Find similar elements in a matrix

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Niki
Niki el 5 de Sept. de 2011
My question is that , I have a matrix, I want to know which element from which columns are similar, which number is the most repeated number for example
X=[1 2 3 3 3
45 7 4 4 4
70 8 5 5 5
88 9 11 11 11
170 205 13 13 13
172 220 14 23 24
194 222 24 24 41
196 224 41 152 67
200 539 62 183 68
250 540 67 184 71
251 1415 68 185 148
255 1426 71 187 151
]
for example here 3 (repeated in columns 3,4 and5) the same for 4, 5, 1, 13
and the most repeated numbers are 3,4,5,11, and 13
  2 comentarios
Grzegorz Knor
Grzegorz Knor el 5 de Sept. de 2011
I've question:
most repeated numbers are 3,4,5,11, and 13
or rather
most repeated numbers are 3,4,5,11, 13 and 24?
Niki
Niki el 5 de Sept. de 2011
you are right
3,4,5,11, 13 and 24

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Respuesta aceptada

Grzegorz Knor
Grzegorz Knor el 5 de Sept. de 2011
[h1 h2] =hist(X(:),unique(X));
h3 = unique(h1);
h3 = h3(end:-1:1);
for k=1:length(h3)
disp(['There is ' num2str(h3(k)) ': [ ' num2str(h2(h1==h3(k))') ' ]'])
end
  2 comentarios
Grzegorz Knor
Grzegorz Knor el 5 de Sept. de 2011
If you think that this answer solved your problem, please accept it :)
Niki
Niki el 5 de Sept. de 2011
Grzegorz, It works , thanks , also could you please take a look at one another question that I have?

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Más respuestas (2)

Daniel Shub
Daniel Shub el 5 de Sept. de 2011
Maybe something like:
[y, z] = hist(X(:), unique(X));
stem(z, y);
  1 comentario
Niki
Niki el 5 de Sept. de 2011
Thanks Daniel, your comments always make a brilliant way,

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Grzegorz Knor
Grzegorz Knor el 5 de Sept. de 2011
My suggestion:
[u, ~, n] = unique(X);
[h1 h2] =hist(n,unique(n));
u(h2(h1==max(h1)))
  3 comentarios
Grzegorz Knor
Grzegorz Knor el 5 de Sept. de 2011
There is 45 zeros:
sum(X(:)==0)
Is this OK?
Niki
Niki el 5 de Sept. de 2011
Yes, exactly
It is okay, but I would like to see the others as well, for example 45 times zero, then 3,4,5,11, 13 and 24

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