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

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 = 
0.468490416082415  -68.0623566640544
0.861091227948802  -62.0460361812610
0.920662009016715  -52.2025156248263
1.08087425638995  -34.5197924821680
1.17439664631856  -48.4780486951264
1.21018219911556  -45.0147047153047
1.24281651424282  -58.9951529447540
1.31409003462488  -62.4048732765717
1.32139564340592  -45.4143879292352
1.32229841475997  -52.0082496386087
1.40064375002948  -49.6318599064140
1.44367554410444  -58.8869137147488
1.45458058653115  -55.1869654737137
1.46737972112234  -49.2348443830192
1.47623284910443  -41.3789920515417
1.48211171204936  -44.6901137312740
1.48464667747544  -45.5836110933174
1.50221080043184  -60.8266330903523
1.51770839240522  -45.3482744837045
1.53398235212043  -45.6985828174899
end

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

if true
0.468490416082415  -68.0623566640544
0.861091227948802  -62.0460361812610
0.920662009016715  -52.2025156248263
1.08087425638995  -34.5197924821680
1.17439664631856  -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.

Answer 1

[Bruno Luong]

AB = [...
0.468490416082415  -68.0623566640544
0.861091227948802  -62.0460361812610
0.920662009016715  -52.2025156248263
1.08087425638995  -34.5197924821680
1.17439664631856  -48.4780486951264
1.21018219911556  -45.0147047153047
1.24281651424282  -58.9951529447540
1.31409003462488  -62.4048732765717
1.32139564340592  -45.4143879292352
1.32229841475997  -52.0082496386087
1.40064375002948  -49.6318599064140
1.44367554410444  -58.8869137147488
1.45458058653115  -55.1869654737137
1.46737972112234  -49.2348443830192
1.47623284910443  -41.3789920515417
1.48211171204936  -44.6901137312740
1.48464667747544  -45.5836110933174
1.50221080043184  -60.8266330903523
1.51770839240522  -45.3482744837045
1.53398235212043  -45.6985828174899 ]

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

Result

Amax =

      1.0809  -34.5198
      1.2102  -45.0147
      1.4762  -41.3790
      1.4821  -44.6901

[Yaser Khojah]

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]

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.

Answer 2

[Matt J]

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);