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.
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I wrote the code like this:
intents = json.loads(open('intents.json').read())
Check your intents.json file is in the same folder on which you python file is.
you can use, for example, the os builf-in module to check on the existence of file and os.path for path manipulation. Check the official doc at https://docs.python.org/3/library/os.path.html
import os
file = 'intents.json'
# location of the current directory
w_dir = os.path.abspath('.'))
if os.path.isfile(os.path.join(w_dir, file)):
with open(file, 'r') as fd:
fd.read()
else:
print('Such file does not exist here "{}"...'.format(w_dir))
You can try opening the file using the normal file operation and then use json.load or json.loads to parse the data as per your needs. I may be unfamiliar with this syntax to the best of my knowledge your syntax is wrong.
You can open the file like this:
f = open(file_name)
Then parse the data:
data = json.load(f)
You can refer to this link for more info and reference
https://www.geeksforgeeks.org/read-json-file-using-python/
I'm looking for a way to extract a specific file (knowing his name) from an archive containing multiple ones, without writing any file on the hard drive.
I tried to use both StringIO and zipfile, but I only get the entire archive, or the same error from Zipfile (open require another argument than a StringIo object)
Needed behaviour:
archive.zip #containing ex_file1.ext, ex_file2.ext, target.ext
extracted_file #the targeted unzipped file
archive.zip = getFileFromUrl("file_url")
extracted_file = extractFromArchive(archive.zip, target.ext)
What I've tried so far:
import zipfile, requests
data = requests.get("file_url")
zfile = StringIO.StringIO(zipfile.ZipFile(data.content))
needed_file = zfile.open("Needed file name", "r").read()
There is a builtin library, zipfile, made for working with zip archives.
https://docs.python.org/2/library/zipfile.html
You can list the files in an archive:
ZipFile.namelist()
and extract a subset:
ZipFile.extract(member[, path[, pwd]])
EDIT:
This question has in-memory zip info. TLDR, Zipfile does work with in-memory file-like objects.
Python in-memory zip library
I finally found why I didn't succeed to do it after few hours of testing :
I was bufferring the zipfile object instead of buffering the file itself and then open it as a Zipfile object, which raised a type error.
Here is the way to do :
import zipfile, requests
data = requests.get(url) # Getting the archive from the url
zfile = zipfile.ZipFile(StringIO.StringIO(data.content)) # Opening it in an emulated file
filenames = zfile.namelist() # Listing all files
for name in filesnames:
if name == "Needed file name": # Verify the file is present
needed_file = zfile.open(name, "r").read() # Getting the needed file content
break
I got the following code from my question on how to convert the tar.gz file to zip file.
import tarfile, zipfile
tarf = tarfile.open(name='sample.tar.gz', mode='r|gz' )
zipf = zipfile.ZipFile.open( name='myzip.zip', mode='a', compress_type=ZIP_DEFLATED )
for m in tarf.getmembers():
f = tarf.extractfile( m )
fl = f.read()
fn = m.name
zipf.writestr( fn, fl )
tarf.close()
zipf.close()
but when I run it I get the error.
What should I change in the code to make it work?
NameError: name 'ZIP_DEFLATED' is not defined
ZIP_DEFLATED is a name defined by the zipfile module; reference it from there:
zipf = zipfile.ZipFile(
'myzip.zip', mode='a',
compression=zipfile.ZIP_DEFLATED)
Note that you don't use the ZipFile.open() method here; you are not opening members in the archive, you are writing to the object.
Also, the correct ZipFile class signature names the 3rd argument compression. compress_type is only used as an attribute on ZipInfo objects and for the ZipFile.writestr() method. The first argument is not named name either; it's file, but you normally would just pass in the value as a positional argument.
Next, you can't seek in a gzip-compressed tarfile, so you'll have issues accessing members in order if you use tarf.getmembers(). This method has to do a full scan to find all members to build a list, and then you can't go back to read the file data anymore.
Instead, iterate directly over the object, and you'll get member objects in order at a point you can still read the file data too:
for m in tarf:
f = tarf.extractfile( m )
fl = f.read()
fn = m.name
zipf.writestr( fn, fl )
I have a zip archive: my_zip.zip. Inside it is one txt file, the name of which I do not know. I was taking a look at Python's zipfile module ( http://docs.python.org/library/zipfile.html ), but couldn't make too much sense of what I'm trying to do.
How would I do the equivalent of 'double-clicking' the zip file to get the txt file and then use the txt file so I can do:
>>> f = open('my_txt_file.txt','r')
>>> contents = f.read()
What you need is ZipFile.namelist() that will give you a list of all the contents of the archive, you can then do a zip.open('filename_you_discover') to get the contents of that file.
import zipfile
# zip file handler
zip = zipfile.ZipFile('filename.zip')
# list available files in the container
print (zip.namelist())
# extract a specific file from the zip container
f = zip.open("file_inside_zip.txt")
# save the extraced file
content = f.read()
f = open('file_inside_zip.extracted.txt', 'wb')
f.write(content)
f.close()
import zipfile
zip=zipfile.ZipFile('my_zip.zip')
f=zip.open('my_txt_file.txt')
contents=f.read()
f.close()
You can see the documentation here. In particular, the namelist() method will give you the names of the zip file members.