How to filter an array?

Dear all,

I have this array

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

I want only one representative for each group of ones. So how to make

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

Any help will be appreciated.

Best, Meshoo

[KSSV]

one represnetative for each group of 1? Question not clear. Can you type expected result into a matrix?

[Stephen23]

And what if there are an even number of ones?

A = [0;1;1;0]

what output do you want?

[Meshooo]

I don't care which right or left to take in case of even number of ones.

A = [0;0;1;0]

or

A = [0;1;0;0]

Any answer is OK.

Answer 1

[Guillaume]

It's very simple:

diff([0; A]) == 1

simply detects all transition from 0 to 1 and return them as 1.

[Guillaume]

Note that this keeps just the first 1 of a block as opposed to the middle 1 of a block, which the other solutions do.

Answer 2

[goerk]

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

dA = diff(A);
ind = 1:length(A);

startInds = ind(dA>0)+1;
endInds = ind(dA<0);

if length(endInds)<length(startInds) %last value is 1
    endInds(end+1) = length(A);
end

midInd = floor((startInds+endInds)/2); % when even choose left
% midInd = ceil((startInds+endInds)/2); % when even choose right

B = zeros(size(A));
B(midInd) = 1;

[A B] % show result, to check input and output

[Guillaume]

Your answer will fail if the first element of A is 1. And if it starts and ends with 1, the start and end offsets will be completely wrong due to the way you detect that the last value is 1.

The best way to solve both is to prepend and append A with 0 before the diff:

 dA = diff([0 A 0]); %guarantees you have the same number of starts and ends.
 midInd = floor((find(dA > 0) + find(dA < 0) - 1) / 2); %much simpler way of calculating mid indices

[goerk]

You are right, for a 1 at the first position my will fail. Thanks for your very nice and short solution. With the correct concatenation it works fine.

 dA = diff([0; A; 0]);

Answer 3

[Stephen23]

>> A = [0;0;0;0;0;0;1;1;1;0;0;0;0;0;0;1;0;0;0;0];
>> X = find(diff([0;A;0]));
>> Z = zeros(size(A));
>> Z(fix((X(1:2:end)+X(2:2:end)-1)/2)) = 1
Z =
     0
     0
     0
     0
     0
     0
     0
     1
     0
     0
     0
     0
     0
     0
     0
     1
     0
     0
     0
     0

Answer 4

[Meshooo]

Thank you all for the kind help and suggestions.