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?

Related: unzipping files recursively https://stackoverflow.com/questions/28339000/unzip-zip-files-in-folders-and-subfolders-with-python & https://stackoverflow.com/questions/36285502/how-to-extract-zip-file-recursively-in-python — Tung, Feb 26 '20 at 18:33
For a one line extraction, see [`shutil.unpack_archive()`](https://docs.python.org/3/library/shutil.html#shutil.unpack_archive). — phoenix, Dec 23 '20 at 13:05
@fonini answer is -- as of 2021 -- the right/best one: https://stackoverflow.com/a/64110098/687896 — Brandt, Jun 07 '21 at 06:06

Rahul · Accepted Answer · 2019-09-24T20:09:08.683

1577

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!

edited Sep 24 '19 at 20:09

answered Aug 10 '10 at 16:23

Rahul

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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? – Irtaza Apr 07 '16 at 16:59
13

@iratzhash I typically create a new temporary directory for the contents using tempfile: https://docs.python.org/3/library/tempfile.html I unzip to the temporary directory and the move / organize the files from there. – Dave Forgac Jul 12 '17 at 17:28
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@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 Feb 13 '18 at 02:00
@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](https://stackoverflow.com/questions/7428318/rename-files-in-zip-folder-using-zipmodule?lq=1) by "bouke" – Debug255 Feb 13 '18 at 02:03
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thnx, note: There is no zipfile library, no need to pip install, zipfile is already there... – ntg Apr 09 '19 at 09:06
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I'd add: `from tempfile import mkdtemp; directory_to_extract_to = mkdtemp()` – Martin Thoma Jul 31 '19 at 15:23
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for passwords use: with ZipFile(zip_file_path, 'r') as zipObj: zipObj.extractall(output_folder, pwd=b'myPassword') – Ryan Loggerythm Aug 29 '21 at 00:42
This minimal code will still do the job without problem: `z = ZipFile(path_to_zip_file)` `z.extractall(directory_to_extract_to)` – Selfcontrol7 Feb 03 '22 at 20:05

score 411 · Answer 2 · edited Sep 13 '18 at 17:32

411

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.

edited Sep 13 '18 at 17:32

simhumileco

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answered Apr 16 '16 at 10:11

user1741137

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ZipFile also works as a context manager in 2.7 or later: https://docs.python.org/2/library/zipfile.html#zipfile.ZipFile – FelixEnescu Jan 22 '17 at 15:01
How to deal with https://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)? – SnowOnion Mar 06 '19 at 08:24
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`zipfile` + `pathlib` = win. mind if i slightly update your answer? – Umar.H Jan 21 '20 at 15:38
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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)`? – Charlie Parker Jan 21 '21 at 20:09
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@CharlieParker The main reason is portability. `system` calls are OS dependent. For example, `tar` would not be available on Windows. – Fareanor May 18 '21 at 16:24
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. – tista3 Oct 25 '21 at 09:53

score 148 · Answer 3 · answered Sep 28 '20 at 21:51

148

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 Sep 28 '20 at 21:51

fonini

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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))`. – msoutopico Mar 03 '21 at 21:00
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@msoutopico you can specify thje format explicitly: `shutil.unpack_archive(filename, extract_dir, format)` – fonini Mar 04 '21 at 19:42
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what is wrong with `os.system(f'tar -xvzf {path2zip} -C {path2unzip}/')`? – Charlie Parker Aug 10 '21 at 18:49
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@CharlieParker you have already asked the same thing in a comment to another answer, and that comment was answered: https://stackoverflow.com/questions/3451111/unzipping-files-in-python/64110098?noredirect=1#comment119468939_36662770 `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. – fonini Aug 10 '21 at 18:58
This solution doesn’t seem to maintain executable bits on the stuff inside the archive as it’s unzipped? – Cody Martin Jan 26 '23 at 00:51

score 49 · Answer 4 · answered Aug 10 '10 at 16:23

49

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

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

answered Aug 10 '10 at 16:23

Dan Breen

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Don't you have to specify a destination (zip.extractall(destination))? – asonnenschein Oct 24 '13 at 18:19
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Not if you're just extracting into the same directory as the zipfile – Dan Gayle Dec 04 '13 at 20:01
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@DanGayle this appears to be extracting the zip file into the current working directory, NOT the location of the zip file – Brian Leishman Jun 09 '17 at 14:41
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for me, ZipFile() didn't work but zipfile.ZipFile() did - after import zipfile – Agile Bean Sep 29 '18 at 07:34
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You need to `zip.close()` at the end if you don't use a `with` statement like the other answers suggest. – Boris Verkhovskiy Jun 22 '20 at 17:52
what is wrong with `os.system(f'tar -xvzf {path2zip} -C {path2unzip}/')`? – Charlie Parker Aug 10 '21 at 18:49
to which directory is this extracted to? – Bionix1441 Mar 18 '23 at 22:16

simhumileco · Answer 5 · 2018-09-13T16:40:18.403

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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.

edited Sep 13 '18 at 16:40

answered Sep 13 '18 at 16:28

simhumileco

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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 – simhumileco Oct 30 '18 at 08:22
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? – simhumileco Jan 28 '20 at 14:09
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@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. – MikeF Feb 26 '20 at 16:38
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@MikeF I had the same problem with Python 3.8.5 but the workaround was to use `from zipfile import ZipFile`. Hope this helps. – Ben Dalling Oct 15 '20 at 09:53

score 14 · Answer 6 · answered Jul 03 '19 at 12:53

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

Nicoolasens · Answer 7 · 2020-07-02T14:31:00.847

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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

edited Jul 02 '20 at 14:31

answered Jul 02 '20 at 14:06

Nicoolasens

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to me it's strictly similar – Nicoolasens Jan 25 '21 at 12:00

score 12 · Answer 8 · answered Aug 05 '20 at 00:20

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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 Aug 05 '20 at 00:20

Indika K

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why not just do `tar -xvzf path_file` in your case? – Charlie Parker Jan 21 '21 at 19:39
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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`) – Perry Jan 22 '21 at 19:00

score 2 · Answer 9 · answered Apr 13 '18 at 10:40

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.

Unzipping files in Python

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