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Converting elements which repeat more than 10 times to 0

조회 수: 2(최근 30일)
Mate 2u
Mate 2u 4 Jul 2013
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
  댓글 수: 1
Sivakumaran Chandrasekaran
i think u can use sum command for 10 ones... if sum==10 then '0' should be placed.

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채택된 답변

Ken Atwell
Ken Atwell 4 Jul 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

추가 답변(1개)

Tom
Tom 4 Jul 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)
  댓글 수: 1
Ken Atwell
Ken Atwell 5 Jul 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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