The goal is to download a file from the internet, and create from it a file object, or a file like object without ever having it touch the hard drive. This is just for my knowledge, wanting to know if its possible or practical, particularly because I would like to see if I can circumvent having to code a file deletion line.
This is how I would normally download something from the web, and map it to memory:
import requests
import mmap
u = requests.get("http://www.pythonchallenge.com/pc/def/channel.zip")
with open("channel.zip", "wb") as f: # I want to eliminate this, as this writes to disk
f.write(u.content)
with open("channel.zip", "r+b") as f: # and his as well, because it reads from disk
mm = mmap.mmap(f.fileno(), 0)
mm.seek(0)
print mm.readline()
mm.close() # question: if I do not include this, does this become a memory leak?
r.raw (HTTPResponse) is already a file-like object (just pass stream=True):
#!/usr/bin/env python
import sys
import requests # $ pip install requests
from PIL import Image # $ pip install pillow
url = sys.argv[1]
r = requests.get(url, stream=True)
r.raw.decode_content = True # Content-Encoding
im = Image.open(r.raw) #NOTE: it requires pillow 2.8+
print(im.format, im.mode, im.size)
In general if you have a bytestring; you could wrap it as f = io.BytesIO(r.content), to get a file-like object without touching the disk:
#!/usr/bin/env python
import io
import zipfile
from contextlib import closing
import requests # $ pip install requests
r = requests.get("http://www.pythonchallenge.com/pc/def/channel.zip")
with closing(r), zipfile.ZipFile(io.BytesIO(r.content)) as archive:
print({member.filename: archive.read(member) for member in archive.infolist()})
You can't pass r.raw to ZipFile() directly because the former is a non-seekable file.
I would like to see if I can circumvent having to code a file deletion line
tempfile can delete files automatically f = tempfile.SpooledTemporaryFile(); f.write(u.content). Until .fileno() method is called (if some api requires a real file) or maxsize is reached; the data is kept in memory. Even if the data is written on disk; the file is deleted as soon as it closed.
Your answer is u.content. The content is in the memory. Unless you write it to a file, it won’t be stored on disk.
This is what I ended up doing.
import zipfile
import requests
import StringIO
u = requests.get("http://www.pythonchallenge.com/pc/def/channel.zip")
f = StringIO.StringIO()
f.write(u.content)
def extract_zip(input_zip):
input_zip = zipfile.ZipFile(input_zip)
return {i: input_zip.read(i) for i in input_zip.namelist()}
extracted = extract_zip(f)
Related
I want to make a in-memory file to use in pygame mixer. I mean something like http://www.pygame.org/docs/ref/music.html#pygame.mixer.music.load which says load() method supports file object.
import requests
from pygame import mixer
r = requests.get("http://example.com/some_small_file.mp3")
in_memory_file = file(r.content) # something like this
mixer.music.init()
mixer.music.load(in_memory_file)
mixer.music.play()
You are probably looking for BytesIO or StringIO classes from Python io package, both available in python 2 and python 3. They provide a file-like interface you can use in your code the exact same way you interact with a real file.
StringIO is used to store textual data:
import io
f = io.StringIO("some initial text data")
BytesIO must be used for binary data:
import io
f = io.BytesIO(b"\x00\x00\x00\x00\x00\x00\x00\x00\x01\x01\x01\x01\x01\x01")
To store MP3 file data, you will probably need the BytesIO class. To initialize it from a GET request to a server, proceed like this:
import requests
from pygame import mixer
import io
r = requests.get("http://example.com/somesmallmp3file.mp3")
inmemoryfile = io.BytesIO(r.content)
mixer.music.init()
mixer.music.load(inmemoryfile)
mixer.music.play()
# This will free the memmory from any data
inmemoryfile.close()
Additional note: as both classes inherit from IOBase, they can be used as context manager with the with statement, so you don't need to manually call the close() method anymore:
import requests
from pygame import mixer
import io
r = requests.get("http://example.com/somesmallmp3file.mp3")
with io.BytesIO(r.content) as inmemoryfile:
mixer.music.init()
mixer.music.load(inmemoryfile)
mixer.music.play()
I'm trying to use Python to read .pdf files from the web directly rather than save them all to my computer. All I need is the text from the .pdf and I'm going to be reading a lot (~60k) of them, so I'd prefer to not actually have to save them all.
