Unzip all zipped files in a folder to that same folder using Python 2.7.5

Question

I would like to write a simple script to iterate through all the files in a folder and unzip those that are zipped (.zip) to that same folder. For this project, I have a folder

Answer 1

You need to construct a ZipFile object with the filename, and then extract it:

    zipfile.ZipFile.extract(item)

is wrong.

    zipfile.ZipFile(item).extractall()

will extract all files from the zip file with the name contained in item.

I think you should more closely read the documentation to zipfile :) but you're on the right track!

Answer 2

I think this is shorter and worked fine for me. First import the modules required:

import zipfile, os

Then, I define the working directory:

working_directory = 'my_directory'
os.chdir(working_directory)

After that you can use a combination of the os and zipfile to get where you want:

for file in os.listdir(working_directory):   # get the list of files
    if zipfile.is_zipfile(file): # if it is a zipfile, extract it
        with zipfile.ZipFile(file) as item: # treat the file as a zip
           item.extractall()  # extract it in the working directory

Answer 3

The accepted answer works great!

Just to extend the idea to unzip all the files with .zip extension within all the sub-directories inside a directory the following code seems to work well:

import os
import zipfile

for path, dir_list, file_list in os.walk(dir_path):
    for file_name in file_list:
        if file_name.endswith(".zip"):
            abs_file_path = os.path.join(path, file_name)

            # The following three lines of code are only useful if 
            # a. the zip file is to unzipped in it's parent folder and 
            # b. inside the folder of the same name as the file

            parent_path = os.path.split(abs_file_path)[0]
            output_folder_name = os.path.splitext(abs_file_path)[0]
            output_path = os.path.join(parent_path, output_folder_name)

            zip_obj = zipfile.ZipFile(abs_file_path, 'r')
            zip_obj.extractall(output_path)
            zip_obj.close()

Answer 4

Below is the code that worked for me:

import os, zipfile

dir_name = 'C:\\SomeDirectory'
extension = ".zip"

os.chdir(dir_name) # change directory from working dir to dir with files

for item in os.listdir(dir_name): # loop through items in dir
    if item.endswith(extension): # check for ".zip" extension
        file_name = os.path.abspath(item) # get full path of files
        zip_ref = zipfile.ZipFile(file_name) # create zipfile object
        zip_ref.extractall(dir_name) # extract file to dir
        zip_ref.close() # close file
        os.remove(file_name) # delete zipped file

Looking back at the code I had amended, the directory was getting confused with the directory of the script.

The following also works while not ruining the working directory. First remove the line

os.chdir(dir_name) # change directory from working dir to dir with files

Then assign file_name as

file_name = dir_name + "/" + item