Stream large binary files with urllib2 to file

Question:

I use the following code to stream large files from the Internet into a local file:

fp = open(file, 'wb')
req = urllib2.urlopen(url)
for line in req:
    fp.write(line)
fp.close()

This works but it downloads quite slowly. Is there a faster way? (The files are large so I don’t want to keep them in memory.)

Asked By: hoju

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Answers:

I used to use mechanize module and its Browser.retrieve() method. In the past it took 100% CPU and downloaded things very slowly, but some recent release fixed this bug and works very quickly.

Example:

import mechanize
browser = mechanize.Browser()
browser.retrieve('http://www.kernel.org/pub/linux/kernel/v2.6/testing/linux-2.6.32-rc1.tar.bz2', 'Downloads/my-new-kernel.tar.bz2')

Mechanize is based on urllib2, so urllib2 can also have similar method… but I can’t find any now.

Answered By: liori

No reason to work line by line (small chunks AND requires Python to find the line ends for you!-), just chunk it up in bigger chunks, e.g.:

# from urllib2 import urlopen # Python 2
from urllib.request import urlopen # Python 3

response = urlopen(url)
CHUNK = 16 * 1024
with open(file, 'wb') as f:
    while True:
        chunk = response.read(CHUNK)
        if not chunk:
            break
        f.write(chunk)

Experiment a bit with various CHUNK sizes to find the “sweet spot” for your requirements.

Answered By: Alex Martelli

You can also use shutil:

import shutil
try:
    from urllib.request import urlopen # Python 3
except ImportError:
    from urllib2 import urlopen # Python 2

def get_large_file(url, file, length=16*1024):
    req = urlopen(url)
    with open(file, 'wb') as fp:
        shutil.copyfileobj(req, fp, length)
Answered By: Tiago

You can use urllib.retrieve() to download files:

Example:

try:
    from urllib import urlretrieve # Python 2

except ImportError:
    from urllib.request import urlretrieve # Python 3

url = "http://www.examplesite.com/myfile"
urlretrieve(url,"./local_file")
Answered By: Aravindh
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