I know how to save a .pdf from the internet using urllib and open it with PyPDF2. (example)
I want to skip the saving-to-file step.
import urllib, PyPDF2
urllib.urlopen('https://bitcoin.org/bitcoin.pdf')
wFile = urllib.urlopen('https://bitcoin.org/bitcoin.pdf')
lFile = PyPDF2.pdf.PdfFileReader(wFile.read())
I get an error that is fairly easy to understand:
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
fil = PyPDF2.pdf.PdfFileReader(wFile.read())
File "C:\Python27\lib\PyPDF2\pdf.py", line 797, in __init__
self.read(stream)
File "C:\Python27\lib\PyPDF2\pdf.py", line 1245, in read
stream.seek(-1, 2)
AttributeError: 'str' object has no attribute 'seek'
Obviously PyPDF2 doesn't like that I'm giving it the urllib.urlopen().read() (which appears to return a string). I know that this string is not the "text" of the .pdf but a string representation of the file. How can I resolve this?
EDIT: NorthCat's solution resolved my error, but when I try to actually extract the text, I get this:
>>> print lFile.getPage(0).extractText()
ˇˆ˘˘˙˘˘˝˘˛˘ˇ˘ˇ˚ˇˇˇ˘ˆ˘˘˘˚ˇˆ˘ˆ˘ˇ˜ˇ˝˚˘˛˘ˇ ˘˘˘ˇ˛˘˚˚ˆˇˇ!
˝˘˚ˇ˘˘˚"˘˘ˇ˘˚ˇ˘˘˚ˇ˘˘˘˙˘˘˘#˘˘˘ˆ˘˛˘˚˛˙ ˘˘˚˚˘˛˙#˘ˇ˘ˇˆ˘˘˛˛˘˘!˘˘˛˘˝˘˘˘˚ ˛˘˘ˇ˘ˇ˛$%&˘ˇ'ˆ˛
$%&˘ˇˇ˘˚ˆ˚˘˘˘˘ ˘ˆ(ˇˇ˘˘˘˘ˇ˘˚˘˘#˘˘˘ˇ˛!ˇ)˘˘˚˘˘˛ ˚˚˘ˇ˘˝˘˚'˘˘ˇˇ ˘˘ˇ˘˛˙˛˛˘˘˚ˇ˘˘ˆ˘˘ˆ˙
$˘˘˘*˘˘˘ˇˆ˘˘ˇˆ˛ˇ˘˝˚˚˘˘ˇ˘ˆ˘"˘ˆ˘ˇˇ˘˛ ˛˛˘˛˘˘˘˘˘˘˛˘˘˚˚˘$ˇ˘ˇˆ˙˘˝˘ˇ˘˘˘ˇˇˆˇ˘ ˘˛ˇ˝˘˚˚#˘˛˘˚˘˘
˘ˇ˘˚˛˛˘ˆ˛ˇˇˇ ˚˘˘˚˘˘ˇ˛˘˙˘˝˘ˇ˘ˆ˘˛˙˘˝˘ˇ˘˘˝˘"˘˛˘˝˘ˇ ˘˘˘˚˛˘˚)˘˘ˆ˛˘˘
˘˛˘˛˘ˆˇ˚˘˘˘˘˚˘˘˘˘˛˛˚˘˚˝˚ˇ˘#˘˘˚ˆ˘˘˘˝˘˚˘ˆˆˇ˘ˆ
˘˘˘ˆ˘˝˘˘˚"˘˘˚˘˚˘ˇ˘ˆ˘ˆ˘˚ˆ˛˚˛ˆ˚˘˘˘˘˘˘˚˛˚˚ˆ#˘ˇˇˆˇ˘˝˘˘ˇ˚˘ˇˇ˘˛˛˚ ˚˘˘˘ˇ˚˘˘ˇ˘˘˚ˆ˘*˘
˘˘ˇ˘˚ˇ˘˙˘˚ˇ˘˘˘˙˙˘˘˚˚˘˘˝˘˘˘˛˛˘ˇˇ˚˘˛#˘ˆ˘˘ˇ˘˚˘ˇˇ˘˘ˇˆˇ˘$%&˘ˆ˘˛˘˚˘,
Try this:
import urllib, PyPDF2
import cStringIO
wFile = urllib.urlopen('https://bitcoin.org/bitcoin.pdf')
lFile = PyPDF2.pdf.PdfFileReader( cStringIO.StringIO(wFile.read()) )
Because PyPDF2 does not work, there are a couple of solutions, however, require saving the file to disk.
