6

I have a data.table in R which I want to use with caret package

set.seed(42)
trainingRows<-createDataPartition(DT$variable, p=0.75, list=FALSE)
head(trainingRows) # view the samples of row numbers

However, I am not able to select the rows with data.table. Instead I had to convert to a data.frame

DT_df <-as.data.frame(DT)
DT_train<-DT_df[trainingRows,]
dim(DT_train)

the data.table alternative

DT_train <- DT[.(trainingRows),] requires the keys to be set.

Any better option other than converting to data.frame?

milos.ai
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James Paul
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2 Answers2

7

Roll you own

inTrain <- sample(MyDT[, .I], floor(MyDT[, .N] * .75))
Train <- MyDT[inTrain]
Test <- MyDT[-inTrain]

Or with Caret function you can just wrap trainingRows with a c().

 trainingRows<-createDataPartition(DT$variable, p=0.75, list=FALSE)
 Train <- DT[c(trainingRows)]
 Test <- DT[c(-trainingRows)]

===

Edit by Matt
Was going to add a comment, but too long.

I've seen sample(.I,...) being used quite a bit recently. This is inefficient because it has to create the (potentially very long) .I vector which is just 1:nrow(DT). This is such a common case that R's sample() doesn't need you to pass that vector. Just pass the length. sample(nrow(DT)) already returns exactly the same result without having to create .I. See ?sample.

Also, it's better to avoid variable name repetition wherever possible. More background here.

So instead of :

inTrain <- sample(MyDT[, .I], floor(MyDT[, .N] * .75))

I'd do :

inTrain <- MyDT[,sample(.N, floor(.N*.75))]
Community
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Bruce Pucci
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6

The reason is that createDataPartition produces integer vector with two dimensions where the second could be losslessly dropped.
You can simply reduce dimension of trainingRows using below:

DT[trainingRows[,1]]

The c() function from Bruce Pucci's answer will reduce dimension too.

This minor difference vs. data.frame was spotted long time ago and recently I've made PR #1275 to fill that gap.

jangorecki
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