Create new matrix based on grid location

Question

eko supriyadi 2021-2-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/744372-create-new-matrix-based-on-grid-location

评论： eko supriyadi 2021-2-13

Suppose I have matrix A:

A=[4 1 0.6;4 1 0.8; 1 4 0.5; 1 3 0.3; 3 2 0.1; 2 1 0.6; 2 4 0.5; 3 2 0.4; 1 1 0.3; 1 2 0.1]

Now, I want to find the mean value in the last column matrix A above and then created in the new matrix. So I want to generate the new matrix A as:

A =
    0.6000         0         0         0
         0    0.2500         0         0
    0.6000         0         0    0.5000
    0.3000    0.1000         0         0

Anyone, can help me? I've tried looping and if command, but i always get errors

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Jan 2021-2-13

Please post your code and a copy of the complete error message.

I do not see, how the output of the 4x4 matrix can be obtained based on the 10x3 input matrix and a "mean of the last column".

eko supriyadi 2021-2-13

在 MATLAB Online 中打开

hello Jan, in matrix A above:

A =
0000    1.0000    0.6000
0000    1.0000    0.8000
0000    4.0000    0.5000
0000    3.0000    0.3000
0000    2.0000    0.1000
0000    1.0000    0.6000
0000    4.0000    0.5000
0000    2.0000    0.4000
0000    1.0000    0.3000
0000    2.0000    0.1000

I want create new matrix with 4x4 based on value of column 1 and 2, you can see that the values ranges from 1 to 4. In short, the column 1 as new row matrix and the column 2 as new column matrix.. the code as far as I'm trying:

result=zeros(length(unique(A(:,1))),length(unique(A(:,2))))
for i = A(:,1)
    for j = A(:,2)
    if A(:,1) == i & A(:,2)==j
    result(i,j)=mean(A(:,3));
    end
    end
end

thx.

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Answer 1

Jan 2021-2-13

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/744372-create-new-matrix-based-on-grid-location#answer_622702

编辑：Jan 2021-2-13

在 MATLAB Online 中打开

A = [4 1 0.6; 4 1 0.8; 1 4 0.5; 1 3 0.3; 3 2 0.1; 2 1 0.6; ...
     2 4 0.5; 3 2 0.4; 1 1 0.3; 1 2 0.1];
result = zeros(max(A(:,1)), max(A(:,2)));
for i = A(:, 1).'
   for j = A(:, 2).'
      index = (A(:,1) == i & A(:,2)==j);
      if any(index)
         result(i, j) = mean(A(index, 3));
      end
   end
end
% result = 
% [0.3, 0.1,  0.3, 0.5; ...
%  0.6, 0,    0,   0.5; ...
%  0,   0.25, 0,   0; ...
%  0.7, 0,    0,   0]

This does not match your wanted output exactly:

A =

0.6000 0 0 0

0 0.2500 0 0

0.6000 0 0 0.5000

0.3000 0.1000 0 0

Problems of your code:

for i = A(:, 1)

This runs a loop with 1 iteration only, where i is the first column of A. Maybe you mean:

for i = A(:, 1).'

Same for the other loop.

Remember that the if command needs a scalar argument. Then:

if A(:,1) == i & A(:,2)==j

is modified internally to:

if all(A(:,1) == i & A(:,2)==j)

which is TRUE because i is the first column of A already (see above).

My code overwrites result(i,j) multiple times with the same value. It is just thought to clarify, what you want to obtain. For a real program you would use the faster:

result = accumarray(A(:, 1:2), A(:, 3), [], @mean)