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?

Solution 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!

Solution 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.

Solution 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.

Solution 4

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

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

Solution 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.

View more solutions

Comments

[asonnenschein]

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

[Dan Gayle]

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

[iratzhash]

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?

[FelixEnescu]

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

[Brian Leishman]

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

[Dave Forgac]

@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.

[Debug255]

@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).

[Debug255]

@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"

[Agile Bean]

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

[simhumileco]

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

[SnowOnion]

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)?

[ntg]

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

[Martin Thoma]

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

[Umar.H]

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

[simhumileco]

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?

[MikeF]

@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.

[Boris Verkhovskiy]

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

[Ben Dalling]

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

[Charlie Parker]

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

[Charlie Parker]

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)?

[Perry]

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)

[Nicoolasens]

to me it's strictly similar

[msoutopico]

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)).

[fonini]

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

[Fareanor]

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

[Charlie Parker]

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

[Charlie Parker]

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

[fonini]

@CharlieParker you have already asked the same thing in a comment to another answer, and that comment was answered: stackoverflow.com/questions/3451111/unzipping-files-in-pytho‌​n/… 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.

[Ryan Loggerythm]

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

[tista3]

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.

[Selfcontrol7]

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