I have the following code but obviously this is not real streaming. It is the best I could find but it reads the whole input file into memory first. I want to stream it to tarfile module without using all my memory when decrypting huge (>100Gb files)
import tarfile, gnupg
gpg = gnupg.GPG(gnupghome='C:/Users/niels/.gnupg')
with open('103330-013.tar.gpg', 'r') as input_file:
decrypted_data = gpg.decrypt(input_file.read(), passphrase='aaa')
# decrypted_data.data contains the data
decrypted_stream = io.BytesIO(decrypted_data.data)
tar = tarfile.open(decrypted_stream, mode='r|')
tar.extractall()
tar.close()
Apparently, you cannot use real streaming using gpnupg module, gnupg module always reads whole output of gnupg into memory.
So to use real streaming, you'll have to run gpg program directly.
Here is a sample code (without proper error handling):
import subprocess
import tarfile
with open('103330-013.tar.gpg', 'r') as input_file:
gpg = subprocess.Popen(("gpg", "--decrypt", "--homedir", 'C:/Users/niels/.gnupg', '--passphrase', 'aaa'), stdin=input_file, stdout=subprocess.PIPE)
tar = tarfile.open(fileobj=gpg.stdout, mode="r|")
tar.extractall()
tar.close()
I dont know how to unzip gz file in python using subprocess.
gzip library is so slow and i was thinking to make the same function above using gnu/linux shell code and subprocess library.
def __unzipGz(filePath):
import gzip
inputFile = gzip.GzipFile(filePath, 'rb')
stream = inputFile.read()
inputFile.close()
outputFile = file(os.path.splitext(filePath)[0], 'wb')
outputFile.write(stream)
outputFile.close()
You can use something like this:
import subprocess
filename = "some.gunzip.file.tar.gz"
output = subprocess.Popen(['tar', '-xzf', filename])
Since there is no much useful output here, You could also use os.system instead of subprocess.Popen like this:
import os
filename = "some.gunzip.file.tar.gz"
exit_code = os.system("tar -xzf {}".format(filename))
I'm looking for a better way to do this, if possible:
import subprocess
f = open('temp.file', 'w+')
f.write('hello world')
f.close()
out = subprocess.check_output(['cat', 'temp.file'])
print out
subprocess.check_output(['rm', 'temp.file'])
In this example I'm creating a file and passing it as input to cat (in reality it's not cat I'm running but some other program that parses an input pcap file).
What I'm wondering is, is there a way in Python I can create a 'file-like object' with some content, and pipe this file-like object as input to a command-line program. If it is possible, I reckon it would be more efficient than writing a file to the disk and then deleting that file.
check_output takes a stdin input argument to specify a file-like object to connect to the process's standard input.
with open('temp.file') as input:
out = subprocess.check_output(['cat'], stdin=input)
Also, there's no need to shell out to run rm; you can remove the file directly from Python:
os.remove('temp.file')
You can write to a TemporaryFile
import subprocess
from tempfile import TemporaryFile
f = TemporaryFile("w")
f.write("foo")
f.seek(0)
out = subprocess.check_output(['cat'],stdin=f)
print(out)
b'foo'
If you just want to write to a file like object and get the content:
from io import StringIO
f = StringIO()
f.write("foo")
print(f.getvalue())
If the program is configured to read from stdin, you can use Popen.communicate:
>>> from subprocess import Popen, PIPE
>>> p = Popen('cat', stdout=PIPE, stdin=PIPE, stderr=PIPE)
>>> out, err = p.communicate(input=b"Hello world!")
>>> out
'Hello world!'
If the command accepts only filenames, if it doesn't read input from its stdin i.e., if you can't use stdin=PIPE + .communicate() or stdin=real_file then you could try /dev/fd/# filenames:
#!/usr/bin/env python3
import os
import subprocess
import threading
def pump_input(pipe):
with pipe:
for i in range(3):
print(i, file=pipe)
r, w = os.pipe()
try:
threading.Thread(target=pump_input, args=[open(w, 'w')]).start()
out = subprocess.check_output(['cat', '/dev/fd/'+str(r)], pass_fds=[r])
finally:
os.close(r)
print('got:', out)
No content touches the disk. The input is passed to the subprocess via the pipe directly.
If you have a file-like object that is not a real file (otherwise, just pass its name as the command-line argument) then pump_input() could look like:
import shutil
def pump_input(pipe):
with pipe:
shutil.copyfileobj(file_like_object, pipe)
I'm in my 2nd week of Python and I'm stuck on a directory of zipped/unzipped logfiles, which I need to parse and process.
Currently I'm doing this:
import os
import sys
import operator
import zipfile
import zlib
import gzip
import subprocess
if sys.version.startswith("3."):
import io
io_method = io.BytesIO
else:
import cStringIO
io_method = cStringIO.StringIO
for f in glob.glob('logs/*'):
file = open(f,'rb')
new_file_name = f + "_unzipped"
last_pos = file.tell()
# test for gzip
if (file.read(2) == b'\x1f\x8b'):
file.seek(last_pos)
#unzip to new file
out = open( new_file_name, "wb" )
process = subprocess.Popen(["zcat", f], stdout = subprocess.PIPE, stderr=subprocess.STDOUT)
while True:
if process.poll() != None:
break;
output = io_method(process.communicate()[0])
exitCode = process.returncode
if (exitCode == 0):
print "done"
out.write( output )
out.close()
else:
raise ProcessException(command, exitCode, output)
which I've "stitched" together using these SO answers (here) and blogposts (here)
However, it does not seem to work, because my test file is 2.5GB and the script has been chewing on it for 10+mins plus I'm not really sure if what I'm doing is correct anyway.
Question:
If I don't want to use GZIP module and need to de-compress chunk-by-chunk (actual files are >10GB), how do I uncompress and save to file using zcat and subprocess in Python?
Thanks!
This should read the first line of every file in the logs subdirectory, unzipping as required:
#!/usr/bin/env python
import glob
import gzip
import subprocess
for f in glob.glob('logs/*'):
if f.endswith('.gz'):
# Open a compressed file. Here is the easy way:
# file = gzip.open(f, 'rb')
# Or, here is the hard way:
proc = subprocess.Popen(['zcat', f], stdout=subprocess.PIPE)
file = proc.stdout
else:
# Otherwise, it must be a regular file
file = open(f, 'rb')
# Process file, for example:
print f, file.readline()
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 = "http://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" ))
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("http://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("http://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('http://mlg.ucd.ie/files/datasets/bbc.zip')
myzip = ZipFile(BytesIO(resp.read()))
myzip.namelist()
# ['bbc.classes', 'bbc.docs', 'bbc.mtx', 'bbc.terms']
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():
[ ... ]
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("http://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("http://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)
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
write to a temporary file which resides in RAM
it turns out the tempfile module ( http://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
Adding on to the other answers using requests:
# download from web
import requests
url = 'http://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.
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")
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("https://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 = "https://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])
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
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 = 'http://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.