how to count non zero elements in a vector and replace this values based on the count values

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Dear all,
I'm a new matlab user and in my case i have a vector let say: v= [0 0 0 0.1 0.2 0.3 0.4 0.5 0 0 0 0 0 0 0.1 0.2] I want to count consecutive non zero values i.e in my vector have first five nonzero values [0.1 0.2 0.3 0.4 0.5] and two last nozeros values[0.1 0.2] what I want is: count the consecutive non zero values and put a condition i.e if the length of nonzeros is greater then 3 (count>3) then the respective values of vector V(i) remain v(i) if the length consecutive values is less than three (count<3) then respective values of v(i) =0 I want to get a new vector let say v1 derivation from vector v where: v1= [0 0 0 0.1 0.2 0.3 0.4 0.5 0 0 0 0 0 0 0 0]
Any help would be appreciated Thank you DM

Respuesta aceptada

Azzi Abdelmalek
Azzi Abdelmalek el 30 de Jul. de 2015
Editada: Azzi Abdelmalek el 30 de Jul. de 2015
v= [0 0 0 0.1 0.2 0.3 0.4 0.5 0 0 0 0 0 0 0.1 0.2]
idx=[0 v~=0 0]
idx1=strfind(idx,[0 1])
idx2=strfind(idx,[1 0])-1
id=idx2-idx1+1
ii1=idx1(id>3)
ii2=idx2(id>3)
jj=cell2mat(arrayfun(@(x,y) x:y,ii1,ii2,'un',0))
out=zeros(size(v))
out(jj)=v(jj)

Más respuestas (2)

Andrei Bobrov
Andrei Bobrov el 30 de Jul. de 2015
Editada: Andrei Bobrov el 30 de Jul. de 2015
v= [0 0 0 0.1 0.2 0.3 0.4 0.5 0 0 0 0 0 0 0.1 0.2]';
ii = v ~= 0;
t = [false;diff(ii)==1];
i1 = cumsum(t).*ii;
N = histcounts(i1,1:(max(i1)+1));
i1(ismember(i1,find(N <= 3))) = 0;
out = v.*i1;
  2 comentarios
D Marini
D Marini el 30 de Jul. de 2015
thank you andwer however unfourtnely this solution didn't work, i got this error:
Undefined function 'histcounts' for input arguments of type 'double'. Error in TESTHeatLossCalcDHWCatoffShortDraw (line 111) N = histcounts(i1,1:(max(i1)+1));
Sean de Wolski
Sean de Wolski el 30 de Jul. de 2015
histcounts is new in R2014b, if you're on an older release either upgrade or use histc.

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D Marini
D Marini el 30 de Jul. de 2015
Thank you Azzi,
The code seems to work OK, Just wonder if would be possible to explain how it works (for bigginers is difficult to understand).
Thank you very much again D,

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