[python] How to download image using requests

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?

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)

my approach was to use response.content (blob) and save to the file in binary mode

img_blob = requests.get(url, timeout=5).content
     with open(destination + '/' + title, 'wb') as img_file:
         img_file.write(img_blob)

Check out my python project that downloads images from unsplash.com based on keywords.

I'm going to post an answer as I don't have enough rep to make a comment, but with wget as posted by Blairg23, you can also provide an out parameter for the path.

 wget.download(url, out=path)

As easy as to import Image and requests

from PIL import Image
import requests

img = Image.open(requests.get(url, stream = True).raw)
img.save('img1.jpg')

Following code snippet downloads a file.

The file is saved with its filename as in specified url.

import requests

url = "http://example.com/image.jpg"
filename = url.split("/")[-1]
r = requests.get(url, timeout=0.5)

if r.status_code == 200:
    with open(filename, 'wb') as f:
        f.write(r.content)

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)

You can do something like this:

import requests
import random

url = "https://images.pexels.com/photos/1308881/pexels-photo-1308881.jpeg? auto=compress&cs=tinysrgb&dpr=1&w=500"
name=random.randrange(1,1000)
filename=str(name)+".jpg"
response = requests.get(url)
if response.status_code.ok:
   with open(filename,'w') as f:
    f.write(response.content)

This is the first response that comes up for google searches on how to download a binary file with requests. In case you need to download an arbitrary file with requests, you can use:

import requests
url = 'https://s3.amazonaws.com/lab-data-collections/GoogleNews-vectors-negative300.bin.gz'
open('GoogleNews-vectors-negative300.bin.gz', 'wb').write(requests.get(url, allow_redirects=True).content)

There are 2 main ways:

1. Using .content (simplest/official) (see Zhenyi Zhang's answer):

   import io  # Note: io.BytesIO is StringIO.StringIO on Python2.
   import requests
   
   r = requests.get('http://lorempixel.com/400/200')
   r.raise_for_status()
   with io.BytesIO(r.content) as f:
       with Image.open(f) as img:
           img.show()
   

2. Using .raw (see Martijn Pieters's answer):

   import requests
   
   r = requests.get('http://lorempixel.com/400/200', stream=True)
   r.raise_for_status()
   r.raw.decode_content = True  # Required to decompress gzip/deflate compressed responses.
   with PIL.Image.open(r.raw) as img:
       img.show()
   r.close()  # Safety when stream=True ensure the connection is released.
   

Timing both shows no noticeable difference.

This is how I did it

import requests
from PIL import Image
from io import BytesIO

url = 'your_url'
files = {'file': ("C:/Users/shadow/Downloads/black.jpeg", open('C:/Users/shadow/Downloads/black.jpeg', 'rb'),'image/jpg')}
response = requests.post(url, files=files)

img = Image.open(BytesIO(response.content))
img.show()

Here is a more user-friendly answer that still uses streaming.

Just define these functions and call getImage(). It will use the same file name as the url and write to the current directory by default, but both can be changed.

import requests
from StringIO import StringIO
from PIL import Image

def createFilename(url, name, folder):
    dotSplit = url.split('.')
    if name == None:
        # use the same as the url
        slashSplit = dotSplit[-2].split('/')
        name = slashSplit[-1]
    ext = dotSplit[-1]
    file = '{}{}.{}'.format(folder, name, ext)
    return file

def getImage(url, name=None, folder='./'):
    file = createFilename(url, name, folder)
    with open(file, 'wb') as f:
        r = requests.get(url, stream=True)
        for block in r.iter_content(1024):
            if not block:
                break
            f.write(block)

def getImageFast(url, name=None, folder='./'):
    file = createFilename(url, name, folder)
    r = requests.get(url)
    i = Image.open(StringIO(r.content))
    i.save(file)

if __name__ == '__main__':
    # Uses Less Memory
    getImage('http://www.example.com/image.jpg')
    # Faster
    getImageFast('http://www.example.com/image.jpg')

The request guts of getImage() are based on the answer here and the guts of getImageFast() are based on the answer above.

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