Counting Specific Number of Consecutive Values in a Matrix

I have a matrix composed of ones and zeros.

I need to idenity where there are three 1s in each COLUMN

1 0 0 1 1
1 0 1 1 1
1 1 1 0 1
1 0 1 1 1
1 1 0 1 1
0 1 1 1 1
1 1 1 0 1

I want to have the code identify if

1. there are three consecutive 1s in the first through third column, and
2. that there are two sets of consecutive 1s in the fourth column.

I then want to compare this matrix with another matrix, and find where they are similar in the exact locations of the three consecutive 1s.

Answer 1

[Image Analyst]

Here's another way using the Image Processing Toolbox to filter out short sequences and count how many are left:

A = [1 0 0 1 1
	1 0 1 1 1
	1 1 1 0 1
	1 0 1 1 1
	1 1 0 1 1
	0 1 1 1 1
	1 1 1 0 1]
A = logical(A); % Make sure it's logical.
[rows, columns] = size(A)
criteria = false(1, 4); % True or false for whether the criteria for that column is met or not
for col = 1 : 3
	% Get rid of any sequences that are not at least 3 long in the first 3 columns.
	A(:, col) = bwareaopen(A(:, col), 3);
	% Set the criteria
	[~, numSequences] = bwlabel(A(:, col)) % Count number of sequences.
	criteria(col) = numSequences >= 1
end
% For the 4th column, make sure there are at least 2 sets of 1s
% that are at least 2 elements long.
col4 = bwareaopen(A(:, 4), 2) % Only sequences 2 or longer survive.
[~, numSequences] = bwlabel(col4)
criteria(4) = numSequences >= 2
% Show results in command window
A
criteria

Answer 2

[KSSV]

A = [1 0 0 1 1
    1 0 1 1 1
    1 1 1 0 1
    1 0 1 1 1
    1 1 0 1 1
    0 1 1 1 1
    1 1 1 0 1];
% First column 
x = A(:,1) ; 
f = find(diff([0 ;x ;0]==1)); 
id = f(1:2:end-1);  % Start indices
N = f(2:2:end)-p;  % Consecutive ones’ counts

Also have a look on ismember, this would be useful to check two different matrices have common eements.