How to download image using requests

Question

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?

Lauro

to use r.raw you need to set stream=True – Metallist

Does this answer your question? Download large file in python with requests – Adjust

Answer 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.

Toughen

With the help of your answer I could able to find data in text file, steps I used are r2 = requests.post(r.url, data); print r2.content. But now I also want to know filename. is their any cleaned way? -- presently I found file name in header -- r2.headers['content-disposition'] that gives me output as: 'attachment; filename=DELS36532G290115.csi' I am parsing this string for filename... is their any cleaner way? – Enounce

@GrijeshChauhan: yes, the content-disposition header is the way to go here; use cgi.parse_header() to parse it and get the parameters; params = cgi.parse_header(r2.headers['content-disposition'])[1] then params['filename']. – Toughen

To get the default 128 byte chunks, you need to iterate over the requests.Response itself: for chunk in r: .... Calling iter_content() without a chunk_size will iterate in 1 byte chunks. – Jeanne

@dtk: thanks, I'll update the answer. Iteration changed after I posted my answer. – Toughen

Wouldn't be much pertinent to use r.ok instead of r.status_code == 200 to check for a valid http code – Trollope

@Trollope two reasons: response.ok was never documented, and it produces true for any 1xx, 2xx or 3xx status, but only a 200 response has a response body. – Toughen

@GrijeshChauhan: to support non-ascii filenames, you could use rfc6266-parser module: rfc6266.parse_requests_response(r).filename_unsafe (it uses url if the filename can't be extracted from Content-Disposition). – Marelda

Answer 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

Toenail

Thank you so much for coming back and answering this. Though the other answer is works, this one is leaps and bounds simpler – Vincenz

It's worth noting that few servers are set to GZIP their images because images already have their own compression. It's counterproductive, wastes CPU cycles with little benefit. So while this may be an issue with text content, specifically with images it's not. – Koo

is there any way we can access the original filename – Indulge

@Koo It's also worth noting that Google PageSpeed reports and does that by default. – Erebus

Should set r.raw.decode_content = True before shutil.copyfileobj(response.raw, out_file) because by default, decode compressed responses (with GZIP or deflate), so you will get a zero-file image. – Quinonez

@SiminJie When you write r.raw.decode_content do you mean response.raw.decode_content? – Shredding

Nice answer, but I'd add if response == 200 before writing to file – Fader

Answer 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)

Romaine

what do you mean with ! f = open("/Users/apple/Desktop/sample.jpg", 'wb') what do you mean with this path !? i want to download image – Ot

That opens a file descriptor in the path specified to which the image file can be written. – Romaine

@AndrewGlazkov I think it would be more Pythonic to use if response.ok: – Hodman

response.ok is True for any 1xx, 2xx or 3xx status, but only a 200 response has a response body as @Martijn Pieters mentioned in the comments above – Crosstree

Answer 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

Eklund

This is because you've not specified the chunk_size parameter which defaults to 1, so iter_content is iterating over the result stream 1 byte at a time. See the documentation python-requests.org/en/latest/api/…. – Kornegay

This also loads the whole response into memory, which you may want to avoid. There is no to use PIL here either, just with open(image_name, 'wb') as outfile: outfile.write(r.content) is enough. – Toughen

PIL is also not in the standard library making this a bit less portable. – Binny

@ZhenyiZhang iter_content is slow because your chunk_size is too small, if you increase it to 100k it will be much faster. – Reptilian

This is the best answer. It isn't always best to read the file into memory, but OP specified "images" meaning the files will usually be less than 4MB, thus having a trivial impact on memory. – Mikkimiko

It appears that from StringIO import StringIO, is now from io import BytesIO according to the requests author http://docs.python-requests.org/en/latest/user/quickstart/#binary-response-content – Trichromatism

Answer 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)

Moonshiner

I used wget without any hassles. Thanks for stating the benefits of using urllib3 – Doralynne

Note that this answer is for Python 2. For Python 3 you need to do urllib.request.urlretrieve("http://example.com", "file.ext"). – Nonparticipation

Thanks @Husky. Updated. – Moonshiner

Can we compress image size here ? @Moonshiner – Veradis

@Veradis No, this is just wget, there is no compression of files. – Moonshiner

In PyCharm I has to do import urllib.request. With import urllib it gave the error Cannot find reference 'request' in '__init__.pyi'. import urllib did work in the console though. – Malefaction

Answer 6

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)

Backswept

Answer 7

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.

Erebus

I tried a bunch of answers, and your 1. answer (using io.BytesIO and Image) was the first one that worked for me on Python 3.6. Don't forget from PIL import Image (and pip install Pillow). – Petrifaction

What's different between .content and .raw? – Semidiurnal

Answer 8

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')

Tammietammuz

Answer 9

TL;DR

Summarizing the great answers from others.

