I'm learning about regular expression. I don't know how to combine different regular expression to make a single generic regular expression.
I want to write a single regular expression which works for multiple cases. I know this is can be done with naive approach by using or " | " operator.
I don't like this approach. Can anybody tell me better approach?
You need to compile all your regex functions. Check this example:
import re re1 = r'\d+\.\d*[L][-]\d*\s[A-Z]*[/]\d*' re2 = '\d*[/]\d*[A-Z]*\d*\s[A-Z]*\d*[A-Z]*' re3 = '[A-Z]*\d+[/]\d+[A-Z]\d+' re4 = '\d+[/]\d+[A-Z]*\d+\s\d+[A-Z]\s[A-Z]*' sentences = [string1, string2, string3, string4] for sentence in sentences: generic_re = re.compile("(%s|%s|%s|%s)" % (re1, re2, re3, re4)).findall(sentence) 
sentence is not defined + findall takes a string not a list. maybe you meant to do for sentence in sentences?
| operator in fact has the lowest precedence (i.e. it has a 'weaker binding force' than any other operator which may be used in any of re1...re4).To findall with an arbitrary series of REs all you have to do is concatenate the list of matches which each returns:
re_list = [ '\d+\.\d*[L][-]\d*\s[A-Z]*[/]\d*', # re1 in question, ... '\d+[/]\d+[A-Z]*\d+\s\d+[A-z]\s[A-Z]*', # re4 in question ] matches = [] for r in re_list: matches += re.findall( r, string) For efficiency it would be better to use a list of compiled REs.
Alternatively you could join the element RE strings using
generic_re = re.compile( '|'.join( re_list) ) I see lots of people are using pipes, but that seems to only match the first instance. If you want to match all, then try using lookaheads.
Example:
>>> fruit_string = "10a11p" >>> fruit_regex = r'(?=.*?(?P<pears>\d+)p)(?=.*?(?P<apples>\d+)a)' >>> re.match(fruit_regex, fruit_string).groupdict() {'apples': '10', 'pears': '11'} >>> re.match(fruit_regex, fruit_string).group(0) '10a,11p' >>> re.match(fruit_regex, fruit_string).group(1) '11' (?= ...) is a look ahead:
Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.
.*?(?P<pears>\d+)p find a number followed a p anywhere in the string and name the number "pears"
find_all). that's an interesting problem to solve. neat solution. though i'm not sure how you got '10a,11p' from group(0). when i ran it it just gave me ''. did you mean groups()?You might not need to compile both regex patterns. Here is a way, let's see if it works for you.
>>> import re >>> text = 'aaabaaaabbb' >>> A = 'aaa' >>> B = 'bbb' >>> re.findall(A+B, text) ['aaabbb'] >>> further read read_doc
If you need to squash multiple regex patterns together the result can be annoying to parse--unless you use P<?> and .groupdict() but doing that can be pretty verbose and hacky. If you only need a couple matches then doing something like the following could be mostly safe:
bucket_name, blob_path = tuple(item for item in matches.groups() if item is not None) 