How to download image using requests

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.

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

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)

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