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

Question

Limei Cheng 2020-8-11

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/578302-find-indicies-of-maximum-and-minmum-elements-of-an-array-in-its-each-segmented-smaller-arrays-plus

编辑： Bruno Luong 2020-8-12

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 ...
4665982   1.0120536   1.0120536   1.0120536   1.0120536   1.0120536 ...
2720058   5.6612258   6.2438807   6.2438807   6.2438807   0.1706306 ...
4740953   7.5749402   8.3634338   8.3634338   8.3634338   0.9215111 ...
1563759   9.1346588   9.5029898   9.5029898   0.3987859   2.7512887 ...
7671609  11.3574429  11.4436398   0.3256813   1.5965981   5.1405444 ...
3805666  13.0157471           0   0.6612921   2.3093321   6.6952753 ...
4918060  12.5332108   0.1177574   1.0389878   3.9089594   8.5320177 ...
2545567  10.2545567   0.0723169   2.8528883   7.3057065   8.5047731  ...
0136646   0.8193274   4.6227546   8.9367990   9.6189337   0.2354149  ...
6503071   5.8358650  10.0257235  10.5574694   0.4393677   2.0512471  ...
2374287  10.5120144  11.1818972   0.4886205   2.0159214   5.8866639  ...
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).

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Bruno Luong 2020-8-11

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/578302-find-indicies-of-maximum-and-minmum-elements-of-an-array-in-its-each-segmented-smaller-arrays-plus#answer_478308

编辑：Bruno Luong 2020-8-11

在 MATLAB Online 中打开

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);

6 个评论
显示 4更早的评论隐藏 4更早的评论

Limei Cheng 2020-8-12

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

Bruno Luong 2020-8-12

编辑：Bruno Luong 2020-8-12

在 MATLAB Online 中打开

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.

请先登录，再进行评论。

Answer 2

Fangjun Jiang 2020-8-11

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/578302-find-indicies-of-maximum-and-minmum-elements-of-an-array-in-its-each-segmented-smaller-arrays-plus#answer_478304

编辑：Fangjun Jiang 2020-8-11

在 MATLAB Online 中打开

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

5 个评论
显示 3更早的评论隐藏 3更早的评论

Limei Cheng 2020-8-12

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 2020-8-12

在 MATLAB Online 中打开

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

请先登录，再进行评论。