Solution 1
You can use ps2ascii (if you are using linux or mac ) or xpdf (Windows). Example of using xpdf:
import os
os.system('C:\\xpdfbin-win-3.03\\bin32\\pdftotext.exe C:\\xpdfbin-win-3.03\\bin32\\bitcoin.pdf bitcoin1.txt')
or
import subprocess
subprocess.call(['C:\\xpdfbin-win-3.03\\bin32\\pdftotext.exe', 'C:\\xpdfbin-win-3.03\\bin32\\bitcoin.pdf', 'bitcoin2.txt'])
Solution 2
You can use one of online pdf to txt converter. Example of using pdf.my-addr.com
import MultipartPostHandler
import urllib2
def pdf2text( absolute_path ):
url = 'http://pdf.my-addr.com/pdf-to-text-converter-tool.php'
params = { 'file' : open( absolute_path, 'rb' ),
'encoding': 'UTF-8',
}
opener = urllib2.build_opener( MultipartPostHandler.MultipartPostHandler )
return opener.open( url, params ).read()
print pdf2text('bitcoin.pdf')
Code of MultipartPostHandler you can find here. I tried to use the cStringIO instead open(), but it did not work.
Maybe it will be helpful for you.
I know this question is old, but I had the same issue and here is how I solved it.
In the newer docs of Py2PDF there is a section about streaming data
The example there looks like this:
from io import BytesIO
# Prepare example
with open("example.pdf", "rb") as fh:
bytes_stream = BytesIO(fh.read())
# Read from bytes_stream
reader = PdfReader(bytes_stream)
Therefore, what I did instead was this:
import urllib
from io import BytesIO
from PyPDF2 import PdfReader
NEW_PATH = 'https://example.com/path/to/pdf/online?id=123456789&date=2022060'
wFile = urllib.request.urlopen(NEW_PATH)
bytes_stream = BytesIO(wFile.read())
reader = PdfReader(bytes_stream)
I have the following view code that attempts to "stream" a zipfile to the client for download:
import os
import zipfile
import tempfile
from pyramid.response import FileIter
def zipper(request):
_temp_path = request.registry.settings['_temp']
tmpfile = tempfile.NamedTemporaryFile('w', dir=_temp_path, delete=True)
tmpfile_path = tmpfile.name
## creating zipfile and adding files
z = zipfile.ZipFile(tmpfile_path, "w")
z.write('somefile1.txt')
z.write('somefile2.txt')
z.close()
## renaming the zipfile
new_zip_path = _temp_path + '/somefilegroup.zip'
os.rename(tmpfile_path, new_zip_path)
## re-opening the zipfile with new name
z = zipfile.ZipFile(new_zip_path, 'r')
response = FileIter(z.fp)
return response
However, this is the Response I get in the browser:
Could not convert return value of the view callable function newsite.static.zipper into a response object. The value returned was .
I suppose I am not using FileIter correctly.
UPDATE:
Since updating with Michael Merickel's suggestions, the FileIter function is working correctly. However, still lingering is a MIME type error that appears on the client (browser):
Resource interpreted as Document but transferred with MIME type application/zip: "http://newsite.local:6543/zipper?data=%7B%22ids%22%3A%5B6%2C7%5D%7D"
To better illustrate the issue, I have included a tiny .py and .pt file on Github: https://github.com/thapar/zipper-fix
FileIter is not a response object, just like your error message says. It is an iterable that can be used for the response body, that's it. Also the ZipFile can accept a file object, which is more useful here than a file path. Let's try writing into the tmpfile, then rewinding that file pointer back to the start, and using it to write out without doing any fancy renaming.
