How to use regex in python in getting a string between two characters?

Question

I have this as my input

content = 'abc.zip'\n

I want to take out abc out of it . How do I do it using regex in python ?

Edit :

No this is not a homework question . I am trying to automate something and I am stuck at a certain point so that I can make the automate generic to any zip file I have .

os.system('python unzip.py -z data/ABC.zip -o data/')

After I take in the zip file , I unzip it . I am planning to make it generic , by getting the filename from the directory the zip file was put in and then provide the file name to the upper stated syntax to unzip it

More details please. Is it always the string 'abc' you want removed? Or is it everything before the dot? Is this a filename, or just a random string? The best tool for the job (almost certainly not regular expressions) depends on these details. — Blair
– Blair, Commented Jun 12, 2011 at 23:13
If this is a homework assignment, you should tag it as homework. — Ryan Ginstrom
– Ryan Ginstrom, Commented Jun 12, 2011 at 23:25
To be clear, is it content = 'abc.zip'\n (with that newline outside the quotes) or input = r"content = 'abc.zip'\n"? — johnsyweb
– johnsyweb, Commented Jun 12, 2011 at 23:50
@Blair I have edited my question . A non re approach can also do . — IamH1kc
– IamH1kc, Commented Jun 12, 2011 at 23:56
@Blair's answer solves the problem of extracting the filename. Calling ZipFile.extractall() sounds like a better plan! — johnsyweb
– johnsyweb, Commented Jun 13, 2011 at 0:21

Blair · Accepted Answer · 2011-06-12 23:36:26Z

4

As I implied in my comment, regular expressions are unlikely to be the best tool for the job (unless there is some artificial restriction on the problem, or it is far more complex than your example). The standard string and/or path libraries provide functions which should do what you are after. To better illustrate how these work, I'll use the following definition of content instead:

>>> content = 'abc.def.zip'

If its a file, and you want the name and extension:

>>> import os.path
>>> filename, extension = os.path.splitext(content)
>>> print filename
abc.def
>>> print extension
.zip

If it is a string, and you want to remove the substring 'abc':

>>> noabc = content.replace('abc', '')
>>> print noabc
.def.zip

If you want to break it up on each occurrence of a period;

>>> broken = content.split('.')
>>> print broken
['abc', 'def', 'zip']

If it has multiple periods, and you want to break it on the first or last one:

>>> broken = content.split('.', 1)
>>> print broken
['abc', 'def.zip']
>>> broken = content.rsplit('.', 1)
>>> print broken
['abc.def', 'zip']

answered Jun 12, 2011 at 23:36

Blair

15.9k7 gold badges48 silver badges56 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

IamH1kc Over a year ago

How do I insert the filename in this statement os.system('python unzip.py -z data/Abc.zip -o data/') . In the place of Abc.zip , that file name

rzetterberg · Accepted Answer · 2011-06-12 23:55:52Z

1

Edit: Changed the regexp to match for "content = 'abc.zip\n'" instead of the string "abc.zip".

import re 

#Matching for "content = 'abc.zip\n'"
matches = re.match("(?P<filename>.*).zip\n'$", "content = 'abc.zip\n'")
matches = matches.groupdict()
print matches

#Matching for "abc.zip"    
matches = re.match("(?P<filename>.*).zip$", "abc.zip")
matches = matches.groupdict()
print matches

Output:

{'filename': 'abc'}

This will print the matches of everything before .zip. You can access everything like a regular dictionary.

edited Jun 12, 2011 at 23:55

answered Jun 12, 2011 at 23:15

rzetterberg

10.3k5 gold badges47 silver badges55 bronze badges

2 Comments

rzetterberg Over a year ago

I thought that content was a variable. And that the string was "abc.zip". I'll edit my answer after that the string is what you wrote.

rzetterberg Over a year ago

@Johnysweb Yeah, it's very unclear :) I'll have to edit my answer to match for both.

Mr Fooz · Accepted Answer · 2011-06-12 23:26:17Z

0

If you're trying to break up parts of a path, you may find the os.path module to be useful. It has nice abstractions with clear semantics that are easy to use.

answered Jun 12, 2011 at 23:26

Mr Fooz

112k7 gold badges77 silver badges106 bronze badges

2 Comments

IamH1kc Over a year ago

Ok then I would tell the bigger problem here . My file structure is this data/Abc.zip . My script doesn't really know the name of the zip file . How do I unzip the file using os.path and also extract the name of the zip file

johnsyweb Over a year ago

@user794916: Please update your question to include all the pertinent information.

Collectives™ on Stack Overflow

How to use regex in python in getting a string between two characters?

3 Answers 3

1 Comment

2 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

2 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related