Extracting a single file from a zip archive without iterating over the entire name list in Python

Question

I have a zip file with a folder in it, like this:

some.zip/
    some_folder/
         some.xml
         ...

I am using the zipfile library. What I want is to open only some.xml file, but I don't now the some_folder name. My solution looks like this:

    def get_xml(zip_file):
        for filename in zip_file.namelist():
            if filename.endswith('some.xml'):
                return zip_file.open(filename)

I would like to know if there is a better solution other than scanning the entire list.

Answer 1

This prints the list of directories inside the test.zip file:

from zipfile import ZipFile


with ZipFile('test.zip', 'r') as f:
    directories = [item for item in f.namelist() if item.endswith('/')]
    print directories

If you know that there is only one directory inside, just take the first item: directories[0].

Answer 2

Do you want to get directory that containing some.xml?

import os
import zipfile

with zipfile.ZipFile('a.zip', 'r') as zf:
    for name in zf.namelist():
        if os.path.basename(name) == 'some.xml':
            print os.path.dirname(name)