How can I create a zip archive of a directory structure in Python?
27 Answers
Use shutil, which is part of python standard library set. Using shutil is so simple(see code below):
- 1st arg: Filename of resultant zip/tar file,
- 2nd arg: zip/tar,
- 3rd arg: dir_name
Code:
import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')
With python 3.9, pathlib & zipfile module you can create a zip files from anywhere in the system.
def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
"""Zip the provided directory without navigating to that directory using `pathlib` module"""
# Convert to Path object
dir = Path(dir)
with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
for entry in dir.rglob("*"):
zip_file.write(entry, entry.relative_to(dir))
It is neat, typed, and has less code.
To retain the folder hierarchy under the parent directory to be archived:
import glob
import os
import zipfile
with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
for fp in glob(os.path.join(parent, "**/*")):
base = os.path.commonpath([parent, fp])
zipf.write(fp, arcname=fp.replace(base, ""))
If you want, you could change this to use pathlib for file globbing.
So many answers here, and I hope I might contribute with my own version, which is based on the original answer (by the way), but with a more graphical perspective, also using context for each zipfile setup and sorting os.walk(), in order to have a ordered output.
Having these folders and them files (among other folders), I wanted to create a .zip for each cap_ folder:
$ tree -d
.
├── cap_01
| ├── 0101000001.json
| ├── 0101000002.json
| ├── 0101000003.json
|
├── cap_02
| ├── 0201000001.json
| ├── 0201000002.json
| ├── 0201001003.json
|
├── cap_03
| ├── 0301000001.json
| ├── 0301000002.json
| ├── 0301000003.json
|
├── docs
| ├── map.txt
| ├── main_data.xml
|
├── core_files
├── core_master
├── core_slave
Here's what I applied, with comments for better understanding of the process.
$ cat zip_cap_dirs.py
""" Zip 'cap_*' directories. """
import os
import zipfile as zf
for root, dirs, files in sorted(os.walk('.')):
if 'cap_' in root:
print(f"Compressing: {root}")
# Defining .zip name, according to Capítulo.
cap_dir_zip = '{}.zip'.format(root)
# Opening zipfile context for current root dir.
with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:
# Iterating over os.walk list of files for the current root dir.
for f in files:
# Defining relative path to files from current root dir.
f_path = os.path.join(root, f)
# Writing the file on the .zip file of the context
new_zip.write(f_path)
Basically, for each iteration over os.walk(path), I'm opening a context for zipfile setup and afterwards, iterating iterating over files, which is a list of files from root directory, forming the relative path for each file based on the current root directory, appending to the zipfile context which is running.
And the output is presented like this:
$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03
To see the contents of each .zip directory, you can use less command:
$ less cap_01.zip
Archive: cap_01.zip
Length Method Size Cmpr Date Time CRC-32 Name
-------- ------ ------- ---- ---------- ----- -------- ----
22017 Defl:N 2471 89% 2019-09-05 08:05 7a3b5ec6 cap_01/0101000001.json
21998 Defl:N 2471 89% 2019-09-05 08:05 155bece7 cap_01/0101000002.json
23236 Defl:N 2573 89% 2019-09-05 08:05 55fced20 cap_01/0101000003.json
-------- ------- --- -------
67251 7515 89% 3 files
To give more flexibility, e.g. select directory/file by name use:
import os
import zipfile
def zipall(ob, path, rel=""):
basename = os.path.basename(path)
if os.path.isdir(path):
if rel == "":
rel = basename
ob.write(path, os.path.join(rel))
for root, dirs, files in os.walk(path):
for d in dirs:
zipall(ob, os.path.join(root, d), os.path.join(rel, d))
for f in files:
ob.write(os.path.join(root, f), os.path.join(rel, f))
break
elif os.path.isfile(path):
ob.write(path, os.path.join(rel, basename))
else:
pass
For a file tree:
.
├── dir
│ ├── dir2
│ │ └── file2.txt
│ ├── dir3
│ │ └── file3.txt
│ └── file.txt
├── dir4
│ ├── dir5
│ └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip
You can e.g. select only dir4 and root.txt:
cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]
with zipfile.ZipFile("selective.zip", "w" ) as myzip:
for f in files:
zipall(myzip, f)
Or just listdir in script invocation directory and add everything from there:
with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
for f in os.listdir():
if f == "listdir.zip":
# Creating a listdir.zip in the same directory
# will include listdir.zip inside itself, beware of this
continue
zipall(myzip, f)
Say you want to Zip all the folders(sub directories) in the current directory.
for root, dirs, files in os.walk("."):
for sub_dir in dirs:
zip_you_want = sub_dir+".zip"
zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
zip_process.write(file_you_want_to_include)
zip_process.close()
print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
Zip a file or a tree (a directory and its sub-directories).
from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED
def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
paths = [Path(tree_path)]
while paths:
p = paths.pop()
if p.is_dir():
paths.extend(p.iterdir())
if skip_empty_dir:
continue
zf.write(p)
To append to an existing archive, pass mode='a', to create a fresh archive mode='w' (the default in the above). So let's say you want to bundle 3 different directory trees under the same archive.
make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')
A solution using pathlib.Path, which is independent of the OS used:
import zipfile
from pathlib import Path
def zip_dir(path: Path, zip_file_path: Path):
"""Zip all contents of path to zip_file"""
files_to_zip = [
file for file in path.glob('*') if file.is_file()]
with zipfile.ZipFile(
zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
for file in files_to_zip:
print(file.name)
zip_f.write(file, file.name)
current_dir = Path.cwd()
zip_dir = current_dir / "test"
tools.zip_dir(
zip_dir, current_dir / 'Zipped_dir.zip')
The obvious way to go would be to go with shutil, Like the second top answer says so, But if you still wish to go with ZipFile for some reason, And if you are getting some trouble doing that (Like ERR 13 in Windows etc), You can use this fix:
import os
import zipfile
def retrieve_file_paths(dirName):
filePaths = []
for root, directories, files in os.walk(dirName):
for filename in files:
filePath = os.path.join(root, filename)
filePaths.append(filePath)
return filePaths
def main(dir_name, output_filename):
filePaths = retrieve_file_paths(dir_name)
zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
main("my_dir", "my_dir_archived")
This one recursively iterates through every sub-folder/file in your given folder, And writes them to a zip file instead of attempting to directly zip a folder.
Well, after reading the suggestions I came up with a very similar way that works with 2.7.x without creating "funny" directory names (absolute-like names), and will only create the specified folder inside the zip.
Or just in case you needed your zip to contain a folder inside with the contents of the selected directory.
def zipDir( path, ziph ) :
"""
Inserts directory (path) into zipfile instance (ziph)
"""
for root, dirs, files in os.walk( path ) :
for file in files :
ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )
def makeZip( pathToFolder ) :
"""
Creates a zip file with the specified folder
"""
zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
zipDir( pathToFolder, zipf )
zipf.close()
print( "Zip file saved to: " + pathToFolder)
makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )
Function to create zip file.
def CREATEZIPFILE(zipname, path):
#function to create a zip file
#Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file
zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
zipf.setpassword(b"password") #if you want to set password to zipfile
#checks if the path is file or directory
if os.path.isdir(path):
for files in os.listdir(path):
zipf.write(os.path.join(path, files), files)
elif os.path.isfile(path):
zipf.write(os.path.join(path), path)
zipf.close()
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