如何从Python CV2，Scikit Image和Mahotas中读取从互联网URL的图像？

Question

如何从Python CV2中的Internet URL读取图像？

这个堆栈溢出答案，

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())

.

不好，因为python向我报告：

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

.

Answer 1

由于CV2图像不是字符串（保存Unicode One，Yucc），但是numpy array， - 使用cv2和numpy来实现它：

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()

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Answer 2

下面将图像直接读入numpy数组：

from skimage import io

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

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Answer 3

在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

.

这是在imutils中实现URL_TO_IMAGE，所以您可以拨打

import imutils
imutils.url_to_image(url)

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