R: for loop with data.table library

Question

I am using library(data.table) and I would like to do a for loop that calculates the mean of columns 1:4, based on col5 values.

colnames(DT) # "col1" "col2" "col3" "col4" "col5" 
for (i in 1:4){
  o=colnames(DT)[[i]]
  l=DT[,mean(o), by=col5]
  print(l)
}

The problem is that DT does not take colnames as character vectors ("col"). Any advice is appreciated.

From your code sample , I think it is better for you to take a close in `data.table` package and understand better how it works. This will help you specially if you want to use R in your phd... — agstudy, Sep 19 '14 at 23:16
possible duplicate of [How to apply same function to every specified column in a data.table](http://stackoverflow.com/questions/16846380/how-to-apply-same-function-to-every-specified-column-in-a-data-table) — bright-star, Sep 20 '14 at 00:32

score 2 · Answer 1 · answered Sep 20 '14 at 09:04

2

Read the data.table vignettes.

library(data.table)
set.seed(42)
DT <- data.table(matrix(rnorm(100), ncol=4))
setnames(DT, paste0("col", 1:4))
DT[, col5 := rep(1:5, 5)]

DT[, lapply(.SD, mean), by = col5]

DT[, lapply(.SD, mean), by = col5, .SDcols = paste0("col", 2:3)]

answered Sep 20 '14 at 09:04

Roland

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:Tx for the syntax and for the advice. – Kizuna Sep 21 '14 at 15:15

score -2 · Answer 2 · edited Apr 17 '20 at 16:46

-2

Try:

DT[,list(mean1=mean(col1), mean2=mean(col2), mean3=mean(col3), mean4=mean(col4)),by=col5]

If columns are too many:

attach(DT)    
for(i in 1:4) print(DT[,mean(get(colnames(DT)[i])),by=DT[,5]])

edited Apr 17 '20 at 16:46

MichaelChirico

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answered Sep 20 '14 at 01:46

rnso

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this works for sure , however this is not time efficient if I will have 100 col to loop.. – Kizuna Sep 20 '14 at 07:42

R: for loop with data.table library

2 Answers2