How to unzip multiple zip files in a folder using python?

Tutorial: Extracting Zip Files

In this tutorial, we will learn how to extract all the Zip files in a directory using Python's zipfile module and the pathlib module. We'll go through the code step by step and discuss different ways to encapsulate this functionality in a function.

Code Explanation

import zipfile from pathlib import Path # Set the target directory p = Path('.') # Iterate over all Zip files in the directory for f in p.glob('*.zip'): # Open the Zip file  with zipfile.ZipFile(f, 'r') as archive: # Extract all contents of the Zip file to a directory with the same name as the file  archive.extractall(path=f'./{f.stem}') # Print a message indicating that the extraction is complete  print(f"Extracted contents from '{f.name}' to '{f.stem}' directory.") 

The code performs the following steps:

1. Import the necessary modules: zipfile for working with Zip files and pathlib for working with file paths.

2. Create a Path object to represent the current directory. You can modify this to specify a different directory if needed.

3. Use the glob method of the Path object to iterate over all the Zip files (*.zip) in the current directory.

4. For each Zip file, open it using zipfile.ZipFile in read mode ('r').

5. Use the extractall method of the Zip file object to extract all the contents of the Zip file to a directory with the same name as the file. The path=f'./{f.stem}' argument specifies the destination directory.

6. After the extraction is complete, print a message indicating that the process is done, using print(f"Extracted contents from '{f.name}' to '{f.stem}' directory.").

Recipe 1: Function with Directory Parameter

import zipfile from pathlib import Path def extract_zip_files(directory): p = Path(directory) for f in p.glob('*.zip'): with zipfile.ZipFile(f, 'r') as archive: archive.extractall(path=f'./{f.stem}') print(f"Extracted contents from '{f.name}' to '{f.stem}' directory.") # Usage example extract_zip_files('.') 

This recipe demonstrates creating a function extract_zip_files that accepts a directory parameter. It performs the same functionality as the previous code but allows you to specify a different directory.

Recipe 2: Function with Output Path

import zipfile from pathlib import Path def extract_zip_files(directory, output_path): p = Path(directory) for f in p.glob('*.zip'): with zipfile.ZipFile(f, 'r') as archive: archive.extractall(path=output_path) print(f"Extracted contents from '{f.name}' to '{output_path}' directory.") # Usage example extract_zip_files('.', './extracted_files') 

This recipe demonstrates creating a function extract_zip_files that accepts both a directory parameter and an output_path parameter. It allows you to specify a custom output directory where the extracted files will be stored.