How to generate unique face id for detected face from image ?
I am extracting images from the video by interval by 1 second, am able to detect all the faces from the image. Now every after one second new image will be generated and i want to find out the way to distinguish the unique faces across images. so i decided to generate the faceID for each faces detected in one image and will do the same thing for consequence image so at the end i can filter out the data based on faceID.
is there any other idea (except faceID) to solve above problem ?
I would prefer some solution based on OpenCV, EmguCV.
There is nothing like faceID - if you find it, you'll be rich :) What you can do is to build (or use an already documented) way to measure similarity between faces - and put a threshold on where two very similay images belong to the same human. Google 'face recognition' to find some inspiration.
OpenCV has a face recognition module, but you should acknowledge that the accuracy will be far from perfect.
thanks sammy. I was just thinking the way we generate the fingerprint ID can't we generate it for the face ?
Fingerprint ID is just a marketing term - completely disconnected from the technical details needed to identify a person by their fingerprint. The process is similar there, however, it's much simpler for fingerprints, due to some specifics of fingerprint scanning & structure.
Sammy, i want to find out the count of people from the video surveillance for given time period. so what i have done so far is i am extracting images from the video upon certain time interval (5 sec) and from the image am counting the faces but there might be a case that one person standing in front of camera for about 1 min so that person will be on every images of that 1 min, i wanted to have unique count of people, is there any other way to filter out duplicate faces ?
Read some papers on face recognition, and you'll have an idea on where to start. Just that it's not going to be a few lines of code, neither a magical
getID(face). It's not something you would answer in a post here.