Converting elements which repeat more than 10 times to 0

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Hi there, I have an array (4048x1) full of 1s and -1s. I want change it so that, if there are more than 10 consecutive 1s or 10 consecutive -1s, then the rest of the consecutive elements after the 10 will become 0 (zeros).
Ie, if we have 1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 -1 1
we would want 1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 0 0 -1 1 -1 1
Thanks
  1 comentario
Sivakumaran Chandrasekaran
Sivakumaran Chandrasekaran el 4 de Jul. de 2013
i think u can use sum command for 10 ones... if sum==10 then '0' should be placed.

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Ken Atwell
Ken Atwell el 4 de Jul. de 2013
I would love to see a vectorized way to do this, but using a loop:
x=[1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 -1 1];
zeroX = false(size(x));
for i =11:numel(x)
xsum = sum(x(i-10:i));
if xsum>10 || xsum <-10
zeroX(i) = true;
end
end
x(zeroX) = 0

Más respuestas (1)

Tom
Tom el 4 de Jul. de 2013
A = [1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 -1 1];
Correct = [1 -1 1 1 -1 -1 1 1 1 1 1 1 1 1 1 1 0 0 -1 1 -1 1];
D = [0 diff(A)]/2; %find split points
C = cumsum(abs(D))+1; %create accumarray subs
M = accumarray(C',A,[], @(x) {x'.*(1:length(x) <= 10)}); %split each section into a cell array and set > 10 to 0
B = [M{:}]; %merge
isequal(Correct,B)
  1 comentario
Ken Atwell
Ken Atwell el 5 de Jul. de 2013
Ah, the magic of accumarray, thanks for sharing.
In this case, the loop is probably the "better choice", as it runs in under half the time and would use far less memory (no cell arrays)

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