Python Parse a List of Paths

Question

I have a list of paths in a .txt file and I'm trying to parse out one folder in the path name using python.

9999\New_folder\A\23818\files\  
9999\New_folder\A\18283_HO\files\  
...

What I'm interested in doing is pulling the string between 9999\New_folder\A\ and \files\ so that I end up with:

23818  
18283_HO

Any help would be appreciated!

EDIT: Thanks a lot everyone! Came up with the following code with your input.

input_text = open('C:\\Python\\textintolist\\Document1.txt', 'r')
output_text = open('output.txt', 'w')

paths =[]


for line in input_text:
    paths.append(line)

for path in paths:
        output_text.write(str(path.split('\\')[3])+"\n")

use regex regex

profitehlolz
– profitehlolz

2012-08-13 21:12:38 +00:00
Commented Aug 13, 2012 at 21:12 — profitehlolz
– profitehlolz, Commented Aug 13, 2012 at 21:12

applicative_functor · Accepted Answer · 2012-08-13 21:08:36Z

1

>>> s = '9999\\New_folder\\A\\23818\\files\\'
>>> s.split('9999\\New_folder\\A\\')[1].split('\\')[0]
'23818'

answered Aug 13, 2012 at 21:08

applicative_functor

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Comments

stranac · Accepted Answer · 2012-08-13 21:10:54Z

0

If your paths are always in this format:

>>> paths
['9999\\New_folder\\A\\23818\\files\\', '9999\\New_folder\\A\\18283_HO\\files']
>>> for path in paths:
...     print path.split('\\')[3]
...
23818
18283_HO

answered Aug 13, 2012 at 21:10

stranac

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Comments

Malhelo · Accepted Answer · 2012-08-13 21:10:41Z

0

There are many solutions. If all paths are like 9999\New_folder\A#number#\files\ then your could simply take a substring by finding the third last and seconds last "\". You can do that by using rfind() (http://docs.python.org/library/string.html#string.rfind)

Another, more general way is the use of regular expressions. http://docs.python.org/library/re.html

answered Aug 13, 2012 at 21:10

Malhelo

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Comments

Vidul · Accepted Answer · 2012-08-13 21:23:38Z

0

#sm.th. like this should work:
file_handler = open("file path")
for line in file_handler:   
    re.search(r'\\(.[^\\]+)\\files', line).groups(0)[0]

answered Aug 13, 2012 at 21:23

Vidul

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