Reputation: 385
I'm new to R. In a data frame, I wanted to create a new column #21 that is equal to the sum of column #1 to #20,row by row.
I knew I could do
df$Col21<-df$Col1+df$Col2+.....+df$Col20
But is there a more concise expression?
Also, can I achieve this if using column names not numbers? Thanks!
Upvotes: 1
Views: 936
Reputation: 3239
There is rowSums:
df$Col21 = rowSums(df[,1:20])
should do the trick, and with names:
df$Col21 = rowSums(df[,paste("Col", 1:20, sep="")])
With leading zeros and 3 digits, try:
df$Col21 = rowSums(df[,sprintf("Col%03d", 1:20, sep="")])
Upvotes: 3
Reputation: 54237
I find the dplyr
functions for column selection very intuitive, like starts_with(), ends_with(), contains(), matches()
and num_range()
:
df <- as.data.frame(replicate(20, runif(10)))
names(df) <- paste0("Col", 1:20)
library(dplyr)
# e.g.
summarise_each(df, funs(sum), starts_with("Col"))
# or
rowSums(select(df, contains("8")))
Upvotes: 0