How can I read an image from an Internet URL in Python cv2, scikit image and mahotas?

Question

How can I read an image from an Internet URL in Python cv2?

This Stack Overflow answer,

import cv2.cv as cv
import urllib2
from cStringIO import StringIO
import PIL.Image as pil
url="some_url"

img_file = urllib2.urlopen(url)
im = StringIO(img_file.read())

is not good because Python reported to me:

TypeError: object.__new__(cStringIO.StringI) is not safe, use cStringIO.StringI.__new__

Answer 1

Since a cv2 image is not a string (save a Unicode one, yucc), but a NumPy array, - use cv2 and NumPy to achieve it:

import cv2
import urllib
import numpy as np

req = urllib.urlopen('http://answers.opencv.org/upfiles/logo_2.png')
arr = np.asarray(bytearray(req.read()), dtype=np.uint8)
img = cv2.imdecode(arr, -1) # 'Load it as it is'

cv2.imshow('lalala', img)
if cv2.waitKey() & 0xff == 27: quit()

Answer 2

The following reads the image directly into a NumPy array:

from skimage import io

image = io.imread('https://raw2.github.com/scikit-image/scikit-image.github.com/master/_static/img/logo.png')

Answer 3

Updated Answer

import urllib
import cv2 as cv2
import numpy as np

url = "https://pyimagesearch.com/wp-content/uploads/2015/01/opencv_logo.png"
url_response = urllib.request.urlopen(url)
img_array = np.array(bytearray(url_response.read()), dtype=np.uint8)
img = cv2.imdecode(img_array, -1)
cv2.imshow('URL Image', img)
cv2.waitKey(0)
cv2.destroyAllWindows()

Answer 4

in python3:

from urllib.request import urlopen
def url_to_image(url, readFlag=cv2.IMREAD_COLOR):
    # download the image, convert it to a NumPy array, and then read
    # it into OpenCV format
    resp = urlopen(url)
    image = np.asarray(bytearray(resp.read()), dtype="uint8")
    image = cv2.imdecode(image, readFlag)

    # return the image
    return image

this is the implementation of url_to_image in imutils, so you can just call

import imutils
imutils.url_to_image(url)

Answer 5

Using requests:

def url_to_numpy(url):                     
  img = Image.open(BytesIO(requests.get(url).content))                                 
  return cv2.cvtColor(np.array(img), cv2.COLOR_RGB2BGR)

Answer 6

For Python 3:

import cv2
from urllib.request import urlopen

image_url = "IMAGE-URL-GOES-HERE"
resp = urlopen(image_url)
image = np.asarray(bytearray(resp.read()), dtype="uint8")
image = cv2.imdecode(image, cv2.IMREAD_COLOR) # The image object

# Optional: For testing & viewing the image
cv2.imshow('image',image)

For Python 3 using Google Colab:

import cv2
from google.colab.patches import cv2_imshow
from urllib.request import urlopen

image_url = "IMAGE-URL-GOES-HERE"
resp = urlopen(image_url)
image = np.asarray(bytearray(resp.read()), dtype="uint8")
image = cv2.imdecode(image, cv2.IMREAD_COLOR) # The image object

# Optional: For testing & viewing the image
cv2_imshow(image)

Answer 7

I used @berak's code as a foundation. If you have a URL that requires authentication (username and password), you can use this method:

import cv2
import numpy as np
import requests


resp = requests.get('http://1.2.3.4/media/cam0/still.jpg?res=max',auth = requests.auth.HTTPDigestAuth("my_username","my_password"))
arr = np.asarray(bytearray(resp.content), dtype=np.uint8)
img = cv2.imdecode(arr, -1)

cv2.imshow('lalala', img)
if cv2.waitKey() & 0xff == 27: quit()