Python string pattern match not work

Question

In this python string pattern matching, I want to filter out s1, it should be like *\2017-01-23\ , date string followed by a '\' . Any idea?

s1="historyData\xx\n3_1010366372_2017-01-25_1126807";
s2="historyData\xx\2017-01-23\n3_1010366372_2017-01-25_1126807";
date_reg_exp = re.compile('\d{4}[-/]\d{2}[-/]\d{2}\\');

 mat = re.match(date_reg_exp, s)
      if mat is not None:
        print("not matched")
      else:
        print("matched")

Is this purely down to you using `re.match` instead of `re.search`? The former only matching the beginning of a string (as if there was an implicit `^` at the start of the pattern) — SpoonMeiser, Jan 26 '17 at 15:48
Probably related t http://stackoverflow.com/questions/4025482/cant-escape-the-backslash-with-regex (as well as match v search) — doctorlove, Jan 26 '17 at 15:52
Also you will most likely need to escape the \ characters in your strings to avoid invalid escape sequence errors — m_callens, Jan 26 '17 at 15:53
@user1615666 please check the answers. Accepting one if it helps would help search point to right answer — Sarath Sadasivan Pillai, Jan 27 '17 at 03:14

score 1 · Answer 1 · answered Jan 26 '17 at 15:52

You will have to use a raw string instead of string. Because \xx is not a recognized character.

a = "\xx" will throw ValueError: invalid \x escape

You can try like so:

import re

s1 = r"historyData\xx\n3_1010366372_2017-01-25_1126807"
s2 = r"historyData\xx\2017-01-23\n3_1010366372_2017-01-25_1126807"

s = r"(?:.*?\\)(\d+-\d+-\d+)(?:\\.*)$"
reg = re.compile(s)

print re.match(reg, s1)
print re.match(reg, s2).group(1)

Output:

None
2017-01-23

Sarath Sadasivan Pillai · Accepted Answer · 2017-01-30T03:11:38.920

1

You have to use search instead of match

Here is what doc says

Python offers two different primitive operations based on regular expressions: re.match() checks for a match only at the beginning of the string, while re.search() checks for a match anywhere in the string (this is what Perl does by default).

The strings provided had invalid \x escape .To use them as row string you may use r"string" .The s1 and s2 variables be written as

s1=r"historyData\xx\n3_1010366372_2017-01-25_1126807"
s2=r"historyData\xx\2017-01-23\n3_1010366372_2017-01-25_1126807"

You may re-write the function as follows.

 import re
 def containsDate(s):
        date_reg_exp = re.compile(r'(\d{4}-\d{2}-\d{2})')
        mat = re.search(date_reg_exp,s)
        return mat is not None

Now the functions may be used as follows

s1=r"historyData\xx\n3_1010366372_2017-01-25_1126807"
s2=r"historyData\xx\2017-01-23\n3_1010366372_2017-01-25_1126807"

if containsDate(s1):
    print "match"
else:
    print "no match"

edited Jan 30 '17 at 03:11

answered Jan 26 '17 at 15:54

Sarath Sadasivan Pillai

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thanks, but I don't want to extract date, just match or not match – user1615666 Jan 26 '17 at 16:34
@user1615666 returning None would be the not match scenario – Sarath Sadasivan Pillai Jan 29 '17 at 09:38
@user1615666 updated the answer by the way you can do the None check in the `getDate` function itself and return `match` or `nomatch` – Sarath Sadasivan Pillai Jan 29 '17 at 09:41
can you modify your script ? almost there. – user1615666 Jan 30 '17 at 00:58

Python string pattern match not work

2 Answers2