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How to filter an array?

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Meshooo
Meshooo on 6 Jan 2016
Commented: Guillaume on 7 Jan 2016
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

  3 Comments

KSSV
KSSV on 6 Jan 2016
one represnetative for each group of 1? Question not clear. Can you type expected result into a matrix?
Stephen Cobeldick
Stephen Cobeldick on 6 Jan 2016
And what if there are an even number of ones?
A = [0;1;1;0]
what output do you want?
Meshooo
Meshooo on 6 Jan 2016
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.

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Accepted Answer

Guillaume
Guillaume on 6 Jan 2016
It's very simple:
diff([0; A]) == 1
simply detects all transition from 0 to 1 and return them as 1.

  1 Comment

Guillaume
Guillaume on 7 Jan 2016
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.

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More Answers (3)

goerk
goerk on 6 Jan 2016
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

  2 Comments

Guillaume
Guillaume on 6 Jan 2016
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
goerk on 6 Jan 2016
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]);

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Stephen Cobeldick
Stephen Cobeldick on 6 Jan 2016
>> 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

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Meshooo
Meshooo on 7 Jan 2016
Thank you all for the kind help and suggestions.

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