find indicies of maximum and minmum elements of an array in its each segmented smaller arrays, plus mean

I have a large array A (1x3,000,000) and it was segmented by an indexA (1x5,000). Part of A (first 100 element) and indexA are as below:

 A=[0           0           0           0           0           0 ...        
    0           0           0           0           0           0 ...
    0           0           0           0           0           0 ... 
    0           0           0           0           0           0 ...
    0           0           0           0           0           0 ...
    0.4665982   1.0120536   1.0120536   1.0120536   1.0120536   1.0120536 ...
    2.2720058   5.6612258   6.2438807   6.2438807   6.2438807   0.1706306 ...
    3.4740953   7.5749402   8.3634338   8.3634338   8.3634338   0.9215111 ...
    5.1563759   9.1346588   9.5029898   9.5029898   0.3987859   2.7512887 ...
    7.7671609  11.3574429  11.4436398   0.3256813   1.5965981   5.1405444 ...
    10.3805666  13.0157471           0   0.6612921   2.3093321   6.6952753 ...
    11.4918060  12.5332108   0.1177574   1.0389878   3.9089594   8.5320177 ...
    10.2545567  10.2545567   0.0723169   2.8528883   7.3057065   8.5047731  ...
    0.0136646   0.8193274   4.6227546   8.9367990   9.6189337   0.2354149  ...
    1.6503071   5.8358650  10.0257235  10.5574694   0.4393677   2.0512471  ...
    6.2374287  10.5120144  11.1818972   0.4886205   2.0159214   5.8866639  ...
    10.4216042  11.6498079   0.3699284   1.7069057];
indexA=[8    16    30    37    43    48    66    71    76    81    86    91    97];

The problem is to find index of max/min of A in each small segemented array based of the indexA, such as the segmented array are A(1:8-1), A(8:16-1),...A(91:97-1), A(97, end).

my solution using for loop is as below:

if indexA(1)>1
    indexA=[1 indexA];
elseif indexA(end)<length(A)
    indexA=[indexA length(A)];
end
Amax=[];idxAmax=[];Amin=[];idxAmin=[];Amean=[];
for i=1:length(indexA)-1
    [val, idx] = max(A(indexA(i):indexA(i+1)-1);
    idxAmax=[idxAmax idx+indexA(i)-1];
    Amax =[Amax val];
    [val, idx] = min(A(indexA(i):indexA(i+1)-1);
    Amin =[Amin val];
    idxAmin=[idxAmin idx+indexA(i)-1];
    Amean=[Amean  mean(A(indexA(i):indexA(i+1)-1)];
end

Is there other solution simpler and faster than using for loop? (The problem is updated. Elements of IndexA is the begining index of each segment in A now. And you can copy A directly to Matlab now).

Answer 1

[Bruno Luong]

Test example

A=1:100;
indexA=[8    16    30    37    43    48    66    71    76    81    86    91    97];

Code

I = zeros(size(A));
I([0 indexA]+1) = 1;
I(length(A)+1:end) = []; % incase indexA(end) contains the last index of A
I = cumsum(I);
[minAI,maxAI] = splitapply(@bounds, A, I);
meanAI = splitapply(@mean, A, I);

[Limei Cheng]

Thx a lot. Since I modify the problem a little bit. Now, the element in indexA is the index of the begining of each segment in A now. (previously it's the end of each segment). Also, minAI and maxAI in your solution is the actual value of the min and max, not the index.

[Fangjun Jiang]

Then [~,maxAI] = splitapply(@max, A, I) and [~,minAI] = splitapply(@min, A, I). Should be faster than using cellfun().

[Limei Cheng]

Let me work on it tomorrow and will get back to you tomorrow. BTW: I posted another question: find indices of each first repeated elements in the array (full of repeated values). Can you help? The link is here: https://www.mathworks.com/matlabcentral/answers/578391-find-indices-of-each-first-repeated-elements-in-the-array-full-of-repeated-values?

[Bruno Luong]

"Now, the element in indexA is the index of the begining of each segment in A now".

Modification for this new request

I = zeros(size(A));
I(indexA) = 1;
I = cumsum(I);
if I(1)==0
    I = I+1;
end
[minAI,maxAI] = splitapply(@bounds, A, I);
meanAI = splitapply(@mean, A, I);

[Limei Cheng]

Thanks Bruno and Fangjun. Both solutions work. Cellfun is actually faster than splitapply after test.

[Bruno Luong]

Yeah as most of the recent MATHWORKS functions (SplitApply) it's slow. If speed is important and you are willing to go with histotical function, they usually faster. In this case the old ACCUMARRAY will do nicely the job and fast (which SPLITAPPLY tries to replace).

%%
A=1:1e6+1;
indexA=8:8:1e6;

tic
NewIndex=diff([0,indexA,numel(A)]);
cellA=mat2cell(A,1,NewIndex);
[~,idxMaxA]=cellfun(@max,cellA);
idxMaxA=idxMaxA+[0,indexA];
toc % Elapsed time is 0.346554 seconds.

tic
I = zeros(size(A));
I([0 indexA]+1) = 1;
I(length(A)+1:end) = []; % incase indexA(end) contains the last index of A
I = cumsum(I);
maxAI = splitapply(@max, A, I);
toc % Elapsed time is 2.071576 seconds.

tic
I = zeros(size(A));
I([0 indexA]+1) = 1;
I(length(A)+1:end) = []; % incase indexA(end) contains the last index of A
I = cumsum(I);
idxmaxAI = accumarray(I(:),A(:),[],@max)';
toc % Elapsed time is 0.120087 seconds.

Answer 2

[Fangjun Jiang]

Here is one way to find the max index without for-loop. Not sure if it is faster for larger data

%%
A=1:100;
indexA=[8    16    30    37    43    48    66    71    76    81    86    91    97];

NewIndex=diff([0,indexA,numel(A)]);
cellA=mat2cell(A,1,NewIndex);
[~,idxMaxA]=cellfun(@max,cellA);
idxMaxA=idxMaxA+[0,indexA]
idxMaxA =
     8    16    30    37    43    48    66    71    76    81    86    91    97   100

[Limei Cheng]

Thanks a lot. There is still some issue with your code since I modify the index of each segment. The element in indexA is the index of the beginning of each segment in A now.

[Limei Cheng]

After modified you code, it works now. Thx. And it's more efficient than for loop when the length of A is huge.

[Fangjun Jiang]

I think Bruno Luong's splitapply() approach has clear advantage. You should figure that out and accept the better answer.

[Limei Cheng]

Thx for your suggestions, but I tested modified solutions of yours and it’s faster than Bruno’s. Also, I use cellfun get mean of each this segment.

[Fangjun Jiang]

Answer for the modified question.

NewIndex=diff([1,indexA,numel(A)+1]);
cellA=mat2cell(A,1,NewIndex);
[~,idxMaxA]=cellfun(@max,cellA);
idxMaxA=idxMaxA+[0,indexA-1]
idxMaxA =
     7    15    29    36    42    47    65    70    75    80    85    90    96   100