Download and extract zip file python

I have managed to get my first python script to work which downloads a list of .ZIP files from a URL and then proceeds to extract the ZIP files and writes them to disk.

I am now at a loss to achieve the next step.

My primary goal is to download and extract the zip file and pass the contents [CSV data] via a TCP stream. I would prefer not to actually write any of the zip or extracted files to disk if I could get away with it.

Here is my current script which works but unfortunately has to write the files to disk.

import urllib, urllister
import zipfile
import urllib2
import os
import time
import pickle

# check for extraction directories existence
if not os.path.isdir['downloaded']:
    os.makedirs['downloaded']

if not os.path.isdir['extracted']:
    os.makedirs['extracted']

# open logfile for downloaded data and save to local variable
if os.path.isfile['downloaded.pickle']:
    downloadedLog = pickle.load[open['downloaded.pickle']]
else:
    downloadedLog = {'key':'value'}

# remove entries older than 5 days [to maintain speed]

# path of zip files
zipFileURL = "//www.thewebserver.com/that/contains/a/directory/of/zip/files"

# retrieve list of URLs from the webservers
usock = urllib.urlopen[zipFileURL]
parser = urllister.URLLister[]
parser.feed[usock.read[]]
usock.close[]
parser.close[]

# only parse urls
for url in parser.urls: 
    if "PUBLIC_P5MIN" in url:

        # download the file
        downloadURL = zipFileURL + url
        outputFilename = "downloaded/" + url

        # check if file already exists on disk
        if url in downloadedLog or os.path.isfile[outputFilename]:
            print "Skipping " + downloadURL
            continue

        print "Downloading ",downloadURL
        response = urllib2.urlopen[downloadURL]
        zippedData = response.read[]

        # save data to disk
        print "Saving to ",outputFilename
        output = open[outputFilename,'wb']
        output.write[zippedData]
        output.close[]

        # extract the data
        zfobj = zipfile.ZipFile[outputFilename]
        for name in zfobj.namelist[]:
            uncompressed = zfobj.read[name]

            # save uncompressed data to disk
            outputFilename = "extracted/" + name
            print "Saving extracted file to ",outputFilename
            output = open[outputFilename,'wb']
            output.write[uncompressed]
            output.close[]

            # send data via tcp stream

            # file successfully downloaded and extracted store into local log and filesystem log
            downloadedLog[url] = time.time[];
            pickle.dump[downloadedLog, open['downloaded.pickle', "wb" ]]

Mel

5,56810 gold badges39 silver badges42 bronze badges

asked Apr 19, 2011 at 2:13

1

Below is a code snippet I used to fetch zipped csv file, please have a look:

Python 2:

from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen

resp = urlopen["//www.test.com/file.zip"]
myzip = ZipFile[StringIO[resp.read[]]]
for line in myzip.open[file].readlines[]:
    print line

Python 3:

from io import BytesIO
from zipfile import ZipFile
from urllib.request import urlopen
# or: requests.get[url].content

resp = urlopen["//www.test.com/file.zip"]
myzip = ZipFile[BytesIO[resp.read[]]]
for line in myzip.open[file].readlines[]:
    print[line.decode['utf-8']]

Here file is a string. To get the actual string that you want to pass, you can use zipfile.namelist[]. For instance,

resp = urlopen['//mlg.ucd.ie/files/datasets/bbc.zip']
myzip = ZipFile[BytesIO[resp.read[]]]
myzip.namelist[]
# ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']

metadaddy

4,0201 gold badge21 silver badges43 bronze badges

answered Apr 19, 2011 at 2:53

VishalVishal

19.3k21 gold badges76 silver badges92 bronze badges

2

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[]:
    [ ... ]

answered Apr 19, 2011 at 2:23

senderlesenderle

140k35 gold badges206 silver badges231 bronze badges

6

I'd like to offer an updated Python 3 version of Vishal's excellent answer, which was using Python 2, along with some explanation of the adaptations / changes, which may have been already mentioned.

