.LOCALE or re.UNICODE
) # end capturing group #1
(\w+) # second capturing group [a-zA-Z0-9_] one or more times if
# no above flags
\W* # see above
\( # match literal paren '('
(.*) # 3rd capturing group NOTE: `*` is greedy `.` matches even ')'
# therefore re.match(r'\((.*)\)', '(a)(b)').group(1) == 'a)(b'
\) # match literal paren ')'
: # match literal ':'
""", re.VERBOSE|re.DEBUG)
re.DEBUG
flag causes the output:
subpattern 1
at at_beginning
max_repeat 0 65535
any None
literal 100
literal 101
literal 102
max_repeat 0 65535
in
category category_not_word
subpattern 2
max_repeat 1 65535
in
category category_word
max_repeat 0 65535
in
category category_not_word
literal 40
subpattern 3
max_repeat 0 65535
any None
literal 41
literal 58
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