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?

Udall

Related: unzipping files recursively #28339500 & #36286002 – Tiannatiara

For a one line extraction, see shutil.unpack_archive(). – Genesa

@fonini answer is -- as of 2021 -- the right/best one: https://mcmap.net/q/53544/-unzipping-files-in-python – Anterior

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!

Coreen

what if the contents of the .zip archive are same, in all .zip archives? how to rename the content before extracting? example: 1.zip 2.zip.. all contain content.txt : extract all like 1content.txt 2content.txt? – Sianna

@iratzhash I typically create a new temporary directory for the contents using tempfile: docs.python.org/3/library/tempfile.html I unzip to the temporary directory and the move / organize the files from there. – Guardado

@3kstc I would from zipfile import ZipFile. When using it, you no longer need to use zipfile.ZipFile, and can use ZipFile(zip_file_name). – Pinup

@iratzhash I realize you commented 1.5 years ago. But just so others know, usually contents within a zip file are read-only. A good answer is here by "bouke" – Pinup

thnx, note: There is no zipfile library, no need to pip install, zipfile is already there... – Multitude

I'd add: from tempfile import mkdtemp; directory_to_extract_to = mkdtemp() – Lorianne

for passwords use: with ZipFile(zip_file_path, 'r') as zipObj: zipObj.extractall(output_folder, pwd=b'myPassword') – Kessia

This minimal code will still do the job without problem: z = ZipFile(path_to_zip_file) z.extractall(directory_to_extract_to) – Perjure

Answer 2

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.

Endoparasite

ZipFile also works as a context manager in 2.7 or later: docs.python.org/2/library/zipfile.html#zipfile.ZipFile – Trueblood

How to deal with docs.python.org/3.6/library/zipfile.html#zipfile.BadZipFile exception? Generally, what is the best practice to use try/except with context manager (with-statement)? – Pisciform

zipfile + pathlib = win. mind if i slightly update your answer? – Aoudad

is there a reason to avoid os.system(f'tar -xvzf {filename}') and instead use zipfile (e.g. zip = ZipFile('file.zip'); zip.extractall() ) or shutil.unpack_archive(filename, extract_dir)? – Stabler

@CharlieParker The main reason is portability. system calls are OS dependent. For example, tar would not be available on Windows. – Ainslee

I do not recommend zipfile. I had problems to extract files added in existing zip file. This can be done with total commander, which addz new file entry and file database at the end of the file. zipfile extracted only the old files. shutil.unpack_archive does not have this problem. – Welton

Answer 3

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.

Schifra

This method doesn't work when the zip file has a custom extension, e.g. (.omt for OmegaT project packages). It gives raise ReadError("Unknown archive format '{0}'".format(filename)). – Zygo

@Zygo you can specify thje format explicitly: shutil.unpack_archive(filename, extract_dir, format) – Schifra

what is wrong with os.system(f'tar -xvzf {path2zip} -C {path2unzip}/')? – Stabler

@CharlieParker you have already asked the same thing in a comment to another answer, and that comment was answered: https://mcmap.net/q/53544/-unzipping-files-in-python/… os.system is not portable, opens up security issues, is harder to use correctly (e.g. your proposal fails when the paths have special characters), and is less readable. – Schifra

This solution doesn’t seem to maintain executable bits on the stuff inside the archive as it’s unzipped? – Valerie

Answer 4

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

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

Faun

Don't you have to specify a destination (zip.extractall(destination))? – Vindicate

Not if you're just extracting into the same directory as the zipfile – Ximenez

@DanGayle this appears to be extracting the zip file into the current working directory, NOT the location of the zip file – Deth

for me, ZipFile() didn't work but zipfile.ZipFile() did - after import zipfile – Goethite

You need to zip.close() at the end if you don't use a with statement like the other answers suggest. – Caesaria

what is wrong with os.system(f'tar -xvzf {path2zip} -C {path2unzip}/')? – Stabler

to which directory is this extracted to? – Voroshilovsk

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.

Stove

Thank you for your attention @MylesHollowed However, this is not a copy from the accepted answer. I agree that they are similar to each other, but they are different. This is also indicated by your comment, because the accepted one is definitely better for you than mine. If it was a copy, it would be the same... For someone my answer may be valuable because it is perhaps more readable and as you noticed import less code... It is because of these differences that I decided to put my answer to give an alternative. Is not that why we can put other answers after accepting one? All the best – Stove

What's wrong with this answer? Why did someone give her a negative point? After all, it is the answer to the question and is distinguished by its simplicity compared to other answers, which may be important for some people who are looking for an answer. Isn't it? – Stove

@MylesHollowed import zipfile.ZipFile generates ModuleNotFoundError: No module named 'zipfile.ZipFile'; 'zipfile' is not a package in 3.6.5. I am open to it being operator error on my part, but I don't know what it is. – Mariettamariette

@Mariettamariette I had the same problem with Python 3.8.5 but the workaround was to use from zipfile import ZipFile. Hope this helps. – Contessacontest

Answer 6

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.

Deepset

why not just do tar -xvzf path_file in your case? – Stabler

ZIP files are not tar files. Unless you have a special version of tar which handles ZIP archives, your command wont work at all. tar with the -z option processes gzipped tar archives (generally files with extensions .tgz or .tar.gz) – Assamese

Answer 7

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

Nidify

Answer 8

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

Oscitant

to me it's strictly similar – Oscitant

Answer 9

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.

Transference