Unzipping files in Python

Question

I read through the zipfile documentation, but couldn’t understand how to unzip a file, only how to zip a file. How do I unzip all the contents of a zip file into the same directory?

Asked By: John Howard

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Source

Answer 1

import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
    zip_ref.extractall(directory_to_extract_to)

That’s pretty much it!

Answered By: Rahul

Answer 2

Use the extractall method, if you’re using Python 2.6+

zip = ZipFile('file.zip')
zip.extractall()

Answered By: Dan Breen

Answer 3

If you are using Python 3.2 or later:

import zipfile
with zipfile.ZipFile("file.zip","r") as zip_ref:
    zip_ref.extractall("targetdir")

You dont need to use the close or try/catch with this as it uses the
context manager construction.

Answered By: user1741137

Answer 4

import os 
zip_file_path = "C:AABB"
file_list = os.listdir(path)
abs_path = []
for a in file_list:
    x = zip_file_path+'\'+a
    print x
    abs_path.append(x)
for f in abs_path:
    zip=zipfile.ZipFile(f)
    zip.extractall(zip_file_path)

This does not contain validation for the file if its not zip. If the folder contains non .zip file it will fail.

Answered By: user3911901

Answer 5

You can also import only ZipFile:

from zipfile import ZipFile
zf = ZipFile('path_to_file/file.zip', 'r')
zf.extractall('path_to_extract_folder')
zf.close()

Works in Python 2 and Python 3.

Answered By: simhumileco

Answer 6

try this :


import zipfile
def un_zipFiles(path):
    files=os.listdir(path)
    for file in files:
        if file.endswith('.zip'):
            filePath=path+'/'+file
            zip_file = zipfile.ZipFile(filePath)
            for names in zip_file.namelist():
                zip_file.extract(names,path)
            zip_file.close()

path : unzip file’s path

Answered By: Done Jin

Answer 7

from zipfile import ZipFile
ZipFile("YOURZIP.zip").extractall("YOUR_DESTINATION_DIRECTORY")

The directory where you will extract your files doesn’t need to exist before, you name it at this moment

YOURZIP.zip is the name of the zip if your project is in the same directory.
If not, use the PATH i.e : C://….//YOURZIP.zip

Think to escape the / by an other / in the PATH
If you have a permission denied try to launch your ide (i.e: Anaconda) as administrator

YOUR_DESTINATION_DIRECTORY will be created in the same directory than your project

Answered By: Nicoolasens

Answer 8

If you want to do it in shell, instead of writing code.

 python3 -m zipfile -e myfiles.zip myfiles/

myfiles.zip is the zip archive and myfiles is the path to extract the files.

Answered By: Indika K

Answer 9

zipfile is a somewhat low-level library. Unless you need the specifics that it provides, you can get away with shutil‘s higher-level functions make_archive and unpack_archive.

make_archive is already described in this answer. As for unpack_archive:

import shutil
shutil.unpack_archive(filename, extract_dir)

unpack_archive detects the compression format automatically from the "extension" of filename (.zip, .tar.gz, etc), and so does make_archive. Also, filename and extract_dir can be any path-like objects (e.g. pathlib.Path instances) since Python 3.7.

Answered By: fonini

Unzipping files in Python

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