Python3: lzma unpack .7z file

You can try using a python library instead, py7zr, which supports 7zip archive compression, decompression, encryption, decryption. https://github.com/miurahr/py7zr

import py7zr
with py7zr.SevenZipFile('sample.7z', mode='r') as z:
    z.extractall()

LZMA and 7z are two very different beasts.

In the simplest of terms LZMA is a lossless compression algorithm. This means that, you feed LZMA some data, it will compress and give you the output. It has no sense of files, folders or how to store them.

7z on the other hand is an archive file format, and this means that 7z is a complete package. You have a few files and folders, feed it to 7z, it will neatly compress them, and store them in a single file (archive). Please note that, 7z uses LZMA and a cocktail of other algorithms to compress and store files in its 7z archive file.

Here is what wikipedia has got to say about the two:

7z is a compressed archive file format that supports several different data compression, encryption and pre-processing algorithms.

The Lempel–Ziv–Markov chain algorithm (LZMA) is an algorithm used to perform lossless data compression. It has been under development either since 1996 or 19983 and was first used in the 7z format of the 7-Zip archiver.

So in short, you cannot use lzma to create or extract 7z files. As far as I know, there is no way to extract a 7z file using python other than: See update below.

import os
os.system( '7z x archive.7z -oPath/to/Name' )

Update: May 2019

Since there some interest about extracting 7z files in python, I thought an update is in order. As of 2019 (perhaps even earlier), libarchive bindings for python do support 7z format. An example for extracting files from 7z archive is given in above link.