Code to count number of heads(people) present in a image

I want a matlab program which will count number of heads present in image and not other things present such as trees,other objects etc.please send me entire code.

[Walter Roberson]

Who will get credit for handing in the assignment ?

[MRahul]

create a folder, put the image in jpg format on which u want to run this code, save the program in the same folder. Run

instead of 15.jpg put your image file name.

%get the input image
I=imread('15.jpg');
imshow(I),title('Image:1');
%change the color space
cform = makecform('srgb2lab');
J=applycform(I,cform);
figure;imshow(J),title('Image:2');
%equalise brightness to get skin area
K=J(:,:,2);% 2nd page of 3-d vector j
figure;imshow(K),title('Image:3');
L=graythresh(J(:,:,2));% find appropriate gray thresh value
BW1=im2bw(J(:,:,2),L);% convert to binary image based on threshold
figure;imshow(BW1),title('Image:4');
bw2=imfill(BW1,'holes');% fill patches with holes
figure;imshow(bw2)
bw3 = bwareaopen(bw2,1890); %opens area greater than 1890
cc=bwconncomp(bw3)% connected comp for finding the density of people in
image
density=cc.NumObjects / (size(bw3,1) * size(bw3,2))
figure;imshow(bw3)
labeledImage = bwlabel(bw3, 8);%same as connected components
figure;imshow(labeledImage)
blobMeasurements = regionprops(labeledImage,'all');%measure all properties of the image
numberOfPeople = size(blobMeasurements, 1)% count the number of people
% draw bounding boxes
imagesc(I);
hold on;
title('Original with bounding boxes');
for k = 1 : numberOfPeople % Loop through all blobs.
% Find the mean of each blob.
% directly into regionprops.
thisBlobsBox = blobMeasurements(k).BoundingBox; 
% Get list of pixels in current blob.
x1 = thisBlobsBox(1);%1st side
y1 = thisBlobsBox(2);%2nd side
x2 = x1 + thisBlobsBox(3);%3rd side
y2 = y1 + thisBlobsBox(4);%4th side
x = [x1 x2 x2 x1 x1];
y = [y1 y1 y2 y2 y1];
%subplot(3,4,2);
plot(x, y, 'LineWidth', 2);
end

[Walter Roberson]

As I glance at that code, I do not see how it handles dark-skinned people.

Answer 1

[Image Analyst]

It should be pretty easy if you have the Computer Vision System Toolbox, as this demo from it shows counted heads: http://www.mathworks.com/products/computer-vision/description4.html

[javed patel]

thank you

Answer 2

[Image Analyst]

See crowd estimation articles here: http://www.visionbib.com/bibliography/motion-f735m1.html#Crowds,%20Tracking%20Multiple%20People,%20Multiple%20Pedestrian%20Tracking

[Iqra]

I need a Matlab code which can count only head in image, This code is not working properly,any demos will also be helpful. Thankyou

[Image Analyst]

Then start calling people who wrote the papers I gave the link to and see if they'll give or sell you their code. Or else start writing, or start looking for someone to hire to write one of the papers for you.

[babar ali]

I need a Matlab code which can count only head in image, This code is not working properly,any demos will also be helpful. Thankyou

[Image Analyst]

What code? Perhaps your image is not what the code was expecting. Did you see the link on crowd estimation that I gave above? If your code doesn't work, try one of the published algorithms.

[babar ali]

yes this code is not working correctly can you suggest me any algorithm for crowd estimation Thanks

[babar ali]

 i am expecting the code for calculating number of people in an image

[Image Analyst]

Again, WHAT "code is not working correctly"? You haven't posted your code or your images or even read this link yet.

And I DID suggest algorithms - Lots of them. Evidently you chose one and coded it up but you haven't said which one or even shared the code, so what can anyone do to help you????

I don't even know how many heads you have in your image. Three or four moving heads is a lot easier and more accurate than tens of thousands of still heads, like from an aerial view.

Keep in mind that this is no 5 minute project and we are not going to develop a turnkey system for you. You're either going to have to write the thing yourself, or hire someone to write it for you.