matching matrices sizes using if statements

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Melissa
Melissa 2013 年 8 月 14 日
Good Afternoon,
I was wondering if someone could assist me on an if condition. I have two data sets A [nx3] and B[mx3]. I would like the rows of each data set to match one another with repeat values of that data set. For example if
A=[ 1 2 3;
4 5 6;
7 8 9]
and
B=[4 5 8;
1 2 7].
Then B would become
[4 5 6;
1 2 7;
4 5 6];
Something like
[ar,ac]=size(A);
[br,bc]=size(B);
if ar = br
A;
B;
if ar > br
if ar < br
Any suggestions?
Thanks!!
[Melissa, I took a little free rein to edit your code and make the question a little clearer. -- the cyclist]
  4 件のコメント
dpb
dpb 2013 年 8 月 14 日
Still me no follow...how did you get the particular rows in new_B again? Need a definitive prescription to be able to write any code...otherwise why isn't
B=[4 5 8;
1 2 7;
1 2 7]
just as valid? In fact, other than size(A,1), what does A have to do with it at all by your description?
Melissa
Melissa 2013 年 8 月 15 日
The B mentioned above would be valid. I just need the number of rows to match both in A and B. If A has more rows than B then New_B will just repeat B's rows until it matches the row size of A. It doesn't matter what rows of B that are used to get the matrix dimensions to match A.
Example: A=[1 2 3; 4 5 6; 7 8 9]; B=[4 5 6; 7 8 9];
New_B can equal
[4 5 6; 7 8 9; 4 5 6] or [4 5 6; 7 8 9; 7 8 9]

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採用された回答

kjetil87
kjetil87 2013 年 8 月 14 日
編集済み: kjetil87 2013 年 8 月 14 日
A=[ 1 2 3;...
4 5 6;...
7 8 9];
B=[4 5 8;...
1 2 7];
[ma,na]=size(A);
[mb,nb]=size(B);
if ma>mb
N=ceil(ma/mb);
B=repmat(B,N,1);
B=B(1:ma,:)
elseif mb>ma
N=ceil(mb/ma);
A=repmat(A,N,1);
A=A(1:mb,:)
end
%else they have equal number of rows, no operation needed.
  5 件のコメント
Melissa
Melissa 2013 年 8 月 15 日
You are right I meant if ma=mb for the last else statement. Thank you for correcting me.
Try A=[1 2 3; 4 5 6; 7 8 9]; B=[3 6 8; 9 1 4];
Wait...Once I fixed that part it worked. Huh. Thanks!!
kjetil87
kjetil87 2013 年 8 月 15 日
編集済み: kjetil87 2013 年 8 月 15 日
Glad to help, but again you are not using the "=" correct.
= will assign right hand side to left hand side.
== will compare left and right side and return logical 1 if they are equal, false if they are not and an error if they are not comparable (say e.g different sizes)
if you confuse them in an if test matlab will return you and error , but as you saw from your earlier code its important to use them correctly.
Glad the code worked. If the two matrices above caused problems, it might be because you ran the code several times and A and B did not have the content you thought when you started.

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