How to open an image from an url with opencv using requests from python

问题

I am trying to open a large list of images using OpenCV on python, because I need to work with them latter.

Actually, I can achieve this goal with pillow like this:

url = r'https://i.imgur.com/DrjBucJ.png'
response = requests.get(url, stream=True).raw
guess = Image.open(response).resize(size)

I am using the library requests from python.

The response looks like this: b'\x89PNG\r\n\x1a\n\x00\x00\x00\rIHDR\x00\x00\x01\xdb\...

And if I am not wrong, those are the values of the pixels from the image of the url, correct?

My question is: how can I do the same with OpenCV?

I have tried it like:

resp = requests.get(url)
image = np.asarray(bytearray(resp.read()), dtype="uint8")
image = cv2.imdecode(image, cv2.IMREAD_COLOR)

And I get this error:

image = np.asarray(bytearray(resp.read()), dtype="uint8")
AttributeError: 'Response' object has no attribute 'read'

I got the code from this web: https://www.pyimagesearch.com/2015/03/02/convert-url-to-image-with-python-and-opencv/

回答1:

You just forgot stream=True and .raw in requests.get

resp = requests.get(url, stream=True).raw

import cv2
import numpy as np
import requests

url = r'https://i.imgur.com/DrjBucJ.png'
resp = requests.get(url, stream=True).raw
image = np.asarray(bytearray(resp.read()), dtype="uint8")
image = cv2.imdecode(image, cv2.IMREAD_COLOR)

# for testing
cv2.imshow('image',image)
cv2.waitKey(0)
cv2.destroyAllWindows()

To answer your question

.raw mean that you want to retrieve the response as a stream of bytes and the response will not evaluated or transformed by any measure (so it will not decode gzip and deflate transfer-encodings) but with .content The gzip and deflate transfer-encodings are automatically decoded for you.

In your case it will be better to use .content over .raw

the following note from Requests package documentation

Note An important note about using Response.iter_content versus Response.raw. Response.iter_content will automatically decode the gzip and deflate transfer-encodings. Response.raw is a raw stream of bytes – it does not transform the response content. If you really need access to the bytes as they were returned, use Response.raw.

References:

https://2.python-requests.org/en/master/user/quickstart/#raw-response-content

https://2.python-requests.org/en/master/user/quickstart/#binary-response-content

回答2:

@Mohamed Saeed's answer solved your problem. Below is an alternative solution to fetch image from a url:

import cv2
import numpy as np
from urllib.request import urlopen

req = urlopen('https://i.imgur.com/DrjBucJ.png')
image = np.asarray(bytearray(req.read()), dtype=np.uint8)
image = cv2.imdecode(image, -1) 

cv2.imshow('image',image)
cv2.waitKey(0)
cv2.destroyAllWindows()

来源：https://stackoverflow.com/questions/57539233/how-to-open-an-image-from-an-url-with-opencv-using-requests-from-python

标签

python

OpenCV

python-requests