mess with regex python

Question

I am trying the next code but it seems that i am doing something wrong.

import re

lista = ["\\hola\\01\\02Jan\\05\\03",
"\\hola\\01\\02Dem\\12",
"\\hola\\01\\02March\\12\\04"]

for l in lista:
    m= re.search("\\\\\d{2,2}\\\\\d{2,2}[a-zA-Z]+\\\\\d{2,2}\s",l)
    if m:
        print (m.group(0))

The result should be second string. I have tried without \s but the result match with all strings.

If you just need to get four sequences, you can do this: ^(?:\\\[^\]+){4}$. But if you care about what text those sequences contain, you will need to elaborate. — lintmouse
– lintmouse, Commented Oct 5, 2015 at 20:10
Tip: use r-strings to avoid escaping many backslashes. — Andrea Corbellini
– Andrea Corbellini, Commented Oct 5, 2015 at 20:21

anubhava · Accepted Answer · 2015-10-05 20:29:51Z

2

You can try this regex:

lista = [r"\hola\01\02Jan\05\03", r"\hola\01\02Dem\12", r"\hola\01\02March\12\04"]

>>> for l in lista:
...     m = re.search(r"\\\d{2,2}\\\d{2,2}[a-zA-Z]+\\\d{2}$", l)
...     if m:
...             print m.group()
...

Output:

\01\02Dem\12

Use r"..." form to declare a regex and input as raw string
Use anchor $ to avoid matching unwanted input

edited Oct 5, 2015 at 20:29

answered Oct 5, 2015 at 20:08

anubhava

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1 Comment

kamboj Over a year ago

thanks. with r i had to use only two backslash (`\`) in regex. please correct the code if it possible. thanks once again i have learned new thing about regex.

Mayur Koshti · Accepted Answer · 2015-10-06 12:56:23Z

0

You can use the following code without regex:

>>> for l in lista:
        totalNo = l.count('\\')
        if totalNo == 4:
            print l

answered Oct 6, 2015 at 12:56

Mayur Koshti

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