Regex in Python does not mach string (but it does when checking on rubular)

Question

I'm writing a program in Python that uses regular expressions. I am having trouble because an expression that I think should match a string doesn't do it. This is the python code that reproduces my problem:

regex = re.compile(r"ord(er)? *0?([1-4])", re.I)
m = regex.match("CMord01")

m evaluates to FALSE. I would very much like to find out why. I've checked on http://rubular.com/ and there the expression does match the string as expected. Thank you!

Use regex.search() instead of regex.match(). See search() vs. match() in the Python docs. — Aya
– Aya, Commented May 28, 2013 at 15:30
Here in you expresion, the start of the string should be ord. It doesn't include CM. You probably want to prepend .* to your regexp. — Marek Kowalski
– Marek Kowalski, Commented May 28, 2013 at 15:30
@MarekKowalski that would obviously work but then all those junk characters are returned as part of the match. Depending on what they plan to do with the match, this might make things dirty/complicated — acattle
– acattle, Commented May 28, 2013 at 15:47

acattle · Accepted Answer · 2013-05-28 15:41:41Z

8

In Python re.match() matches from the beginning of the string. The first letter in CMord01 is C, not O, thus it does not match.

Most language's regular expression modules don't have this limitation so they match the string no problem. To have this kind of behaviour in Python you need to use re.search() instead. See "search() vs. match()" for more information on the difference between the two functions.

edited May 28, 2013 at 15:41

answered May 28, 2013 at 15:30

acattle

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2 Comments

Guio Over a year ago

Cheers, problem solved! Wasn't aware of search() vs match(), now I know.

acattle Over a year ago

@Guio It's an idiosyncrasy of Python. I remember I had to exact same problem.

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