matching all characters in any order in regex

Question

I'm a regex newbie, but I understand how to match any characters in a regex query in order (ex. [abc] will match any of a, b or c. Also, I believe "abc" will match abc exactly).

However, how do I construct a regex query that will match all the characters abc in any order? So for example, I want it to match "cab" or "bracket". I'm using Python as my scripting language (not sure if this matters or not).

Answer 1

In Python, I wouldn't use a regualar expression for this purpose, but rather a set:

>>> chars = set("abc")
>>> chars.issubset("bracket")
True
>>> chars.issubset("fish")
False
>>> chars.issubset("bad")
False

Regular expressions are useful, but there are situations where different tools are more appropriate.

Answer 2

This can be done with lookahead assertions:

^(?=.*a)(?=.*b)(?=.*c)

matches if your string contains at least one occurrence of a, b and c.

But as you can see, that's not really what regexes are good at.

I would have done:

if all(char in mystr for char in "abc"):
    # do something

Checking for speed:

>>> timeit.timeit(stmt='chars.issubset("bracket");chars.issubset("notinhere")',
... setup='chars=set("abc")')
1.3560583674019995
>>> timeit.timeit(stmt='all(char in "bracket" for char in s);all(char in "notinhere" for char in s)', 
... setup='s="abc"')
1.4581878714681409
>>> timeit.timeit(stmt='r.match("bracket"); r.match("notinhere")', 
... setup='import re; r=re.compile("(?=.*a)(?=.*b)(?=.*c)")')
1.0582279123082117

Hey, look, the regex wins! This even holds true for longer search strings:

>>> timeit.timeit(stmt='chars.issubset("bracketed");chars.issubset("notinhere")', 
... setup='chars=set("abcde")')
1.4316702294817105
>>> timeit.timeit(stmt='all(char in "bracketed" for char in s);all(char in "notinhere" for char in s)', 
... setup='s="abcde"')
1.6696223364866682
>>> timeit.timeit(stmt='r.match("bracketed"); r.match("notinhere")', 
... setup='import re; r=re.compile("(?=.*a)(?=.*b)(?=.*c)(?=.*d)(?:.*e)")')
1.1809254199004044