Need help vertically/horizontally scanning a binary matrix with reference to an indexed point!

So I have a binary matrix (just zeros and ones) with dimensions 400 by 500. I want to perform vertical and horizontal scans (that simply count how many zeros are to the left, right, top, bottom) through the matrix centered at a specific point.

Let's call my Matrix BW. I have an index such that BW(i,j). To find how many units are to the left, right, top, bottom I calculate these variables:

tothetop=i-1;

tothebottom=400-tothetop;

totheleft=j-1;

totheright=500-totheleft;

I tried to do a scan up and down from the point (i,j) with the following loops where TopCount & BotCount keep track of how many 0's are detected above and below the point at i,j:

    for p=1:1:tothetop
                   if BW(p,j)==0
                       TopCount=TopCount+1;
                   end
               end
    for q=tothebottom:1:rows

                   if BW(q,j)==0
                       BotCount=BotCount+1;
                   end
               end

However, I find that my counts are not accurate, namely, my scan downwards looks very similar to my scan upwards. If there is something wrong, I can't seem to figure it out, so please help!

[Paolo]

I wrote a few lines which I believe could help you solve your problem. It creates a random 5 by 5 matrix with zeros and ones. I defined the center point as point P(3,3). It uses the position of the center point to look for the zeros you require.

 tmp = randi([0 1],5,5);

 xcenter = 3;
 ycenter = 3;

 left =  numel(find(~(tmp(:,1:(xcenter-1)))));
 up   =  numel(find(~(tmp(1:(xcenter-1),ycenter))));
 down =  numel(find(~(tmp((xcenter+1):end,ycenter))));
 right = numel(find(~(tmp(:,((ycenter+1:end))))));

 tmp = [1     0     1     0     1
        1     0     1     0     1
        1     0     1     0     0
        0     1     1     1     1
        1     1     1     1     1]

Left = 4. Up = 0. Down = 0. Right = 4.

To confirm, is this the output you require?

[Guillaume]

@Paolo,

a faster and simpler way of obtaining

numel(find(~x))

is

sum(~x)

[Aydin Akyol]

@Paolo, I am trying to count the number of zeros to the left, right, above, and below of the defined center. Meaning that starting at the center I want to look at all elements to the left, and know how many zeroes there are. And then I want to look at all all of the elements to the right and know how many zeroes there are to the right. Given your definition of tmp and a center at (3,3), left should be 1 right should be 2, up and down should both be zero. @Guillaume sum may work. I'll look into it.

[Aydin Akyol]

I think I'll just reverse my matrix (make all the 0's into 1 and vice versa) and then I was able to adapt the code used before to do what I wanted, with sum. So here is my scanning/summing code below:

    clear
  clc
  close all

xcenter = 3; 
ycenter = 3;
tmp = randi([0 1],5,5);

[rows,columns]=size(tmp);
left = sum(tmp((ycenter),(1:(xcenter-1)))); 
up = sum(tmp(1:ycenter-1,xcenter)); 
down = sum(tmp(ycenter+1:rows,xcenter));
right = sum(tmp((ycenter),(xcenter+1:columns)));
tmp
left
up
down
right

Thanks for the help all!

[Guillaume]

Actually, even simpler than sum, is nnz.