import os
import zipfile
import tempfile
from pyramid.response import FileIter
def zipper(request):
_temp_path = request.registry.settings['_temp']
fp = tempfile.NamedTemporaryFile('w+b', dir=_temp_path, delete=True)
## creating zipfile and adding files
z = zipfile.ZipFile(fp, "w")
z.write('somefile1.txt')
z.write('somefile2.txt')
z.close()
# rewind fp back to start of the file
fp.seek(0)
response = request.response
response.content_type = 'application/zip'
response.app_iter = FileIter(fp)
return response
I changed the mode on NamedTemporaryFile to 'w+b' as per the docs to allow the file to be written to and read from.
current Pyramid version has 2 convenience classes for this use case- FileResponse, FileIter. The snippet below will serve a static file. I ran this code - the downloaded file is named "download" like the view name. To change the file name and more set the Content-Disposition header or have a look at the arguments of pyramid.response.Response.
from pyramid.response import FileResponse
#view_config(name="download")
def zipper(request):
path = 'path_to_file'
return FileResponse(path, request) #passing request is required
docs:
http://docs.pylonsproject.org/projects/pyramid/en/latest/api/response.html#
hint: extract the Zip logic from the view if possible
I am trying to download and open a zipped file and seem to be having trouble using a file type handle with zipfile. I'm getting the error "AttributeError: addinfourl instance has no attribute 'seek'" when running this:
import zipfile
import urllib2
def download(url,directory,name):
webfile = urllib2.urlopen('http://www.sec.gov'+url)
webfile2 = zipfile.ZipFile(webfile)
content = zipfile.ZipFile.open(webfile2).read()
localfile = open(directory+name, 'w')
localfile.write(content)
localfile.close()
return()
download(link.get("href"),'./fails_data', link.text)
Putting things together, the following retrieves the content of the first file within a zipped file from a website:
import urllib.request
import zipfile
url = 'http://www.gutenberg.lib.md.us/4/8/8/2/48824/48824-8.zip'
filehandle, _ = urllib.request.urlretrieve(url)
zip_file_object = zipfile.ZipFile(filehandle, 'r')
first_file = zip_file_object.namelist()[0]
file = zip_file_object.open(first_file)
content = file.read()
As of 2020, you can use dload to download and unzip a file, i.e.:
import dload
dload.save_unzip("https://file-examples.com/wp-content/uploads/2017/02/zip_2MB.zip")
By default it extracts to a dir on the script path with the zip file name, but you can specify the extract location:
dload.save_unzip("https://file-examples.com/wp-content/uploads/2017/02/zip_2MB.zip", "/extract/here")
install using pip install dload
You can't seek on a urllib2.urlopened file. The methods it supports are listed here: http://docs.python.org/library/urllib.html#urllib.urlopen.
You'll have to retrieve the file (possibly with urllib.urlretrieve, http://docs.python.org/library/urllib.html#urllib.urlretrieve), then use zipfile on it.
Alternatively, you could read() the urlopened file, then put it into a StringIO, then use zipfile on that, if you wanted the zipped data in memory. Also check out the extract and extract_all methods of zipfile if you just want to extract the file, instead of using read.
I do not have enough rep to comment but regarding Marius's answer above please note that for Python3 there is a slight modification needed regarding import and urlretrieve call, since urllib has been split into several modules.
import urllib
Becomes:
import urllib.request
And
filehandle, _ = urllib.urlretrieve(url)
Becomes
filehandle, _ = urllib.request.urlretrieve(url)
Iterating on #Marius answer (which reads a single file directly from the zip), if you want to extract all files to a directory, do this:
import urllib
import zipfile
url = "http://www.gutenberg.lib.md.us/4/8/8/2/48824/48824-8.zip"
extract_dir = "example"
zip_path, _ = urllib.request.urlretrieve(url)
with zipfile.ZipFile(zip_path, "r") as f:
f.extractall(extract_dir)
This stores the zip file in a temporary dir. If you want to keep it around, you can pass a filename to urlretrieve, e.g. urllib.request.urlretrieve(url, "my_zip_file.zip").
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.