Compressing directory using shutil.make_archive() while preserving directory structure

I have written a wrapper function myself because shutil.make_archive is too confusing to use.

Here it is http://www.seanbehan.com/how-to-use-python-shutil-make_archive-to-zip-up-a-directory-recursively-including-the-root-folder/

And just the code..

import os, shutil
def make_archive(source, destination):
        base = os.path.basename(destination)
        name = base.split('.')[0]
        format = base.split('.')[1]
        archive_from = os.path.dirname(source)
        archive_to = os.path.basename(source.strip(os.sep))
        shutil.make_archive(name, format, archive_from, archive_to)
        shutil.move('%s.%s'%(name,format), destination)

make_archive('/path/to/folder', '/path/to/folder.zip')

I think, I am able to improve seanbehan’s answer by removing the file moving:

def make_archive(source, destination):
    base_name = '.'.join(destination.split('.')[:-1])
    format = destination.split('.')[-1]
    root_dir = os.path.dirname(source)
    base_dir = os.path.basename(source.strip(os.sep))
    shutil.make_archive(base_name, format, root_dir, base_dir)

There are basically 2 approaches to using shutil: you may try to understand the logic behind it or you may just use an example. I couldn’t find an example here so I tried to create my own.

;TLDR. Run shutil.make_archive('dir1_arc', 'zip', root_dir='dir1') or shutil.make_archive('dir1_arc', 'zip', base_dir='dir1') or just shutil.make_archive('dir1_arc', 'zip', 'dir1') from temp.

Suppose you have ~/temp/dir1:

temp $ tree dir1
dir1
├── dir11
│   ├── file11
│   ├── file12
│   └── file13
├── dir1_arc.zip
├── file1
├── file2
└── file3

How can you create an archive of dir1? Set base_name='dir1_arc', format='zip'. Well you have a lot of options:

• cd into dir1 and run shutil.make_archive(base_name=base_name, format=format); it will create an archive dir1_arc.zip inside dir1; the only problem you’ll get a strange behavior: inside your archive you’ll find file dir1_arc.zip;
• from temp run shutil.make_archive(base_name=base_name, format=format, base_dir='dir1'); you’ll get dir1_arc.zip inside temp that you can unzip into dir1; root_dir defaults to temp;
• from ~ run shutil.make_archive(base_name=base_name, format=format, root_dir='temp', base_dir='dir1'); you’ll again get your file but this time inside ~ directory;
• create another directory temp2 in ~ and run inside it: shutil.make_archive(base_name=base_name, format=format, root_dir='../temp', base_dir='dir1'); you’ll get your archive in this temp2 folder;

Can you run shutil without specifying arguments? You can. Run from temp shutil.make_archive('dir1_arc', 'zip', 'dir1'). This is the same as run shutil.make_archive('dir1_arc', 'zip', root_dir='dir1'). What can we say about base_dir in this case? From documentation not so much. From source code we may see that:

if root_dir is not None:
  os.chdir(root_dir)

if base_dir is None:
        base_dir = os.curdir 

So in our case base_dir is dir1. And we can keep asking questions.

I was having issues with path split on some paths with ‘.’ periods in them and i found having an optional format which defaults to ‘zip’ is handy and still allows you to override for other formats and is less error prone.

import os
import shutil
from shutil import make_archive

def make_archive(source, destination, format='zip'):
    import os
    import shutil
    from shutil import make_archive
    base, name = os.path.split(destination)
    archive_from = os.path.dirname(source)
    archive_to = os.path.basename(source.strip(os.sep))
    print(f'Source: {source}nDestination: {destination}nArchive From: {archive_from}nArchive To: {archive_to}n')
    shutil.make_archive(name, format, archive_from, archive_to)
    shutil.move('%s.%s' % (name, format), destination)

make_archive('/path/to/folder', '/path/to/folder.zip')

Special thanks to seanbehan’s original answer or i would have been lost in the sauce much longer.

This solution builds off the responses from irudyak and seanbehan and uses Pathlib. You need to pass source and destination as Path objects.

from pathlib import Path
import shutil

def make_archive(source, destination):
    base_name = destination.parent / destination.stem
    format = (destination.suffix).replace(".", "")
    root_dir = source.parent
    base_dir = source.name
    shutil.make_archive(base_name, format, root_dir, base_dir)

This is a variation on @nick’s answer that uses pathlib, type hinting and avoids shadowing builtins:

from pathlib import Path
import shutil

def make_archive(source: Path, destination: Path) -> None:
    base_name = destination.parent / destination.stem
    fmt = destination.suffix.replace(".", "")
    root_dir = source.parent
    base_dir = source.name
    shutil.make_archive(str(base_name), fmt, root_dir, base_dir)

Usage:

make_archive(Path("/path/to/dir/"), Path("/path/to/output.zip"))

You can use Pathlib and shutil:

from pathlib import Path
import shutil
shutil.make_archive(
   *dest_path.split('.'), 
   root_dir=Path(src_path).parent, 
   base_dir=Path(src_path).name)
)

• src_path is the path of the source directory.
• dest_path is the path of the destination archive to be created.