# How can i detect max value and his index ?

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benghenia aek on 22 Jan 2019
Commented: benghenia aek on 23 Jan 2019
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]
##### 2 CommentsShowHide 1 older comment
benghenia aek on 23 Jan 2019
thank you very much for your assistance

Rik on 22 Jan 2019
Edited: Rik on 22 Jan 2019
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)

Edited: TADA on 22 Jan 2019
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);
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
##### 2 CommentsShowHide 1 older comment
benghenia aek on 23 Jan 2019
thank you very much for your assistance

Kevin Phung on 22 Jan 2019
Edited: Kevin Phung on 22 Jan 2019
[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.
##### 2 CommentsShowHide 1 older comment
benghenia aek on 23 Jan 2019
thank you very much for your assistance

Brian Hart on 22 Jan 2019
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 on 23 Jan 2019
thank you very much for your assistance

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