- swapping a matrix

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Anusha
Anusha 2014 年 4 月 23 日
コメント済み: Geoff Hayes 2014 年 4 月 24 日
a =
1 3 5 6 7
2 3 2 3 2
5 6 7 8 7
b=
5 2 1 3 7
6 8 2 3 1
5 7 8 6 3
how to swap b to a.
ans=
1 3 5 6 7
2 3 2 3 2
5 6 7 8 7

採用された回答

Andrew Newell
Andrew Newell 2014 年 4 月 24 日
If I understand your latest comment, you begin by finding the unique numbers:
an = unique(a,'stable').'
bn = unique(b,'stable').'
an =
1 3 5 6 7 2 8
bn =
5 2 1 3 7 6 8
Now you take all the 5's in b and replace them by 1's, all the 2's by 3's, and so on, all at once (otherwise in a later step all the 1's will be changed back to 5's). You can do this by initializing a new matrix and putting all the numbers in it:
c = zeros(size(b));
for i=1:length(an)
idx = (b==bn(i));
c(idx) = an(i);
end
c
c =
1 5 7 8 6
2 3 5 8 7
1 6 3 2 8
This seems to be what you're saying, although the result is completely different from the answer you provided.

その他の回答 (1 件)

Andrew Newell
Andrew Newell 2014 年 4 月 23 日
I have two theories for what you mean:
b=a
just makes b equal to a. Or
z=b
b=a
a=z
exchanges the two matrices. I think you need to look at a Matlab tutorial or two.
  8 件のコメント
Anusha
Anusha 2014 年 4 月 24 日
I use following comments to take a valu an=unique(a,'stable'); bn=unique(b,'stable');
Now swap an and bn in b value. this is useful to get a?
Geoff Hayes
Geoff Hayes 2014 年 4 月 24 日
Was there some sort of suggestion (or part of the problem description) that to do the swap you can use the unique values of a and b (retrieved via stable)?

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