Python: Open File in Zip Without Temporarily Extracting It

Python: Open file in zip without temporarily extracting it

Vincent Povirk's answer won't work completely;

import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgfile = archive.open('img_01.png')
...

You have to change it in:

import zipfile
archive = zipfile.ZipFile('images.zip', 'r')
imgdata = archive.read('img_01.png')
...

For details read the ZipFile docs here.

Open file from zip without extracting it in Python?

Try this:

import pandas as pd
import requests
from io import BytesIO, StringIO
from zipfile import ZipFile

response = requests.get(url)
zip_file = ZipFile(BytesIO(response.content))
files = zip_file.namelist()
with zip_file.open(files[0]) as csvfile:   
    print(pd.read_csv(csvfile, encoding='utf8', sep=","))

Downloading and unzipping a .zip file without writing to disk

My suggestion would be to use a StringIO object. They emulate files, but reside in memory. So you could do something like this:

# get_zip_data() gets a zip archive containing 'foo.txt', reading 'hey, foo'

import zipfile
from StringIO import StringIO

zipdata = StringIO()
zipdata.write(get_zip_data())
myzipfile = zipfile.ZipFile(zipdata)
foofile = myzipfile.open('foo.txt')
print foofile.read()

# output: "hey, foo"

Or more simply (apologies to Vishal):

myzipfile = zipfile.ZipFile(StringIO(get_zip_data()))
for name in myzipfile.namelist():
    [ ... ]

In Python 3 use BytesIO instead of StringIO:

import zipfile
from io import BytesIO

filebytes = BytesIO(get_zip_data())
myzipfile = zipfile.ZipFile(filebytes)
for name in myzipfile.namelist():
    [ ... ]

unzip file without creating temporary files

To only extract certain files, you can pass a list to extractall:

with zipfile.ZipFile('/tmp/DataPump_10000838.zip', 'r') as zip_ref:
    lstNEW = list(filter(lambda x: not x.startswith("__MACOSX/"), zip_ref.namelist()))
    zip_ref.extractall('/tmp/', members=lstNEW)

The files are not temporary files, but rather macOS's way of representing resource forks in zip files that don't normally support this.

How to unzip without writing to disk?

Using ZipFile.open on the already opened archive, we can do just that:

import zipfile
import pandas as pd

with zipfile.ZipFile("archive.zip") as archive:
    with archive.open("testing.txt") as csv:
        df = pd.read_csv(csv)

print(df)

Unzipping files in Python

import zipfile
with zipfile.ZipFile(path_to_zip_file, 'r') as zip_ref:
    zip_ref.extractall(directory_to_extract_to)

That's pretty much it!

How to extract file from zip without maintaining directory structure in Python?

You can use zipfile.ZipFile.open:

import shutil
import zipfile

with zipfile.ZipFile('/path/to/my_file.apk') as z:
    with z.open('/res/drawable/icon.png') as zf, open('temp/icon.png', 'wb') as f:
        shutil.copyfileobj(zf, f)

Or use zipfile.ZipFile.read:

import zipfile

with zipfile.ZipFile('/path/to/my_file.apk') as z:
    with open('temp/icon.png', 'wb') as f:
        f.write(z.read('/res/drawable/icon.png'))

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