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Checking a specific column in a matrix with a specific row,column in another matrix

조회 수: 1 (최근 30일)
Greetings,
I want to check an (entire column) of B with a specific (row,column) in A and then extract values in B that are lower/higher than that value in A. For example; B(:,1) with A(1,1). I have already done this by creating a function file and calling it in my script.
My question now is how do I go from B(:,1) with A(1,1) to B(:,2) with A(2,2) without using this.
[Higher, Lower] = check(A(1,1), B(:,1))
[Higher1, Lower1] = check(A(1,2), B(:,2))
I want something that will automatically insert the desired matrix position. I'm asking this because my actual data set for matrices A = 30*12 and B = 62*12, thus making my method below inefficient. I have also included the expected results below.
clc; clear;
A = [1 2 1 2 5 2 1 2; 3 4 5 6 7 8 2 1];
B = [1 2;0 1; 2 4; 3 7; 0 5; 1 3; 2 4; 3 3; 9 8];
% Check B with A (I HAVE PROBLEM WITH THIS)
[Higher, Lower] = check(A(1,1), B(:,1))
[Higher1, Lower1] = check(A(1,2), B(:,2))
My function file:
function [Higher, Lower] = check(x,y)
% This function checks the y against x and extracts the respective higher/lower values.
% Set initial count to 0
con1 = 0;
count1 = 0;
for j = 1
for i = 1:9
if y(i) > x(j)
con1 = con1 + 1;
Higher(con1,j) = y(i);
end
end
con1 = 0;
end
for j = 1
for i = 1:9
if y(i) < x(j)
count1 = count1 + 1;
Lower(count1,j) = y(i);
end
end
count1 = 0;
end
end
Expected results:
Higher = [2;3;2;3;9]
Lower = [0;0]
Higher1 = [4;7;5;3;4;3;8]
Lower1 = [1]
All help is greatly appreciated. Thanks.
  댓글 수: 1
Guillaume
Guillaume 2019년 3월 29일
Jan has already shown how to write a much simpler check function and you should use that. However, even if you were going to use a loop as you have done, never hardcode the bounds of the loops. You wrote a function that accept a x of any size, yet your loop assumes that x is 9 elements. The whole point of functions is that they adapt to the input, so just changing
for i = 1:9
to
for i = 1:numel(x)
would mean that your current code would work regardless of the size of x (instead of erroring if x has less than 9 elements and producing the wrong result if it has more).
Also,
for j = 1
Well, that's absurd, it's the same as
j = 1;

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채택된 답변

Jan
Jan 2019년 3월 28일
편집: Jan 2019년 3월 28일
I've simplified the code at first:
A = [1 2 1 2 5 2 1 2; 3 4 5 6 7 8 2 1];
B = [1 2;0 1; 2 4; 3 7; 0 5; 1 3; 2 4; 3 3; 9 8];
[Higher, Lower] = check(A(1,1), B(:,1))
[Higher1, Lower1] = check(A(1,2), B(:,2))
function [Higher, Lower] = check(x,y)
% This function checks the y against x and extracts the respective higher/lower values.
Higher = y(y > x);
Lower = y(y < x);
end
This produces the wanted output. So what is the remaining problem? What does "how do I go from B(:,1) with A(1,1) to B(:,2) with A(2,2) without explicitly mentioning it" mean?
Maybe you want this:
A = rand(30, 12);
B = rand(62, 12);
Higher = cell(size(A));
Lower = cell(size(A));
for i1 = 1:size(A, 1)
for i2 = 1:size(A, 2)
[Higher{i1, i2}, Lower{i1, i2}] = check(A(i1, i2), B(:, i2));
end
end
I'm not sure, what the wanted output is. But for storing arrays of different sizes, cell arrays are perfect.
  댓글 수: 9
Jan
Jan 2019년 3월 29일
편집: Jan 2019년 3월 29일
@Elizabeth: This is not the complete message. The part, which mentions what the problem is, is missing.
I guess that now either A is the complete matrix, or B contains something other than mentioned before. Note that in all othercode before, check was called with 1 element of A only, A(i1,i2), but now you provide the complete matrix.
Elizabeth Yeap
Elizabeth Yeap 2019년 3월 29일
편집: Elizabeth Yeap 2019년 3월 29일
@Jan @Guillaume
I have managed to solve the issue. Thank you both for your help.

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