Find a value using an index obtained from a different matrix with same dimensions

Hello, I am working with huge amount of data which require to find a specific index in a matrix using a condition, and then use the indexes to extract another value using another matrix with the same dimension. For example: Let us say that I am iterating the same calculation varying the angle in a specfic range: x=0:1:5 degrees and then calculate for each angle a specific case like this.

x=[0 1 2 3 4 5] angle matrix;

a1= [2 3 5 2 1 0] % one case scenario, each value it has been obtanined for each angles 0,1 2 3 4 5 as the matrix "x"

if I repeat the proces using different conditions like a2 and a3 with dfifferent angles

a2=[0 1 3 1 0 0];

a3=[3 2 1 0 0 0]

The total matrix "a" will be:

a=[a1;a2;a3]=[2 3 5 2 1 0;0 1 3 1 0 0;3 2 1 0 0 0];

if I want to find the maximun value for each case a1, a2 and a3 (or each row of a) and obtain the angles at which the maximun occurs I did the following:

Max_a=max(a,[],2);

ind=find(a==Max_a); or [row,column]=find(a==Max_a);

then I can use the index to locate the angle creating a new matrix of angles with the same length of matrix a according to the different cases like this:

x_total=[x;x;x];=[0 1 2 3 4 5;0 1 2 3 4 5;0 1 2 3 4 5]

Then, I can use the indexes (ind or row and column) to extract the angles at which the maximum occurs for each a1, a2 and a3 with the following:

x_maxa = a(ind); or ind1 = sub2ind(size(x), row, column); x_maxa(ind1);

x_maxa=[2;2;0];

The problem is when I used huge amount of data like varying angles from 0:1:180 and cases from 1 to 100, I realized that these commands do not provide an accurate estimations of the angles at which the maximum occurs for each case. I would appreciate the help. I provide as well two txt.files where you find the total angle matrix and total values matrix for my real case.

[Bruno Luong]

"The problem is when I used huge amount of data like varying angles from 0:1:180 and cases from 1 to 100, I realized that these commands do not provide an accurate estimations of the angles at which the maximum occurs for each case. "

I'm not sure why you say that. The commands you post should work fine. Please post your code with all commands put together. May be the problem uis when there is tie values in a row. Not sure if you want to handle the tie case. The two answers bellow do not.

[Jorge Luis]

Hi, thank you for your response. The code is like this

% loading data

TableRotDpp_SA_NR= load('Total_matrix_cases.txt');

TableRotDpp_Rot_NR= load('Total_matrix_rotations.txt');

Max_RotD_SA=max(TableRotDpp_SA_NR,[],2);

idx_Max_SA = find(TableRotDpp_SA_NR==Max_RotD_SA);

Max_Rotation_RotD_SA = TableRotDpp_Rot_NR(idx_Max_SA);

Kind regards

[Aquatris]

The issue in your code was I think, it was not giving you the correct values for each row. So your expectation was Max_Rotation_RotD_SA would represent the max values of each row, but it was not.

If you make this simple change as @Stephen23 showed in his answer, your code also works properly:P

[Max_RotD_SA,idx_Max_SA]=max(TableRotDpp_SA_NR,[],2,'linear');

But @Jorge Luis is correct that this code only checks a single instance of max values and neglects other values that might also exist in the same row that result in the same max value for that row.

Answer 1

[Aquatris]

Something like this maybe:

%% your matrices
val = load('Total_matrix_cases.txt');
x = load('Total_matrix_rotations.txt');

% find maximum values in val for each row
[valMax,idx] = max(val,[],2); % idx is column index for each row

s = [];
for i = 1:size(val,1)
    s(i,:) = [i x(i,idx(i)) val(i,idx(i)) valMax(i)];
end
% s = [rowNumber x val(x) valMax] for each row
disp(s)

    1.0000   17.0000   10.6700   10.6700
    2.0000   17.0000   11.6398   11.6398
    3.0000  152.0000   16.2679   16.2679
    4.0000   89.0000   25.3473   25.3473
    5.0000   50.0000   42.3446   42.3446
    6.0000  154.0000   39.9599   39.9599
    7.0000  159.0000   40.2815   40.2815
    8.0000  125.0000   47.3734   47.3734
    9.0000   82.0000   34.9502   34.9502
   10.0000   57.0000   34.9621   34.9621
   11.0000  107.0000    7.4710    7.4710
   12.0000  113.0000    5.8002    5.8002
   13.0000  126.0000    2.3651    2.3651
   14.0000  124.0000    1.2982    1.2982
   15.0000   96.0000    0.7132    0.7132
   16.0000   95.0000    0.5445    0.5445
   17.0000   68.0000    0.5043    0.5043
   18.0000   45.0000    0.4373    0.4373
   19.0000   55.0000    0.3334    0.3334
   20.0000   50.0000    0.2636    0.2636

[Jorge Luis]

Thank you very much, really appreciated.

Answer 2

[Stephen23]

The efficient MATLAB approach:

tmc = readmatrix('Total_matrix_cases.txt');
tmr = readmatrix('Total_matrix_rotation.txt');
[tmx,idx] = max(tmc,[],2,'linear');
nmr = 1:numel(tmx);
out = [nmr(:),tmr(idx),tmx]

out = 20x3

1.0000 17.0000 10.6700 2.0000 17.0000 11.6398 3.0000 152.0000 16.2679 4.0000 89.0000 25.3473 5.0000 50.0000 42.3446 6.0000 154.0000 39.9599 7.0000 159.0000 40.2815 8.0000 125.0000 47.3734 9.0000 82.0000 34.9502 10.0000 57.0000 34.9621

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[Jorge Luis]

Thank you very much, :).