Last data update: 2014.03.03

R: k-Fold Partitioning
kfoldR Documentation

k-Fold Partitioning

Description

This function computes the k-fold partitioning of a vector. Each record in a vector is randomly assigned to a group. Group numbers are between 1 and k.

Usage

kfold(constrain,k=10)

Arguments

constrain

a vector of labels.

k

number of groups.

Value

A vector with group assignmnets.

Author(s)

Stefano Cacciatore and Leonardo Tenori

Examples

kfold(1:100, k=10)  	
kfold(rep(1:20,each=5), k=10)

Results


R version 3.3.1 (2016-06-21) -- "Bug in Your Hair"
Copyright (C) 2016 The R Foundation for Statistical Computing
Platform: x86_64-pc-linux-gnu (64-bit)

R is free software and comes with ABSOLUTELY NO WARRANTY.
You are welcome to redistribute it under certain conditions.
Type 'license()' or 'licence()' for distribution details.

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Type 'contributors()' for more information and
'citation()' on how to cite R or R packages in publications.

Type 'demo()' for some demos, 'help()' for on-line help, or
'help.start()' for an HTML browser interface to help.
Type 'q()' to quit R.

> library(KODAMA)
Loading required package: e1071
Loading required package: plsgenomics
Loading required package: MASS
Loading required package: boot
Loading required package: parallel
Loading required package: class

Attaching package: 'KODAMA'

The following object is masked from 'package:plsgenomics':

    transformy

> png(filename="/home/ddbj/snapshot/RGM3/R_CC/result/KODAMA/kfold.Rd_%03d_medium.png", width=480, height=480)
> ### Name: kfold
> ### Title: k-Fold Partitioning
> ### Aliases: kfold
> ### Keywords: partitioning
> 
> ### ** Examples
> 
> kfold(1:100, k=10)  	
  [1]  7  8  2  9  1  6  5  7  5  9  7  9  1 10 10 10  3  8  8  8  3  6  6  9 10
 [26]  2  6  3  1  7  3 10  7  7  3  8  3  7  5  2  2  2  1 10  1  6  1  6  1  4
 [51] 10  5 10  6  4  3  8  1  2  1  2  9  5  8  8  9  5  3  2  8  8  6  4  4  2
 [76]  4  6  9  5  5  3  7  4  4 10  7  9  6  3  1  4 10  4  4  5  2  9  9  7  5
> kfold(rep(1:20,each=5), k=10)
  [1]  7  7  7  7  7  8  8  8  8  8  3  3  3  3  3  8  8  8  8  8  4  4  4  4  4
 [26]  9  9  9  9  9  5  5  5  5  5  1  1  1  1  1  5  5  5  5  5 10 10 10 10 10
 [51]  2  2  2  2  2  7  7  7  7  7  6  6  6  6  6  2  2  2  2  2 10 10 10 10 10
 [76]  1  1  1  1  1  9  9  9  9  9  4  4  4  4  4  3  3  3  3  3  6  6  6  6  6
> 
> 
> 
> 
> 
> dev.off()
null device 
          1 
>