Using find in a loop

Hello,

I have a problem with the code below. I am trying to get the maximum values of the vector I and its indices. I find the maximum values without problems and store them in Maxi. Finding the indices does not work. The code below gives a 1+15 double filled with [3,2,1,3,3,3,2,1,1,1,3,3,3,2,1] instead of [3, 3, 3, 4,7, 8,8, 8, 9, 10, 13, 14, 15, 15, 15]. Could anyone please let me know what I am doing wrong?

I = [1, 2, 4, 3,3,3,5, 10,8,7, 2,4, 5, 6, 8,6,5 ];
for ii = 1:15
  Maxi(ii) = max(I(ii:(ii+2)));
  I_Maxi(ii) = find(I(ii:(ii+2)) == max(I(ii:(ii+2))));
end

I want only to find the first index within the window ii:ii+2 of the maximum value of I and store it. Is this somehow possible?

Thank you for your help.

Answer 1

[Azzi Abdelmalek]

Use find(expression,1)

I = [1, 2, 4, 3,3,3,5, 10,8,7, 2,4, 5, 6, 8,6,5 ];
for ii = 1:15
  Maxi(ii) = max(I(ii:(ii+2)));
  I_Maxi(ii) = find(I(ii:(ii+2)) == max(I(ii:(ii+2))),1);
end

[Silke]

Hi Azzi,

Thank you for your reply. Unfortunately, the code that you have posted does not find the correct values. It finds the same values for I_Maxi then as with my code. Do you have another idea?

Cheers,

[Azzi Abdelmalek]

I = [1, 2, 4, 3,3,3,5, 10,8,7, 2,4, 5, 6, 8,6,5 ];
for ii = 1:15
  [b,c] = max(I(ii:(ii+2)));
  maxi(ii)=b;
  idxi(ii)=c+ii-1
end

%or

I = [1, 2, 4, 3,3,3,5, 10,8,7, 2,4, 5, 6, 8,6,5 ];
n=numel(I)-2
for ii = 1:n
  [maxi(ii),idxi(ii)] = max(I(ii:(ii+2)));
end
idxi=idxi+(0:n-1)

[Silke]

Thanks a lot. It is working now fine.

Answer 2

[Andrei Bobrov]

I = [1, 2, 4, 3,3,3,5, 10,8,7, 2,4, 5, 6, 8,6,5 ];
for ii = 15:-1:1
    [Maxi(ii,1),jj]= max(I(ii:(ii+2)));
    I_Maxi(ii,1) =  jj + ii -1;
end

OR

ii = (0:numel(I)-3)';
[Maxi,jj] = max(I(bsxfun(@plus,1:3,ii)),[],2);
I_Maxi = jj + ii;

[Silke]

Hi Andrei,

Thanks for the answer. Unfortunately, your code gives the same result as mine and Azzi's. Is there no way to find the correct indices from I and not the indices of I(1:3) that are always between 1 and 3?

Thank you, Silke

[Andrei Bobrov]

Corrected.