How can I quantify the number of cells between specific values?

I have a 100x1500 array [A] filled with zeros and ones.

For example, A(1,:) = 0 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1....

For each row, I would like to find the average distance, or average number of zeros, between ones. Is there a simple way to do this?

[the cyclist]

In a given row, do you want to ignore leading zeros and trailing zeros, or count them as part of the average?

For example, if

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

do you want the mean of [3 1 2 3], or just the mean of [1 2]?

[omidm]

I'd like to ignore the leading and trailing zeros, so just the mean of [1 2] in your example. Thanks for clarifying.

[Voss]

@omidm: How should consecutive ones be treated? That is, should they be considered as having a run of zero zeros in between them? Consider:

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

Is that a sequence of 1 zero followed by a sequence of 3 zeros (average length 2), or is it a sequence of 1 zero followed by a sequence of 0 zeros followed by a sequence of 3 zeros (average length 4/3)?

Answer 1

[Voss]

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

[r,c] = find(A);
groupsummary(c,r,@mean_diff)

ans = 10×1

4.0000 2.0000 1.5000 2.0000 2.5000 2.0000 1.0000 1.0000 1.0000 1.0000

function out = mean_diff(x)
m = diff(x)-1;
out = mean(m(m > 0));
end

[omidm]

This worked great, thank you. Indeed I did not want consecutive ones to have a value of zero between them. Thanks again!

Answer 2

[Eric Delgado]

I don't know if my solution is the "simple way" that you request! :)

randomMatrix = randi([0,1],10)

randomMatrix = 10×10

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

meanZeros = zeros(height(randomMatrix), 1);

for ii = 1:height(randomMatrix)
    meanZeros(ii) = mean(diff(find(randomMatrix(ii,:) == 1))-1);
end

meanZeros

meanZeros = 10×1

2.0000 1.3333 0.7500 1.0000 1.2500 0.3333 1.0000 0.5000 0.6667 0.4000

[Voss]

Note that this method treats consecutive 1's as containing zero 0's in between them, e.g., for the first row of A above, this method gives the average number of zeros as 2 = (4+0)/2 rather than 4 = 4/1.

It's not clear whether this is what OP intends.