Why is this regex not matched in python

Question

I need to find "7.1/10" in "7.1/10&nb" with the following regex:

\d{1}\.?\d{0,2}\/10

But the following code does not match anything:

rating= "7.1/10&nb"
p = re.compile(re.escape("\d{1}\.?\d{0,2}\/10"))
m = p.match(rating)
if m:
    print("rating: {}".format(m.group()))
else:
    print("no match found in {}".format(rating))

What's the problem with my code?

you need to use raw string literal r instead of re.escape p = re.compile(r"\d{1}\.?\d{0,2}\/10") — rock321987
– rock321987, Commented Jun 4, 2016 at 5:55
Change p = re.compile(re.escape("\d{1}\.?\d{0,2}\/10")) to p = re.compile(r"\d{1}\.?\d{0,2}\/10") it will work. — Rahul K P
– Rahul K P, Commented Jun 4, 2016 at 5:57
also, i suspect that you may have problem later while using match..because match matches only in starting of string.. — rock321987
– rock321987, Commented Jun 4, 2016 at 5:58

J.Joe · Accepted Answer · 2016-06-04 06:02:32Z

2

p = re.compile(r"\d\.?\d{0,2}/10")

There are several problems in you re:

add 'r' for raw string or you will have to escape all '\': re.compile("\\d\\.?\\d{0,2}/10")
\d{1} can be \d
\/ can be /, no need to escape

answered Jun 4, 2016 at 6:02

J.Joe

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Comments

Rahul K P · Accepted Answer · 2016-06-04 05:59:34Z

1

Change only one line in your code.

p = re.compile(re.escape("\d{1}\.?\d{0,2}\/10")) to re.compile(r"\d{1}\.?\d{0,2}\/10")

It will work smoothly.

answered Jun 4, 2016 at 5:59

Rahul K P

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