from io import BytesIO
from zipfile import ZipFile
import urllib.request
    
url = urllib.request.urlopen["//www.unece.org/fileadmin/DAM/cefact/locode/loc162txt.zip"]

with ZipFile[BytesIO[url.read[]]] as my_zip_file:
    for contained_file in my_zip_file.namelist[]:
        # with open[["unzipped_and_read_" + contained_file + ".file"], "wb"] as output:
        for line in my_zip_file.open[contained_file].readlines[]:
            print[line]
            # output.write[line]

Necessary changes:

• There's no StringIO module in Python 3 [it's been moved to io.StringIO]. Instead, I use io.BytesIO]2, because we will be handling a bytestream -- Docs, also this thread.
• urlopen:
  • "The legacy urllib.urlopen function from Python 2.6 and earlier has been discontinued; urllib.request.urlopen[] corresponds to the old urllib2.urlopen.", Docs and this thread.

Note:

• In Python 3, the printed output lines will look like so: b'some text'. This is expected, as they aren't strings - remember, we're reading a bytestream. Have a look at Dan04's excellent answer.

A few minor changes I made:

• I use with ... as instead of zipfile = ... according to the Docs.
• The script now uses .namelist[] to cycle through all the files in the zip and print their contents.
• I moved the creation of the ZipFile object into the with statement, although I'm not sure if that's better.
• I added [and commented out] an option to write the bytestream to file [per file in the zip], in response to NumenorForLife's comment; it adds "unzipped_and_read_" to the beginning of the filename and a ".file" extension [I prefer not to use ".txt" for files with bytestrings]. The indenting of the code will, of course, need to be adjusted if you want to use it.
  • Need to be careful here -- because we have a byte string, we use binary mode, so "wb"; I have a feeling that writing binary opens a can of worms anyway...
• I am using an example file, the UN/LOCODE text archive:

What I didn't do:

• NumenorForLife asked about saving the zip to disk. I'm not sure what he meant by it -- downloading the zip file? That's a different task; see Oleh Prypin's excellent answer.

Here's a way:

import urllib.request
import shutil

with urllib.request.urlopen["//www.unece.org/fileadmin/DAM/cefact/locode/2015-2_UNLOCODE_SecretariatNotes.pdf"] as response, open["downloaded_file.pdf", 'w'] as out_file:
    shutil.copyfileobj[response, out_file]

answered Feb 8, 2017 at 16:49

ZuboZubo

1,3532 gold badges18 silver badges25 bronze badges

2

I'd like to add my Python3 answer for completeness:

from io import BytesIO
from zipfile import ZipFile
import requests

def get_zip[file_url]:
    url = requests.get[file_url]
    zipfile = ZipFile[BytesIO[url.content]]
    files = [zipfile.open[file_name] for file_name in zipfile.namelist[]]
    return files.pop[] if len[files] == 1 else files

        

answered Jan 18, 2016 at 19:58

lababidilababidi

2,4841 gold badge20 silver badges14 bronze badges

write to a temporary file which resides in RAM

it turns out the tempfile module [ //docs.python.org/library/tempfile.html ] has just the thing:

tempfile.SpooledTemporaryFile[[max_size=0[, mode='w+b'[, bufsize=-1[, suffix=''[, prefix='tmp'[, dir=None]]]]]]]

This function operates exactly as TemporaryFile[] does, except that data is spooled in memory until the file size exceeds max_size, or until the file’s fileno[] method is called, at which point the contents are written to disk and operation proceeds as with TemporaryFile[].

The resulting file has one additional method, rollover[], which causes the file to roll over to an on-disk file regardless of its size.

The returned object is a file-like object whose _file attribute is either a StringIO object or a true file object, depending on whether rollover[] has been called. This file-like object can be used in a with statement, just like a normal file.

New in version 2.6.

or if you're lazy and you have a tmpfs-mounted /tmp on Linux, you can just make a file there, but you have to delete it yourself and deal with naming

answered Apr 19, 2011 at 2:16

ninjageckoninjagecko

85.2k24 gold badges134 silver badges143 bronze badges

1

Adding on to the other answers using requests:

 # download from web

 import requests
 url = '//mlg.ucd.ie/files/datasets/bbc.zip'
 content = requests.get[url]

 # unzip the content
 from io import BytesIO
 from zipfile import ZipFile
 f = ZipFile[BytesIO[content.content]]
 print[f.namelist[]]

 # outputs ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']

Use help[f] to get more functions details for e.g. extractall[] which extracts the contents in zip file which later can be used with with open.

