comparing two matrices of different dimensions

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ahmad Saad
ahmad Saad 2023-8-20
评论： Bruno Luong 2023-8-21
i have two matrices,A with dimension (23,2) and B with dimension (50465,2) .
i compare the first col of each matrix.
i need to keep values such that : abs(A(:,1)-B(:,1)) < 0.5.
C has four col : A(:,1) B(:,1) A(:,2) B(:,2)
my trial :
num_rows1=size(A(:,1));
num_rows2=size(B(:,1));
for i=1:1:num_rows1
for j=1:1:num_rows2
if abs(A(j,1)-B(i,1))<=0.5
%if data1(i,1)==0
% data1(i,1)=[];
% end
c1(j,[1:4])=[A(i,1),B(j,1), A(j,2), B(i,2)] ;
end
end
end
++++
ANY HELP
  2 个评论
Bruno Luong
Bruno Luong 2023-8-20
c1(j,[1:4])=[A(i,1),B(j,1), A(j,2), B(i,2)] ;
You take an element from row #j (and 2nd column) of A (third element of rhs) and j supposes to be up to 50465?
What did you said? A has 23 rows?

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采纳的回答

Bruno Luong
Bruno Luong 2023-8-21
编辑：Bruno Luong 2023-8-21
Ran in:
Fix several bugs of your for-loop code
load('data1.mat');
load('data2.mat');
A=data1;
B=data2;
num_rows1=size(A(:,1),1); % BUG here
num_rows2=size(B(:,1),1); % BUG here
c1 = nan([num_rows2,4]);
for i=1:1:num_rows1
for j=1:1:num_rows2
if abs(A(i,1)-B(j,1))<=0.5 % BUG here
c1(j,[1:4])=[A(i,1),B(j,1),A(i,2),B(j,2)] ; % BUG here
end
end
end
c1
c1 = 50465×4
0.1050 0.0083 1.5745 2.3700 0.1050 0.0167 1.5745 2.3600 0.5222 0.0250 2.0570 2.3500 0.5222 0.0333 2.0570 2.3400 0.5222 0.0417 2.0570 2.3100 0.5222 0.0500 2.0570 2.3300 0.5222 0.0583 2.0570 2.3000 0.5222 0.0667 2.0570 2.2700 0.5222 0.0750 2.0570 2.2700 0.5222 0.0833 2.0570 2.3000
  4 个评论
显示 2更早的评论
ahmad Saad
ahmad Saad 2023-8-21
Bruno Luong : yes. it works..
if you please.. i need to classify c1sorted to be hourly-based data.
for example:
for 0 < col2 <=1 get the median of corresponding col3 (and col4)
for 1 < col2 <=2 get the median of corresponding col3 (and col4)
.
.
.
.
for 23 < col2 <=24 get the median of corresponding col3 (and col4)
so, i get a matrix of three columns:
c1final= [i median(col3) median(col4)];
where i =1:24

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更多回答（1 个）

Image Analyst
Image Analyst 2023-8-20
Here is an alternate way using pdist2 in the Statistics and Machine Learning Toolbox.
% Create simple, sample integer data.
A = randi(9, 15, 2) % [x, y]
B = randi(9, 25, 2)
% Find distances between each each point and every other point.
distances = pdist2(A, B)
% Find map of where distances are less than some threshold
threshold = 3; % Whatever closeness value you want.
closeDistances = distances <= threshold
[rowsOfA, rowOfB] = find(closeDistances)
% Get in form of xA, xB, yA, yB
C = [A(rowsOfA, 1), B(rowOfB, 1), A(rowsOfA, 2), B(rowOfB, 2)]
  2 个评论
ahmad Saad
ahmad Saad 2023-8-20
Image Analyst : Thanks for kind attention.
unfortunately, i havnt install Machine Learning Toolbox.
So, " for loop "is perferable.
Image Analyst
Image Analyst 2023-8-21
Then we're not sure what you're asking when you tersely say "Any help". You have a for loop, which seems to be your "preferable" way. So what's the problem?

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