how to exchange matrix rows depending on distance

1 vue (au cours des 30 derniers jours)
janny
janny le 16 Nov 2014
Modifié(e) : janny le 16 Nov 2014
hi guys, i have the following matrix A with distance D1 and matrix B with distance D2.
i want a code to exchange A rows with B rows which have higher distance, as moving row 2 in B to matrix A row 2 and remove the old row 2 in A. the output should be:
how to do this guys?

Réponse acceptée

Guillaume
Guillaume le 16 Nov 2014
Modifié(e) : Guillaume le 16 Nov 2014
Hum, I'm sure I've seen a very similar question (same illustration) in the past. Is this homework?
find the index of the rows you want to exchange:
idx = find(D2 > D1);
And use that to copy the rows of B in A:
A(idx, :) = B(idx, :);
Or, using logical indexing:
A(D2>D1, :) = B(D2>D1);
  3 commentaires
Guillaume
Guillaume le 16 Nov 2014
I can't test your code, I don't have the stats toolbox (no pdist) and you haven't given A and B.
In any case,
A = [1 2 3 4]'
B = [11 12 13 14]'
D1 = [0 14 15 16]' %as in your example
D2 = [0 16 19 14]' %as in your example
A(D2>D1) = B(D2>D1)
returns
A =
1
12
13
4
which is what you asked.
Whatever problem you're having, it's not to do with my bit of code.
janny
janny le 16 Nov 2014
Modifié(e) : janny le 16 Nov 2014
thank you very much... its working fine now,,,,

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Plus de réponses (1)

MA
MA le 16 Nov 2014
clear all
close all
clc;
A=[11 22 33 44;0 14 15 16]'
B=[66 77 88 99;0 16 19 14]'
for i=1:4
if A(i,2)<B(i,2)
A(i,1)=B(i,1);
end
end
A
  1 commentaire
Guillaume
Guillaume le 16 Nov 2014
M.A., Your loop is equivalent to:
idx = A(:,2)<B(:,2);
A(idx, :) = B(idx, :);
No need for a loop.

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