How to download image using requests
Solution 1
You can either use the response.raw
file object, or iterate over the response.
To use the response.raw
file-like object will not, by default, decode compressed responses (with GZIP or deflate). You can force it to decompress for you anyway by setting the decode_content
attribute to True
(requests
sets it to False
to control decoding itself). You can then use shutil.copyfileobj()
to have Python stream the data to a file object:
import requests
import shutil
r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
with open(path, 'wb') as f:
r.raw.decode_content = True
shutil.copyfileobj(r.raw, f)
To iterate over the response use a loop; iterating like this ensures that data is decompressed by this stage:
r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
with open(path, 'wb') as f:
for chunk in r:
f.write(chunk)
This'll read the data in 128 byte chunks; if you feel another chunk size works better, use the Response.iter_content()
method with a custom chunk size:
r = requests.get(settings.STATICMAP_URL.format(**data), stream=True)
if r.status_code == 200:
with open(path, 'wb') as f:
for chunk in r.iter_content(1024):
f.write(chunk)
Note that you need to open the destination file in binary mode to ensure python doesn't try and translate newlines for you. We also set stream=True
so that requests
doesn't download the whole image into memory first.
Solution 2
Get a file-like object from the request and copy it to a file. This will also avoid reading the whole thing into memory at once.
import shutil
import requests
url = 'http://example.com/img.png'
response = requests.get(url, stream=True)
with open('img.png', 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
del response
Solution 3
How about this, a quick solution.
import requests
url = "http://craphound.com/images/1006884_2adf8fc7.jpg"
response = requests.get(url)
if response.status_code == 200:
with open("/Users/apple/Desktop/sample.jpg", 'wb') as f:
f.write(response.content)
Solution 4
I have the same need for downloading images using requests. I first tried the answer of Martijn Pieters, and it works well. But when I did a profile on this simple function, I found that it uses so many function calls compared to urllib
and urllib2
.
I then tried the way recommended by the author of requests module:
import requests
from PIL import Image
# python2.x, use this instead
# from StringIO import StringIO
# for python3.x,
from io import StringIO
r = requests.get('https://example.com/image.jpg')
i = Image.open(StringIO(r.content))
This much more reduced the number of function calls, thus speeded up my application. Here is the code of my profiler and the result.
#!/usr/bin/python
import requests
from StringIO import StringIO
from PIL import Image
import profile
def testRequest():
image_name = 'test1.jpg'
url = 'http://example.com/image.jpg'
r = requests.get(url, stream=True)
with open(image_name, 'wb') as f:
for chunk in r.iter_content():
f.write(chunk)
def testRequest2():
image_name = 'test2.jpg'
url = 'http://example.com/image.jpg'
r = requests.get(url)
i = Image.open(StringIO(r.content))
i.save(image_name)
if __name__ == '__main__':
profile.run('testUrllib()')
profile.run('testUrllib2()')
profile.run('testRequest()')
The result for testRequest
:
343080 function calls (343068 primitive calls) in 2.580 seconds
And the result for testRequest2
:
3129 function calls (3105 primitive calls) in 0.024 seconds
Solution 5
This might be easier than using requests
. This is the only time I'll ever suggest not using requests
to do HTTP stuff.
Two liner using urllib
:
>>> import urllib
>>> urllib.request.urlretrieve("http://www.example.com/songs/mp3.mp3", "mp3.mp3")
There is also a nice Python module named wget
that is pretty easy to use. Found here.
This demonstrates the simplicity of the design:
>>> import wget
>>> url = 'http://www.futurecrew.com/skaven/song_files/mp3/razorback.mp3'
>>> filename = wget.download(url)
100% [................................................] 3841532 / 3841532>
>> filename
'razorback.mp3'
Enjoy.
Edit: You can also add an out
parameter to specify a path.
>>> out_filepath = <output_filepath>
>>> filename = wget.download(url, out=out_filepath)
Comments
-
shkschneider over 2 years
I'm trying to download and save an image from the web using python's
requests
module.Here is the (working) code I used:
img = urllib2.urlopen(settings.STATICMAP_URL.format(**data)) with open(path, 'w') as f: f.write(img.read())
Here is the new (non-working) code using
requests
:r = requests.get(settings.STATICMAP_URL.format(**data)) if r.status_code == 200: img = r.raw.read() with open(path, 'w') as f: f.write(img)
Can you help me on what attribute from the response to use from
requests
?