How can i detect max value and his index ?

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benghenia aek
benghenia aek il 22 Gen 2019
Commentato: benghenia aek il 23 Gen 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 Commenti
Rik
Rik il 22 Gen 2019
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
benghenia aek il 23 Gen 2019
thank you very much for your assistance

Accedi per commentare.

Risposte (4)

Rik
Rik il 22 Gen 2019
Modificato: Rik il 22 Gen 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)

TADA
TADA il 22 Gen 2019
Modificato: TADA il 22 Gen 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);
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
  2 Commenti
TADA
TADA il 22 Gen 2019
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
benghenia aek il 23 Gen 2019
thank you very much for your assistance

Accedi per commentare.


Kevin Phung
Kevin Phung il 22 Gen 2019
Modificato: Kevin Phung il 22 Gen 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 Commenti
Rik
Rik il 22 Gen 2019
The odd notation is the same as in his other question, where we implicitly assumed he wanted a cell array.
benghenia aek
benghenia aek il 23 Gen 2019
thank you very much for your assistance

Accedi per commentare.


Brian Hart
Brian Hart il 22 Gen 2019
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.

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