Method	Needs requests	Needs PIL	Needs ...
requests.get -> shutil	Yes	No	-
requests.get -> open(mode="wb")	Yes	No	-
requests.get -> ByteIO -> Image.save	Yes	Yes	-
urllib	-	-	-
wget	No	No	wget
requests.get -> PIL.Image -> np.save	Yes	Yes	numpy

---

Use shutil and output the decoded raw content from requests.get

Original answer modified from https://mcmap.net/q/73859/-how-to-download-image-using-requests

import shutil
import requests

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

response = requests.get(img_url, stream=True)        
with open('dpreview.jpg', 'wb') as fout:
    response.raw.decode_content = True
    shutil.copyfileobj(response.raw, fout)             

Writing the binary directly into file I/O

import requests

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

response = requests.get(img_url, stream=True) 

with open('dpreview.jpg', 'wb') as fout:
    for chunk in response:
        fout.write(chunk)

Stream content into io.BytesIO into PIL.Image object and save it

from io import BytesIO

import requests
from PIL import Image

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

# Stream to BytesIO
response = requests.get(img_url, stream=True)
img = Image.open(BytesIO(response.content))
img.save('dpreview.jpg')


# Using raw content
response = requests.get(img_url, stream=True)
img = Image.open(response.raw)
img.save('dpreview.jpg')

Use urllib

Original answer from https://mcmap.net/q/73859/-how-to-download-image-using-requests

import urllib

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

urllib.request.urlretrieve(img_url, "dpreview.jpg")

And if a specific user-agent is needed for the request, from https://mcmap.net/q/73859/-how-to-download-image-using-requests

import urllib

opener=urllib.request.build_opener()
opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
urllib.request.install_opener(opener)

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

urllib.request.urlretrieve(img_url, "dpreview.jpg")

Use wget

import wget

img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

wget.download(img_url, out='dpreview.jpg')

Saving PIL.Image as numpy array

import requests
from PIL import Image

import numpy as np


img_url = 'https://techcrunch.com/wp-content/uploads/2023/03/dpreview.jpg'

response = requests.get(img_url, stream=True) 
img = Image.open(response.raw)

# Converts and save image into numpy array.
np.save('dpreview.npy', np.asarray(img))

# Loads a npy file to Image
img_arr = np.load('dpreview.npy')
img = Image.fromarray(img_arr.astype(np.uint8))

Sungod

Answer 10

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()

Latria

Answer 11

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.

Raleigh

Answer 12

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)

Pizzeria

Answer 13

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.

Corybantic

Answer 14

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)

Broadcloth

Nice! It has even an implicit .close(). This is the best answer as of 2019 I guess. – Bouffard

Answer 15

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)

Camouflage

Answer 16

Agree with Blairg23 that using urllib.request.urlretrieve is one of the easiest solutions.

One note I want to point out here. Sometimes it won't download anything because the request was sent via script (bot), and if you want to parse images from Google images or other search engines, you need to pass user-agent to request headers first, and then download the image, otherwise, the request will be blocked and it will throw an error.

Pass user-agent and download image:

opener=urllib.request.build_opener()
opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
urllib.request.install_opener(opener)

urllib.request.urlretrieve(URL, 'image_name.jpg')

Code in the online IDE that scrapes and downloads images from Google images using requests, bs4, urllib.requests.

---

Alternatively, if your goal is to scrape images from search engines like Google, Bing, Yahoo!, DuckDuckGo (and other search engines), then you can use SerpApi. It's a paid API with a free plan.

The biggest difference is that there's no need to figure out how to bypass blocks from search engines or how to extract certain parts from the HTML or JavaScript since it's already done for the end-user.

Example code to integrate:

import os, urllib.request
from serpapi import GoogleSearch

params = {
  "api_key": os.getenv("API_KEY"),
  "engine": "google",
  "q": "pexels cat",
  "tbm": "isch"
}

search = GoogleSearch(params)
results = search.get_dict()

print(json.dumps(results['images_results'], indent=2, ensure_ascii=False))

# download images 
for index, image in enumerate(results['images_results']):

    # print(f'Downloading {index} image...')
    
    opener=urllib.request.build_opener()
    opener.addheaders=[('User-Agent','Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/70.0.3538.102 Safari/537.36 Edge/18.19582')]
    urllib.request.install_opener(opener)

    # saves original res image to the SerpApi_Images folder and add index to the end of file name
    urllib.request.urlretrieve(image['original'], f'SerpApi_Images/original_size_img_{index}.jpg')

-----------
'''
]
  # other images
  {
    "position": 100, # 100 image
    "thumbnail": "https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQK62dIkDjNCvEgmGU6GGFZcpVWwX-p3FsYSg&usqp=CAU",
    "source": "homewardboundnj.org",
    "title": "pexels-helena-lopes-1931367 - Homeward Bound Pet Adoption Center",
    "link": "https://homewardboundnj.org/upcoming-event/black-cat-appreciation-day/pexels-helena-lopes-1931367/",
    "original": "https://homewardboundnj.org/wp-content/uploads/2020/07/pexels-helena-lopes-1931367.jpg",
    "is_product": false
  }
]
'''

Disclaimer, I work for SerpApi.

Schnitzler

Answer 17

Here is a very simple code

import requests

response = requests.get("https://i.imgur.com/ExdKOOz.png") ## Making a variable to get image.

file = open("sample_image.png", "wb") ## Creates the file for image
file.write(response.content) ## Saves file content
file.close()

Borgerhout

Answer 18

for download Image

import requests
Picture_request = requests.get(url)

Pitapat

It would be great if everything was that simple. Unfortunately, the code in your example doesn't save image. It can open the image and that's it. – Schnitzler