One way with zipfile is to filter the files returned from namelist() and then use that list with extractall(). archive = 'archive.zip' directory = './' extensions = ('.txt', '.pdf') zip_file = …

Sep 7, 2017 · Simply pass the zipfile object to the extractor as param. You shouldn't try to parse the filepath out of string representation of the list - that is most likely what causes problem. Try …

If you want to list all files with the specified extension in a certain directory and its subdirectories you could do: import os def filterFiles(path, extension): return [file for root, dirs, files in …

Nov 21, 2022 · Let’s break down how to extract files conditionally from a zip file in Python: Open the zip file using the read method; We then get a list of all the files in the zip file using the …

Jul 22, 2021 · To work on zip files using python, we will use an inbuilt python module called zipfile. 1. Extracting a zip file. print('Extracting all the files now...') print('Done!') The above program …

Open a ZIP file, where file can be a path to a file (a string), a file-like object or a path-like object. The mode parameter should be 'r' to read an existing file, 'w' to truncate and write a new file, …

Feb 15, 2025 · If you only want to extract specific files from a ZIP archive, you can use the extract() method. This method takes the name of the file to extract as an argument. Here is an …

Nov 23, 2024 · Step 1: List the Files in the Zip Archive. The first step involves using the ZipFile.namelist() function from the zipfile library to list all files contained within the zip archive.

Feb 11, 2025 · To unzip a file in Python, you need to open the zip file and extract its contents. Let’s assume you have a zip file named data_usa.zip that you want to extract. Example: Open …

Apr 6, 2023 · The example code opens the specified ZIP file, prints the list of file names in the archive, extracts all files to the current working directory, and extracts specific files to a …