Convert image from PIL to openCV format

I'm trying to convert image from PIL to OpenCV format. I'm using OpenCV 2.4.3. here is what I've attempted till now.

>>> from PIL import Image
>>> import cv2 as cv
>>> pimg = Image.open('D:\\traffic.jpg')                           #PIL Image
>>> cimg = cv.cv.CreateImageHeader(pimg.size,cv.IPL_DEPTH_8U,3)    #CV Image
>>> cv.cv.SetData(cimg,pimg.tostring())
>>> cv.cv.NamedWindow('cimg')
>>> cv.cv.ShowImage('cimg',cimg)
>>> cv.cv.WaitKey()

But I think the image is not getting converted to CV format. The Window shows me a large brown image. Where am I going wrong in Converting image from PIL to CV format?

Also , why do i need to type cv.cv to access functions?

Solution 1:

use this:

pil_image = PIL.Image.open('Image.jpg').convert('RGB') 
open_cv_image = numpy.array(pil_image) 
# Convert RGB to BGR 
open_cv_image = open_cv_image[:, :, ::-1].copy() 

Solution 2:

This is the shortest version I could find,saving/hiding an extra conversion:

pil_image = PIL.Image.open('image.jpg')
opencvImage = cv2.cvtColor(numpy.array(pil_image), cv2.COLOR_RGB2BGR)

If reading a file from a URL:

import cStringIO
import urllib
file = cStringIO.StringIO(urllib.urlopen(r'http://stackoverflow.com/a_nice_image.jpg').read())
pil_image = PIL.Image.open(file)
opencvImage = cv2.cvtColor(numpy.array(pil_image), cv2.COLOR_RGB2BGR)