pos*_*res 22 python opencv image-processing scikit-image mahotas
如何从Python cv2中的Internet URL读取图像?
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())
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不好,因为Python向我报告:
TypeError: object.__new__(cStringIO.StringI) is not safe, use cStringIO.StringI.__new__
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ber*_*rak 25
由于cv2图像不是字符串(保存Unicode,yucc),而是NumPy数组, - 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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Ton*_* Yu 22
以下将图像直接读入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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Kel*_*ang 15
在python3中:
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
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这是 imutils 中 url_to_image 的实现,所以你可以调用
import imutils
imutils.url_to_image(url)
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