# zip – Unzipping files in Python

## The Question :

506 people think this question is useful

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?

The Question Comments :

## The Answer 1

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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!

## The Answer 2

324 people think this answer is useful

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.

## The Answer 3

43 people think this answer is useful

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

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



## The Answer 4

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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.

## The Answer 5

11 people think this answer is useful

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.

## The Answer 6

10 people think this answer is useful

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

## The Answer 7

4 people think this answer is useful
import os
zip_file_path = "C:\AA\BB"
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.

## The Answer 8

4 people think this answer is useful
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

## The Answer 9

3 people think this answer is useful

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.