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.
So an update, I found my compile issue was that I needed to change my notebook to a py file and choosing save as doesn't do that. So I had to run a different script turn my notebook to a py file. And part of my exe issue was I was using the fopen command that apparently isn't useable when compiled into a exe. So I redid the code to what is above. But now I get a write error when trying to run the script. I can not find anything on write functions with os is there somewhere else I should look?
Original code:
import requests
import json
import pandas as pd
import csv
from pathlib import Path
response = requests.get('url', headers={'CERT': 'cert'}, stream=True).json()
json2 = json.dumps(response)
f = open('data.json', 'r+')
f.write(json2)
f.close()
Path altered code:
import requests
import json
import pandas as pd
import csv
from pathlib import Path
response = requests.get('url', headers={'CERT': 'cert'}, stream=True).json()
json2 = json.dumps(response)
filename = 'data.json'
if '_MEIPASS2' in os.environ:
filename = os.path.join(os.environ['_MEIPASS2'], filename)
fd = open(filename, 'r+')
fd.write(json2)
fd.close()
The changes to the code allowed me to get past the fopen issue but created a write issue. Any ideas?
If you want to write to a file, you have to open it as writable.
fd = open(filename, 'wb')
Although I don't know why you're opening it in binary if you're writing text.
I am doing a data science project.
I am using google notebook for my job
My dataset is residing at here which I want to access directly at python Notebook.
I am using following line of code to get out of it.
df = pd.read_csv('link')
But Command line is throwing an error like below
What should I do?
Its difficult to answer exactly as there lack of data but here you go for this kind of request..
you have to import ZipFile & urlopen in order to get data from url and extract the data from Zip and the use the csv file for pandas processings.
from zipfile import ZipFile
from urllib.request import urlopen
import pandas as pd
import os
URL = 'https://he-s3.s3.amazonaws.com/media/hackathon/hdfc-bank-ml-hiring-challenge/application-scorecard-for-customers/05d2b4ea-c-Dataset.zip'
# open and save the zip file onto computer
url = urlopen(URL)
output = open('05d2b4ea-c-Dataset.zip', 'wb') # note the flag: "wb"
output.write(url.read())
output.close()
# read the zip file as a pandas dataframe
df = pd.read_csv('05d2b4ea-c-Dataset.zip') zip files
# if keeping on disk the zip file is not wanted, then:
os.remove(zipName) # remove the copy of the zipfile on disk
Use urllib module to download into memory the zip file which returns a file-like object that you can read(), pass it to ZipFile(standard package).
Since here there are multiple files like
['test_data/AggregateData_Test.csv', 'test_data/TransactionData_Test.csv', 'train_data/AggregateData_Train.csv', 'train_data/Column_Descriptions.xlsx', 'train_data/sample_submission.csv', 'train_data/TransactionData_Train.csv']
Load it to a dict of dataframes with filename as the key. Altogether the code will be.
from urllib.request import urlopen
from zipfile import ZipFile
from io import BytesIO
zip_in_memory = urlopen("https://he-s3.s3.amazonaws.com/media/hackathon/hdfc-bank-ml-hiring-challenge/application-scorecard-for-customers/05d2b4ea-c-Dataset.zip").read()
z = ZipFile(BytesIO(zip_in_memory))
dict_of_dfs = {file.filename: pd.read_csv(z.open(file.filename))\
for file in z.infolist()\
if file.filename.endswith('.csv')}
Now you can access dataframes of each csv like dict_of_dfs['test_data/AggregateData_Test.csv'].
Ofcourse all of this is unnecessary if you will just download the zip from the link and pass it as a zipfile.
import tempfile
import shutil
tmp_f = tempfile.NamedTemporaryFile(delete=False)
tmp_f.write("hello world")
shutil.copy(tmp_f.name, "/etc/exports")
When I read /etc/exports, it is a completely empty file. what is wrong?
You need to close the file:
tmp_f.write("hello world")
tmp_f.close()
In addition to closing the file, it You can use .flush() to write the buffer to disk. This may be beneficial in cases when you want to move the file after writing to it, but also want to take advantage of a context manager and delete=True (the default mode) to automatically clean up the temp file. For example:
import os
import shutil
import tempfile
destination_path = "hello.txt"
content = "hello world"
with tempfile.NamedTemporaryFile() as temp:
temp.write(content.encode("utf-8"))
temp.flush()
shutil.copy(temp.name, destination_path)
with open(destination_path) as f:
print(f.read())
os.remove(destination_path)
I want to get the content of a remote file with fabric, without creating a temporary file.
from StringIO import StringIO
from fabric.api import get
fd = StringIO()
get(remote_path, fd)
content=fd.getvalue()
With Python 3 (and fabric3), I get this fatal error when using io.StringIO: string argument expected, got 'bytes', apparently because Paramiko writes to the file-like object with bytes. So I switched to using io.BytesIO and it works:
from io import BytesIO
def _read_file(file_path, encoding='utf-8'):
io_obj = BytesIO()
get(file_path, io_obj)
return io_obj.getvalue().decode(encoding)
import tempfile
from fabric.api import get
with tempfile.TemporaryFile() as fd:
get(remote_path, fd)
fd.seek(0)
content=fd.read()
See: http://docs.python.org/2/library/tempfile.html#tempfile.TemporaryFile
and: http://docs.fabfile.org/en/latest/api/core/operations.html#fabric.operations.get