Don't know whats wrong there (except if you're using the eclipse R plugin on a
mac like me and the window for choosing the download site doesn't pop up.. did
it?^^)
Anyway, you could just split all of your data into 2 datasets, one that has all
the data labeled 0, the other for all labeled 1, then take a random 80% of
both, put them back together into the 80% data, and put the rest back together
to form the 20%.
Since i don't know your data, heres an example:
M<-cbind(c(rep(0,10),rep(1,10)),1:20)
> M
[,1] [,2]
[1,] 0 1
[2,] 0 2
[3,] 0 3
[4,] 0 4
[5,] 0 5
[6,] 0 6
[7,] 0 7
[8,] 0 8
[9,] 0 9
[10,] 0 10
[11,] 1 11
[12,] 1 12
[13,] 1 13
[14,] 1 14
[15,] 1 15
[16,] 1 16
[17,] 1 17
[18,] 1 18
[19,] 1 19
[20,] 1 20
index1<-which(M[,1]==1)
> index1
[1] 11 12 13 14 15 16 17 18 19 20
> M1<-M[index1,]
> M1
[,1] [,2]
[1,] 1 11
[2,] 1 12
[3,] 1 13
[4,] 1 14
[5,] 1 15
[6,] 1 16
[7,] 1 17
[8,] 1 18
[9,] 1 19
[10,] 1 20
> M0<-M[-index1,]
> M0
[,1] [,2]
[1,] 0 1
[2,] 0 2
[3,] 0 3
[4,] 0 4
[5,] 0 5
[6,] 0 6
[7,] 0 7
[8,] 0 8
[9,] 0 9
[10,] 0 10
> s1<-sample(1:dim(M1)[1],0.8*dim(M1)[1])
> s1
[1] 10 3 5 9 2 6 8 7
> s0<-sample(1:dim(M0)[1],0.8*dim(M0)[1])
> s0
[1] 8 10 9 3 7 4 2 1
> data80<-rbind(M1[s1,],M0[s0,])
> data80
[,1] [,2]
[1,] 1 20
[2,] 1 13
[3,] 1 15
[4,] 1 19
[5,] 1 12
[6,] 1 16
[7,] 1 18
[8,] 1 17
[9,] 0 8
[10,] 0 10
[11,] 0 9
[12,] 0 3
[13,] 0 7
[14,] 0 4
[15,] 0 2
[16,] 0 1
> data20<-rbind(M1[-s1,],M0[-s0,])
> data20
[,1] [,2]
[1,] 1 11
[2,] 1 14
[3,] 0 5
[4,] 0 6
which is probably not how you really do things efficiently, but it should work.
greetings
Jessi
Am 25.04.2012 um 17:18 schrieb Dwaipayan Dasgupta:
> Thank you so much for replying. I tried what you said but it still throws the
> same error i.e could not find function "sample.split"
> Might be because of the version of R I am running (R version 2.12.2).i do not
> have admin rights to upgrade to the newest version.
> Is there anything else I can try? Im trying to split my data into 80:20
> keeping the ratio of 0,1 in the Y variable(binary) constant.
>
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Jessica Streicher
> Sent: Wednesday, April 25, 2012 7:17 PM
> To: r-help@r-project.org
> Subject: Re: [R] Splitting data into test and train (80:20) kepping
> attributes similar
>
> Well, it throws an error, because there is no such function in default R. A
> bit of googling showed it might be the one in the caTools package.
>
> execute this:
> install.packages("caTools")
> library(caTools)
>
> before executing your code
>
>
> Am 25.04.2012 um 12:39 schrieb Dwaipayan Dasgupta:
>
>> Hi,
>> Could someone help me with this please , im trying to use
>> Y = Attrition_data[,1] # extract labels from the data
>> msk = sample.split (Y, SplitRatio=3/4)
>> table(Y,msk)
>> to do the splitting but it keeps throwing up and error
>> Error: could not find function "sample.split"
>> Could you please help
>>
>> Thanks in advance
>> doy
>>
>>
>> -----Original Message-----
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
>> Behalf Of Dwaipayan Dasgupta
>> Sent: Tuesday, April 24, 2012 9:08 PM
>> To: r-help@r-project.org
>> Subject: [R] Splitting data into test and train (80:20) kepping attributes
>> similar
>>
>> Hi,
>> I am trying to do some predictive modeling around attrition and want to
>> split the dataset into test and train (80:20) and keep the ratio of
>> attritees:non attrites same.
>> In my dataset the attrition indicator is coded as 0(for non-attritees) and 1
>> (for attritees) and I want to keep the ratio of 0's to 1 similar.
>> I apologize for this trivial question but this is my second week with R.
>>
>> Thanks,
>> Doy
>>
>>
>>
>>
>>
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