How can i detect max value and his index ?

I have vector a=[1 2 5 NAN NAN 9 0 23 12 NAN NAN NAN 6 2 8]

how to detect max value and it index ??

a(1)=[5] and index index(1)=[3]

a(2)=[23] and index index(2)=[8]

a(3)=[8] and index index(n)=[15]

[Rik]

Please use meaningfull tags. Using the names of contributors will not magically attract them to your question. Also, it would have been better to provide a link to your previous question for a bit more context.

[benghenia aek]

thank you very much for your assistance

Answer 1

[Rik]

Loop through the elements of your cell vector a, use the max function to find the maximum value for a, and the second output of max to find which index in index you need to keep.

Using the strategy by Jan (with a minor edit):

a = [1 2 5 NaN NaN 9 0 23 12 NaN NaN NaN 6 2 8];
m   = [true, isnan(a), true];
ini = strfind(m, [true, false])-1;
fin = strfind(m, [false, true])-1;
n   = numel(ini);
Result = cell(1, n);
Index  = cell(1, n);
Result_max = cell(1, n);
Index_max  = cell(1, n);
for k = 1:n
    Index{k}  = ini(k)+1:fin(k);
    Result{k} = a(ini(k)+1:fin(k));
    [Result_max{k},idx]=max(Result{k});
    Index_max{k}=Index{k}(idx);
end

celldisp(Result)
celldisp(Index)
celldisp(Result_max)
celldisp(Index_max)

Answer 2

[TADA]

If you meant to split the array into subarrays according to locations of NaN, then find maximal values of each subarray but the index of each maximal value in the original array:

a=[1 2 5 nan nan 9 0 23 12 nan nan nan 6 2 8];

numIdx = find(~isnan(a));
values = a(numIdx);
startAt = numIdx([true diff(numIdx) > 1]);
ii = find(diff(numIdx) > 1);
splittingMask = [ii(1) diff(ii) numel(values)-ii(end)];
c = mat2cell(values, 1, splittingMask);
maxVal = [zeros(1, numel(c));startAt-1];
for i = 1:numel(c)
    [mv, mi] = max(c{i});
    maxVal(:,i) = maxVal(:,i) + [mv; mi];
end

maxVal

maxVal =

     5    23     8
     3     8    15

[TADA]

or alternatively:

maxVal1 = {};
currMax = [nan;0];
for i = 1:numel(a)
    if isnan(a(i))
        if ~isnan(currMax(1))
            maxVal1{numel(maxVal1)+1} = currMax;
            currMax = [nan;0];
        end
        continue;
    end
    
    if isnan(currMax(1)) || a(i) > currMax(1)
        currMax = [a(i);i];
    end
end
if ~isnan(currMax(1))
    maxVal1{numel(maxVal1)+1} = currMax;
    currMax = [nan;0];
end

maxVal2 = cell2mat(maxVal1);
maxVal2

maxVal2 =

     5    23     8
     3     8    15

[benghenia aek]

thank you very much for your assistance

Answer 3

[Kevin Phung]

[max_val idx] = max(a)

max_val would be the max value in your array a,

idx would be the index.

i have no idea what

'a(1)=[5] and index index(1)=[3]

a(2)=[23] and index index(2)=[8]

a(3)=[8] and index index(n)=[15]'

means though.

[Rik]

The odd notation is the same as in his other question, where we implicitly assumed he wanted a cell array.

[benghenia aek]

thank you very much for your assistance

Answer 4

[Brian Hart]

Let's say your vector is

a =[1     2     5   NaN   NaN     9     0    23    12   NaN   NaN   NaN     6     2     8]

Then the max value is given by:

>> max(a)

ans =

    23

and the index of that value is given by:

    >> find(a==max(a))

ans =

     8

I'm not sure what you mean by a(1)=[5] (it would be "1"), a(2)=[23] (really "2"), etc.

[benghenia aek]

thank you very much for your assistance