Comparing Elements of two matrices if loop

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Esther Roosenbrand 2017-9-20

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评论： OCDER 2017-9-22

Hi, I want to compare two matrices one with size m*n and the other 2*k. The second matrix (2*k) indicates values which need to be satisfied by the m*n matrix. Basically, I calculated a value in the second matrix for 360 degrees of a circle. In the first matrix measurements of these values for a circle. In the end I want to know how many times both values, the 360 degrees and the values are met.

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OCDER 2017-9-20

编辑：OCDER 2017-9-20

在 MATLAB Online 中打开

Hi Esther, we want to help but it's still confusing. In MATLAB, m*n means m rows by n columns. You have azimuth + elevation in "separate columns" (2 columns) for 24 hrs (24 rows) for 365 days (but there're 730 columns?) Did you mean a (24*365) row by 2 column matrix?

Then you have 2*360 matrix, which Matlab users interpret as 2 rows by 360 columns, where the "first column" (2 data points) is the "required sun altitude for 360 degrees" (360 data points). Did you mean 360x2 matrix? Also, what does the second column store?

Lastly, about the sun being at a correct azimuth and elevation, I'm not sure what is "correct". Must they have the exact, minimum, maximum value of degree and elevation?

Since you are comparing two matrices, there is a MATLAB function called intersect that lets you find the location and values where 1st matrix row equals 2nd matrix row. Is this what you need? Ex:

[AB, Aidx, Bidx] = intersect(A, B, 'rows')
AB = matrix that stores rows of A and B that are the same (intersect)
Aidx = row index in A that's the same as AB ( AB = A(Aidx, :) )
Bidx = row index in B that's the same as AB ( AB = B(Bidx, :) )

Esther Roosenbrand 2017-9-21

在 MATLAB Online 中打开

A=

 -60  340  -60  340
 -61  8  -61  8
 -57  34  -57  34
 -51  55  -51  55
 -43  72  -42  71
 -34  85  -33  85
 -24  97  -24  97
 -15  108  -15  108
 -7  119  -7  119
 1  130  1  130
 7  143  7  143
 12  156  12  156

So the first column here is the elevation angle and the second column is the corresponding azimuth for this angle. The third and forth follow the same pattern but for day 2. A is extended to 24*730 matrix.

B=

So the first column of b is the elevation and the second the azimuth required for the location. B is extended to a 360*2 matrix.

I want to determine how many times both the azimuth and elevation of matrix B are met by matrix A.

Cedric 2017-9-21

编辑：Cedric 2017-9-21

By counting, do you mean checking if there was one match?

For a given day you have on row of B, which is a single pair of elevation and azimuth. Then for the same day in A you have 24 pairs. How is it possible that you get twice (or more) the same pair of values from A when pair are given hourly?

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Answer 1

OCDER 2017-9-21

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在 MATLAB Online 中打开

Hi Esther, this is one of many ways to do this.

%Reshape A into 2-columns matrix (called Ar) to avoid having to use nested for loops later
Ar = reshape([A(:, 1:2:end) A(:, 2:2:end)], numel(A)/2, 2);
%R will store value = 1 if conditions are met, value = 0 if conditions fail
R = zeros(size(Ar, 1), 1); 
for k = 1:size(Ar, 1)
    %if azimuth and elevation in Ar is greater than ANY from B, change R(k) to 1
    if any( Ar(k, 1) >= B(:, 1) & Ar(k, 2) >= B(:, 2)) 
        R(k) = 1;
    end
end
%Number of times conditions are met per hour (row) for all day (col)
R = reshape(R, size(A, 1), size(A, 2)/2);
R =
     0     0
     0     0
     0     0
     0     0
     0     0
     0     0
     0     0
     0     0
     0     0
     0     0
     0     0
%Number of times conditions are met per day
Rday = sum(R, 1);
Rday = 
     0     0
%Number of times conditions are met per year
Ryear = sum(Rday);
Ryear = 
     0

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Esther Roosenbrand 2017-9-22

Thanks so much! Exactly what I was looking for!

OCDER 2017-9-22

You're welcome!

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Answer 2

Cedric 2017-9-21

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编辑：Cedric 2017-9-21

在 MATLAB Online 中打开

Or you can operate in 3D:

 A = [-10, 2, -8, 4; ...
       -4, 6, -3, 7; ...
        2, 5,  3, 6] ;
 B = [ 3, 6; ...
      -8, 4] ;
 dayHasMatch = squeeze(any(all(reshape(A, size(A, 1), 2, []) - permute(B, [3,2,1]) == 0, 2)))

with that you get

 dayHasMatch =
  2×1 logical array
   0
   1

which indicates that there was a match on day 2.

PS: and if you have an old version of MATLAB, the expansion must be performed using BSXFUN, and you can use a test of equality directly instead of checking that the difference is null:

 dayHasMatch = squeeze(any(all(bsxfun(@eq, reshape(A, size(A, 1), 2, []), permute(B, [3,2,1])), 2)))

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