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__
Answers:
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()
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')
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)
If you’re using requests, you can use this
import requests
import numpy as np
from io import BytesIO
from PIL import Image
def url_to_img(url, save_as=''):
img = Image.open(BytesIO(requests.get(url).content))
if save_as:
img.save(save_as)
return np.array(img)
img = url_to_img('https://xxxxxxxxxxxxxxxxxx')
img = url_to_img('https://xxxxxxxxxxxxxxxxxx', 'sample.jpg')
cv2.imshow(img)
Using requests:
def url_to_numpy(url):
img = Image.open(BytesIO(requests.get(url).content))
return cv2.cvtColor(np.array(img), cv2.COLOR_RGB2BGR)
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()
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__
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()
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')
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)
If you’re using requests, you can use this
import requests
import numpy as np
from io import BytesIO
from PIL import Image
def url_to_img(url, save_as=''):
img = Image.open(BytesIO(requests.get(url).content))
if save_as:
img.save(save_as)
return np.array(img)
img = url_to_img('https://xxxxxxxxxxxxxxxxxx')
img = url_to_img('https://xxxxxxxxxxxxxxxxxx', 'sample.jpg')
cv2.imshow(img)
Using requests:
def url_to_numpy(url):
img = Image.open(BytesIO(requests.get(url).content))
return cv2.cvtColor(np.array(img), cv2.COLOR_RGB2BGR)
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()