NightOwl888

54.3k23 gold badges132 silver badges206 bronze badges

answered Mar 7, 2018 at 11:00

AksonAkson

6718 silver badges8 bronze badges

1

All of these answers appear too bulky and long. Use requests to shorten the code, e.g.:

import requests, zipfile, io
r = requests.get[zip_file_url]
z = zipfile.ZipFile[io.BytesIO[r.content]]
z.extractall["/path/to/directory"]

answered Dec 2, 2020 at 10:38

AlexAlex

2,1462 gold badges26 silver badges39 bronze badges

Vishal's example, however great, confuses when it comes to the file name, and I do not see the merit of redefing 'zipfile'.

Here is my example that downloads a zip that contains some files, one of which is a csv file that I subsequently read into a pandas DataFrame:

from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen
import pandas

url = urlopen["//www.federalreserve.gov/apps/mdrm/pdf/MDRM.zip"]
zf = ZipFile[StringIO[url.read[]]]
for item in zf.namelist[]:
    print["File in zip: "+  item]
# find the first matching csv file in the zip:
match = [s for s in zf.namelist[] if ".csv" in s][0]
# the first line of the file contains a string - that line shall de ignored, hence skiprows
df = pandas.read_csv[zf.open[match], low_memory=False, skiprows=[0]]

[Note, I use Python 2.7.13]

This is the exact solution that worked for me. I just tweaked it a little bit for Python 3 version by removing StringIO and adding IO library

Python 3 Version

from io import BytesIO
from zipfile import ZipFile
import pandas
import requests

url = "//www.nseindia.com/content/indices/mcwb_jun19.zip"
content = requests.get[url]
zf = ZipFile[BytesIO[content.content]]

for item in zf.namelist[]:
    print["File in zip: "+  item]

# find the first matching csv file in the zip:
match = [s for s in zf.namelist[] if ".csv" in s][0]
# the first line of the file contains a string - that line shall de     ignored, hence skiprows
df = pandas.read_csv[zf.open[match], low_memory=False, skiprows=[0]]

answered Oct 10, 2017 at 21:35

It wasn't obvious in Vishal's answer what the file name was supposed to be in cases where there is no file on disk. I've modified his answer to work without modification for most needs.

from StringIO import StringIO
from zipfile import ZipFile
from urllib import urlopen

def unzip_string[zipped_string]:
    unzipped_string = ''
    zipfile = ZipFile[StringIO[zipped_string]]
    for name in zipfile.namelist[]:
        unzipped_string += zipfile.open[name].read[]
    return unzipped_string

answered Jun 7, 2015 at 5:00

plowmanplowman

13.3k8 gold badges50 silver badges52 bronze badges

1

Use the zipfile module. To extract a file from a URL, you'll need to wrap the result of a urlopen call in a BytesIO object. This is because the result of a web request returned by urlopen doesn't support seeking:

from urllib.request import urlopen

from io import BytesIO
from zipfile import ZipFile

zip_url = '//example.com/my_file.zip'

with urlopen[zip_url] as f:
    with BytesIO[f.read[]] as b, ZipFile[b] as myzipfile:
        foofile = myzipfile.open['foo.txt']
        print[foofile.read[]]

If you already have the file downloaded locally, you don't need BytesIO, just open it in binary mode and pass to ZipFile directly:

from zipfile import ZipFile

zip_filename = 'my_file.zip'

with open[zip_filename, 'rb'] as f:
    with ZipFile[f] as myzipfile:
        foofile = myzipfile.open['foo.txt']
        print[foofile.read[].decode['utf-8']]

Again, note that you have to open the file in binary ['rb'] mode, not as text or you'll get a zipfile.BadZipFile: File is not a zip file error.

It's good practice to use all these things as context managers with the with statement, so that they'll be closed properly.

answered Jun 23, 2020 at 21:25

How do I download and save a ZIP file in Python?

How do I extract a ZIP file in Python?

How do I download a zip file?

How do I extract multiple ZIP files in Python?