Compile a regex with one pattern per line

Question

I have multiple patterns to check against. Say hello and bye, but will be many more, so I chose to use re.compile() to store them and then being able to check this regex:

import re

mypatt = re.compile(r'(hello|bye)', re.IGNORECASE)
url = ["bye bye", "hello how are you", "i am fine", "ok byeee"]
for u in url:
    if mypatt.search(u):
        print "yes --> %s" %(u)

Upon running this code I get the desired output:

yes --> bye bye
yes --> hello how are you
yes --> ok byeee

However, since there are multiple patterns I would like to write one per line, with something like:

mypatt = re.compile(r'(\
    hello|\
    bye\
    )', re.IGNORECASE)

However this does not work and I cannot understand why. What is the way to write such statement, writing every pattern in a different line?

You need the re.VERBOSE flag to ignore unescaped whitespace, I think. — jonrsharpe
– jonrsharpe, Commented Dec 30, 2015 at 15:39
@jonrsharpe good one! re.compile(r"""( ... )""", re.IGNORECASE |re.VERBOSE) made it. Feel free to post it as an answer if you want. — fedorqui
– fedorqui, Commented Dec 30, 2015 at 15:41
Using the verbose flag even allows one to enter comments per line. — user707650
– user707650, Commented Dec 30, 2015 at 15:44
@Evert interesting, thanks. I just found a nice example in the re.VERBOSE reference. — fedorqui
– fedorqui, Commented Dec 30, 2015 at 15:45
Yup, the reference works well here. I see I was mistaken about quoting each line though. — user707650
– user707650, Commented Dec 30, 2015 at 15:46

jonrsharpe · Accepted Answer · 2015-12-30 15:55:27Z

3

You're creating a pattern that includes the whitespace at the start of each line. To avoid this, either:

Use textwrap.dedent, which removes common leading whitespace from each line in a multiline string; or
Add the re.VERBOSE (or re.X) flag to ignore unescaped whitespace and allow the addition of inline comments.

edited Dec 30, 2015 at 15:55

answered Dec 30, 2015 at 15:44

jonrsharpe

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