Counting zeros which are lying between nonzero elements by considering consecutive zeros as a single element in a matrix

I have a (4x8) matrix as

A=[1 0 1 1 1 0 0 1;0 1 0 1 1 0 1 0;0 0 1 0 0 0 0 0;1 0 1 1 0 0 1 1]

A = 4×8

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

I want to calculate the number of only those zeros that are lying between nonzero elements i.e 1 in each row in such a way that I want to consider only the consecutive zeros as a single element. Single zeros will be considered as a separate element. Desired Output is (2 2 0 2). How to do so?

Answer 1

[Mahdi Hayati]

Hi.

you can use diff() function to find number of times that elements of each row, turn from 1 to 0. for example for the first row we can have:

D = diff(A(1,:));
temp = size(find(D == -1));
number_of_turns = temp(1);
if D(7) == -1
    number_of_turns = number_of_turns - 1;
end

number_of_turns

in this code, I found how many times zeros are stuck between ones. The 'if' statement is because if the last element of the row is 0, it must not be count.

I hope it was useful

Answer 2

[Bruno Luong]

A=[1 0 1 1 1 0 0 1;0 1 0 1 1 0 1 0;0 0 1 0 0 0 0 0;1 0 1 1 0 0 1 1]

A = 4×8

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

d=diff(A~=0,1,2);
[~,i,v]=find(d');
vl=zeros(size(A,1),1);
vl(i)=v;
sum(d==-1,2)-(vl==-1)

ans = 4×1

2 2 0 2

Answer 3

[Bruno Luong]

Always helpful the old for-loop

A=[1 0 1 1 1 0 0 1;0 1 0 1 1 0 1 0;0 0 1 0 0 0 0 0;1 0 1 1 0 0 1 1]

A = 4×8

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

[m,n] = size(A);
count = zeros(m,1);
for i = 1:m
    s0startded = false;
    c = 0;
    isprevnull = A(i,1) == 0;
    for j = 2:n
        isnull = A(i,j) == 0;
        if s0startded
            if ~isnull
                c = c+1;
                s0startded = false;
            end
        else
            s0startded = isnull && ~isprevnull;
        end
        isprevnull = isnull;
    end
    count(i) = c;
end

count

count = 4×1

2 2 0 2

Answer 4

[Matt J]

A=[1 0 1 1 1 0 0 1;0 1 0 1 1 0 1 0;0 0 1 0 0 0 0 0;1 0 1 1 0 0 1 1]

A = 4×8

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

d=diff(A,1,2);
result= max(0,  min( sum(d==-1,2) - ~A(:,end), sum(d==1,2) - ~A(:,1)  ))

result = 4×1

2 2 0 2

[Bruno Luong]

is there a typo

~A(:,end)

[Matt J]

Yep. I fixed it.

[Bruno Luong]

It seems not correct with A contains only 0

A = [0 0 0 0]

A = 1×4

0 0 0 0

d=diff(A,1,2);
result= min( sum(d==-1,2) - ~A(:,end), sum(d==1,2) - ~A(:,1)  )

result = -1

[Matt J]

Yeah. I added a threshold step.

Answer 5

[Bruno Luong]

A=[1 0 1 1 1 0 0 1;0 1 0 1 1 0 1 0;0 0 1 0 0 0 0 0;1 0 1 1 0 0 1 1]

A = 4×8

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

max(sum(diff(~~A,1,2)==1,2)-~A(:,1),0)

ans = 4×1

2 2 0 2

% If A is binary you can simplify to
% max(sum(diff(A,1,2)==1,2)-~A(:,1),0)