There is no way you could do this 100% (i would say maybe 1-5% would be plausible) with nowdays knowledge. You would get much better result (than those 1-5%) just checking the image-names for sex-related-words :).
#!python
import os, glob
from PIL import Image
def get_skin_ratio(im):
im = im.crop((int(im.size[0]*0.2), int(im.size[1]*0.2), im.size[0]-int(im.size[0]*0.2), im.size[1]-int(im.size[1]*0.2)))
skin = sum([count for count, rgb in im.getcolors(im.size[0]*im.size[1]) if rgb[0]>60 and rgb[1]<(rgb[0]*0.85) and rgb[2]<(rgb[0]*0.7) and rgb[1]>(rgb[0]*0.4) and rgb[2]>(rgb[0]*0.2)])
return float(skin)/float(im.size[0]*im.size[1])
for image_dir in ('porn','clean'):
for image_file in glob.glob(os.path.join(image_dir,"*.jpg")):
skin_percent = get_skin_ratio(Image.open(image_file)) * 100
if skin_percent>30:
print "PORN {0} has {1:.0f}% skin".format(image_file, skin_percent)
else:
print "CLEAN {0} has {1:.0f}% skin".format(image_file, skin_percent)
我听说过一些工具,它们使用非常简单,但是非常有效的算法。该算法计算颜色值接近某些预定义的“皮肤”颜色的像素的相对数量。如果数量高于某个预定义的值,那么图像被认为是色情/色情内容。当然,这种算法会给出假阳性结果的特写脸照片和许多其他事情。
Since you are writing about social networking there will be lots of "normal" photos with high amount of skin colour on it, so you shouldn't use this algorithm to deny all pictures with positive result. But you can use it provide some help for moderators, for example flag these pictures with higher priority, so if moderator want to check some new pictures for pornographic content he can start from these pictures.
Like someone above said, it cannot be done 100% percent. There will be cases where such algorithms fail. The actual precision will be determined by your training data, the structure of your neuronal networks and how you will choose to cluster the training data (penises, vaginas, breasts, etc, and combinations of such). In any case I am very confident that this can be achieved with high accuracy for explicit porn imagery.