Splitting a matrix into group based into the first column and select the maximum value from the second column for each interval

Question

Yaser Khojah 2018-10-17

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https://ww2.mathworks.cn/matlabcentral/answers/424575-splitting-a-matrix-into-group-based-into-the-first-column-and-select-the-maximum-value-from-the-seco

评论： Bruno Luong 2018-10-17

I have a big matrix,x, with size of 21,6000 x 2. The first Column is A and The second one is B. I'm looking for away to split the first column into interval and find the maximum value from each interval in B. For example;

if true
Original Matrix = 
468490416082415  -68.0623566640544
861091227948802  -62.0460361812610
920662009016715  -52.2025156248263
08087425638995  -34.5197924821680
17439664631856  -48.4780486951264
21018219911556  -45.0147047153047
24281651424282  -58.9951529447540
31409003462488  -62.4048732765717
32139564340592  -45.4143879292352
32229841475997  -52.0082496386087
40064375002948  -49.6318599064140
44367554410444  -58.8869137147488
45458058653115  -55.1869654737137
46737972112234  -49.2348443830192
47623284910443  -41.3789920515417
48211171204936  -44.6901137312740
48464667747544  -45.5836110933174
50221080043184  -60.8266330903523
51770839240522  -45.3482744837045
53398235212043  -45.6985828174899
end

After splitting them into 5 group, the first group will look like

if true
468490416082415  -68.0623566640544
861091227948802  -62.0460361812610
920662009016715  -52.2025156248263
08087425638995  -34.5197924821680
17439664631856  -48.4780486951264
end

From this first group, I would select the maximum in B and create a new vector which has:

if true
  1.08087425638995  -34.5197924821680
end

I tried this code but did not work for me.

if true
d1 = sortrows(x,[1 2])
subs=discretize(d1(:,1),3)
intervalMax=accumarray(subs(:),d1(:,2),[],@max)
end

Can anyone please help. thank you in advance.

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Answer 1

Bruno Luong 2018-10-17

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https://ww2.mathworks.cn/matlabcentral/answers/424575-splitting-a-matrix-into-group-based-into-the-first-column-and-select-the-maximum-value-from-the-seco#answer_341951

编辑：Bruno Luong 2018-10-17

在 MATLAB Online 中打开

AB = [...
468490416082415  -68.0623566640544
861091227948802  -62.0460361812610
920662009016715  -52.2025156248263
08087425638995  -34.5197924821680
17439664631856  -48.4780486951264
21018219911556  -45.0147047153047
24281651424282  -58.9951529447540
31409003462488  -62.4048732765717
32139564340592  -45.4143879292352
32229841475997  -52.0082496386087
40064375002948  -49.6318599064140
44367554410444  -58.8869137147488
45458058653115  -55.1869654737137
46737972112234  -49.2348443830192
47623284910443  -41.3789920515417
48211171204936  -44.6901137312740
48464667747544  -45.5836110933174
50221080043184  -60.8266330903523
51770839240522  -45.3482744837045
53398235212043  -45.6985828174899 ]
B = reshape(AB(:,2),5,[]);
[~,loc] = max(B,[],1);
Amax = AB(loc+5*(0:size(B,2)-1),:)

Result

Amax =
0809  -34.5198
2102  -45.0147
4762  -41.3790
4821  -44.6901

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Yaser Khojah 2018-10-17

Thanks Bruno for your help. I have tried your code with my data and i'm wondering if I can control the size of the interval.

Bruno Luong 2018-10-17

在 MATLAB Online 中打开

The size of interval is 5, so change the two "5" in my code.

p = 5;
B = reshape(AB(:,2),p,[]);
[~,loc] = max(B,[],1);
Amax = AB(loc+p*(0:size(B,2)-1),:)

If the length of AB is not divisible by the block size (p) you must Pad AB by NaN like this

AB(end+1:ceil(size(AB,1)/p)*p,2) = NaN;

before the rest of the code.

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Answer 2

Matt J 2018-10-17

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https://ww2.mathworks.cn/matlabcentral/answers/424575-splitting-a-matrix-into-group-based-into-the-first-column-and-select-the-maximum-value-from-the-seco#answer_341952

在 MATLAB Online 中打开

You could use MAT2TILES (Download) as follows

C=mat2tiles(OriginalMatrix, [5,2]);
for i=1:numel(C)
     A=C{i}(:,1);
     B=C{i}(:,2);
    [~,j]=max(B);
    C{i}=[A(j), B(j)];
end
 intervalMax=cell2mat(C);