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.
I have written some code to read the contents from a specific url:
import requests
import os
def read_doc(doc_ID):
filename = doc_ID + ".txt"
if not os.path.exists(filename):
my_url = encode_url(doc_ID) #this is a call to another function that would encode the url
my_response = requests.get(my_url)
if my_response.status_code == requests.codes.ok:
return my_response.text
return None
This checks if there's a file named doc_ID.txt (where doc_ID could be any name provided). And if there's no such file, it would read the contents from a specific url and would return them. What I would like to do is to store those returned contents in a file called doc_ID.txt. That is, I would like to finish my function by creating a new file in case it didn't exist at the beginning.
How can I do that? I tried this:
my_text = my_response.text
output = os.rename(my_text, filename)
return output
but then, the actual contents of the file would become the name of the file and I would get an error saying the filename is too long.
So the issue I think I'm seeing is that you want to put the contents of your request's response into the file, rather than naming the file with the contents. The code below should create a file with the filename you want, and insert the text from your response!
import requests
import os
def read_doc(doc_ID):
filename = doc_ID + ".txt"
if not os.path.exists(filename):
my_url = encode_url(doc_ID) #this is a call to another function that would encode the url
my_response = requests.get(my_url)
if my_response.status_code == requests.codes.ok:
with open(filename, "w") as file:
file.write(my_response.text)
return file
return None
To write the response text to the file, you can simply use python file object, https://docs.python.org/3/tutorial/inputoutput.html#reading-and-writing-files
with open(filename, "w") as file:
file.write(my_text)
I have a long list of .json files that I need to download to my computer. I want to download them as .json files (so no parsing or anything like that at this point).
I have some code that works for small files, but it is pretty buggy. Also it doesn't handle multiple links well.
Appreciate any advice to fix up this code:
import os
filename = 'test.json'
path = "C:/Users//Master"
fullpath = os.path.join(path, filename)
import urllib2
url = 'https://www.premierlife.com/secure/index.json'
response = urllib2.urlopen(url)
webContent = response.read()
f = open(fullpath, 'w')
f.write(webContent)
f.close
It's creating a blank file because the f.close at the end should be f.close().
I took your code and made a little function and then called it on a little loop to go through a .txt file with the list of urls called "list_of_urls.txt" having 1 url per line (you can change the delimiter in the split function if you want to format it differently).
def save_json(url):
import os
filename = url.replace('/','').replace(':','')
# this replaces / and : in urls
path = "C:/Users/Master"
fullpath = os.path.join(path, filename)
import urllib2
response = urllib2.urlopen(url)
webContent = response.read()
f = open(fullpath, 'w')
f.write(webContent)
f.close()
And then the loop:
f = open('list_of_urls.txt')
p = f.read()
url_list = p.split('\n') #here's where \n is the line break delimiter that can be changed
for url in url_list:
save_json(url)
I have made this simple download manager, but the problem is it wont work on complex urls, when pages are redirected.
def str(d):
for i in range(len(d)):
if d[-i] == '/':
x=-i
break
s=[]
l=len(d)+x+1
print d[l],d[len(d)-1]
s=d[l:]
return s
import urllib2
url=raw_input()
filename=str(url)
webfile = urllib2.urlopen(url)
data = webfile.read()
fout =open(filename,"w")
fout.write(data)
fout.close()
webfile.close()
it wouldn't work for http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&ved=0CG0QFjAI&url=http%3A%2F%2Fwww.iasted.org%2Fconferences%2Fformatting%2FPresentations-Tips.ppt&ei=clfWTpjZEIblrAfC8qWXDg&usg=AFQjCNEIgqx6x4ULHFXzzYDzCITuUJOczA&sig2=0VtKXPvoDnIq-lIR4S9LEQ
while it would work for http://www.iasted.org/conferences/formatting/Presentations-Tips.ppt
and both links are for the same file.
How to solve the problem of redirection?
I think redirection is not a problem here:
Since urllib2 already follows redirect automatically, google redirects to a page in case of error.
Try this script :
url1 = 'http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&ved=0CG0QFjAI&url=http%3A%2F%2Fwww.iasted.org%2Fconferences%2Fformatting%2FPresentations-Tips.ppt&ei=clfWTpjZEIblrAfC8qWXDg&usg=AFQjCNEIgqx6x4ULHFXzzYDzCITuUJOczA&sig2=0VtKXPvoDnIq-lIR4S9LEQ'
url2 = 'http://www.iasted.org/conferences/formatting/Presentations-Tips.ppt'
from urlparse import urlsplit
from urllib2 import urlopen
for url in [url1, url2]:
split = urlsplit(url)
filename = split.path[split.path.rfind('/')+1:]
if not filename:
filename = split.query[split.query.rfind('/')+1:]
f = open(filename, 'w')
f.write(urlopen(url).read())
f.close()
# Yields 2 files : url and Presentations-Tips.ppt [Both are ppt files]
The above script works every time.
In general, you handle redirection by using urllib2.HTTPRedirectHandler, like this:
import urllib2
opener = urllib.build_opener(urllib2.HTTPRedirectHandler)
res = open.open('http://example.com/some/url/')
However, it doesn't like like this will work for the Google URL you've given in your example, because rather than including a Location header in the response, the Google result looks like this:
<script>window.googleJavaScriptRedirect=1</script><script>var a=parent,b=parent.google,c=location;if(a!=window&&b){if(b.r){b.r=0;a.location.href="http://www.iasted.org/conferences/formatting/Presentations-Tips.ppt";c.replace("about:blank");}}else{c.replace("http://www.iasted.org/conferences/formatting/Presentations-Tips.ppt");};</script><noscript><META http-equiv="refresh" content="0;URL='http://www.iasted.org/conferences/formatting/Presentations-Tips.ppt'"></noscript>
...which is to say, it uses a Javascript redirect, which substantially complicates your life. You could use Python's re module to extract the correct location from this block.
I have a design question. I have a function loadImage() for loading an image file. Now it accepts a string which is a file path. But I also want to be able to load files which are not on physical disk, eg. generated procedurally. I could have it accept a string, but then how could it know the string is not a file path but file data? I could add an extra boolean argument to specify that, but that doesn't sound very clean. Any ideas?
It's something like this now:
def loadImage(filepath):
file = open(filepath, 'rb')
data = file.read()
# do stuff with data
The other version would be
def loadImage(data):
# do stuff with data
How to have this function accept both 'filepath' or 'data' and guess what it is?
You can change your loadImage function to expect an opened file-like object, such as:
def load_image(f):
data = file.read()
... and then have that called from two functions, one of which expects a path and the other a string that contains the data:
from StringIO import StringIO
def load_image_from_path(path):
with open(path, 'rb') as f:
load_image(f)
def load_image_from_string(s):
sio = StringIO(s)
try:
load_image(sio)
finally:
sio.close()
How about just creating two functions, loadImageFromString and loadImageFromFile?
This being Python, you can easily distinguish between a filename and a data string. I would do something like this:
import os.path as P
from StringIO import StringIO
def load_image(im):
fin = None
if P.isfile(im):
fin = open(im, 'rb')
else:
fin = StringIO(im)
# Read from fin like you would from any open file object
Other ways to do it would be a try block instead of using os.path, but the essence of the approach remains the same.