indices to column vectros of a matrix using “crossvalind”

6 vues (au cours des 30 derniers jours)
laplace laplace
laplace laplace le 7 Juin 2013
i got a 3x50 matrix. i would like to obtain the 50 3x1 column vectors and assign to them indices using the crossvalind command seen above
Indices = crossvalind('Kfold',N, 10);
so lets hypotheticaly say that i had a 3x3 matrix named "input" :
input=[1 2 3;4 5 6;7 8 9]
how can I imply an index generation for the vector columns of my "input" matrix using crossvalind?
A=[1;4;7]
B=[2;5;9]
C=[3;6;9]
**the reason i wanna do it is so i can after apply cross validation while having as train/test set these column vectors

Réponses (3)

Azzi Abdelmalek
Azzi Abdelmalek le 7 Juin 2013
  3 commentaires
Azzi Abdelmalek
Azzi Abdelmalek le 7 Juin 2013
What are your feeling about this?
A=[1;4;7]
Indices = arrayfun(@(x) crossvalind('Kfold',x, 10),A,'un',0);
laplace laplace
laplace laplace le 10 Juin 2013
can you explain me the command you wrote :D maybe its correct and i am doing smth wrong :(

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laplace laplace
laplace laplace le 10 Juin 2013
Modifié(e) : laplace laplace le 10 Juin 2013
inputs is the 3x50 matrix
C = num2cell(inputs,1);
Indices = cellfun(@(C) crossvalind('Kfold',C, 10),C,'un',0);
Indices =
Columns 1 through 9
[3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double]
Columns 10 through 18
[3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double]
Columns 19 through 27
[3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double]
Columns 28 through 36
[3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double]
Columns 37 through 45
[3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double]
Columns 46 through 50
[3x1 double] [3x1 double] [3x1 double] [3x1 double] [3x1 double]
what i wanted to get was just indices for each column vector... any thoughts?

laplace laplace
laplace laplace le 16 Juin 2013
noone?:D

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