create a matrix of maximums

Hey guys, thanks in advance,

I have this code that correlates in time two signals. The correlation is presented in a matrix(range_compressed_matrix) with 1960(time_compression_cut) x 400(waypoints).

This matrix has 400 columns, and each column is filled with 1960 values of a signal in time domain. So the x of this matrix is a position of a plane=1:400. And the y of this matrix is 1960 values of time. Which means that for each column, or position of a plane, I have a value among that 1960 values that is a maximum.

I want to get the values(xmax ,ymax) where the correlation is maximum for each column. Meaning that I will have a xmax=400(position of plane) and ymax=400(time).

In the end I want to save this two variables in two matrixes, I have this code, but I m not getting what I want, and dont know why.

[rows,columns]=size(range_compressed_matrix);
for col = 1 : columns
    thisColumn = range_compressed_matrix(:, col);
    maxValue = max(thisColumn(:));
    [rows2, columns] = find(thisColumn == maxValue);
    x_max(:,col) = waypoints(col);
    y_max(rows,:) = time_compression_cut(rows2);
    y_max(rows,:)= y_max(:,col)*c;
end

[Pooja Kumari]

what is c in your code?

[Miguel Albuquerque]

its the speed of light

[Johan]

It's not very clear to me what you want to achieve here, a clearer explaination might help you and us solving your problem. I put some comments in the code snippets you shared:

%Random values
range_compressed_matrix = rand(1960,400);
waypoints = rand(400,1);
time_compression_cut = rand(1960,1);
[rows,columns]=size(range_compressed_matrix)

rows = 1960

columns = 400

c = 1;
for col = 1 : columns
    thisColumn = range_compressed_matrix(:, col);
    maxValue = max(thisColumn(:));
    %You are setting columns as an output, its already used as your loop limit,
    %moreover your not using that variable so you can replace it by ~
%     [rows2, columns] = find(thisColumn == maxValue);
     [rows2, ~] = find(thisColumn == maxValue);
     
    %This is equivalent to x_max = waypoints'
    x_max(:,col) = waypoints(col);
    
    %rows = 1960 so this line sets the 1960 value of y_max to time_compression_cut(rows2);
    y_max(rows,:) = time_compression_cut(rows2);
    
    %y_max is single columned so this doesn't work
%     y_max(rows,:)= y_max(:,col)*c;
end
all(x_max == waypoints')

ans = logical
   1

Answer 1

[Voss]

% get the maximum value of each column, along with the row it is located in
% (if multiple elements in a column are at the same maximum value, only the
% first occurrence of the maximum is stored):
[max_correlation, row_idx] = max(range_compressed_matrix, [], 1);

% store the values of time_compression_cut where the max values occur:
y_max = time_compression_cut(row_idx);

% store the value of waypoints for each column:
x_max = waypoints;

Answer 2

[Rohit Kulkarni]

To get maximum value for each column and the corresponding indices you can do the following:

max_values = max(range_compressed_matrix)
%idx will give logical indices
idx = (range_compressed_matrix == max_values)
[rows, cols] = find(idx)

[dpb]

More direct is simply

[max_values,rows] = max(range_compressed_matrix);
cols=1:size(range_compressed_matrix,2);

You've already searched the array once to find the maxima, why on earth search it again to locate what has already been located????

find is rarely needed; as here, when not needed specifically, it's just wasted motion.

Answer 3

[dpb]

You didn't say there were values anywhere excepting in one column out of this array before -- you don't need no steenkin' loops here, either, max() is, like virtually all builtin MATLAB functions, vectorized and operates by column by default.

All you need is as I showed before is to return the optional second argument of max; here it'll give you the rows automagically because you're not looking for a global max across an entire array as in other...

Example:

>> r=rand(3,6)          % some dummy data for your long-winded variable
r =
    0.0020    0.3362    0.6381    0.7472    0.0574    0.7932
    0.3105    0.5159    0.9211    0.7499    0.0476    0.4048
    0.1045    0.7225    0.8400    0.2613    0.0946    0.0230
>> [mx,rmx]=max(r)       % get maxima by column, row location
mx =
    0.3105    0.7225    0.9211    0.7499    0.0946    0.7932
rmx =
     2     3     2     2     3     1
>> 

The column is simply 1:size(r,2) of course.

The other arrays(?) in your code are undefined; we've no way to know what to do with those.