I have a data frame in R. One column of this data frame takes values from 1 to 6. I have another data frame that maps this column. There is some way to substitute the values in this column without using loops (using some function)?
[,1] [,2]
[1,] "1" "A"
[2,] "2" "B"
[3,] "3" "C"
[4,] "4" "D"
[1] 4 2 2 3 4 1 4 4
Exists a function that return the vector below without using loops?
[1] D B B C D A D D
You can try
v1 <- c(4, 2, 2, 3, 4, 1, 4, 4)
LETTERS[v1]
#[1] "D" "B" "B" "C" "D" "A" "D" "D"
Suppose if your mapping dataset 2nd column is not just "LETTERS"
d1 <- data.frame(Col1=c(1,3,4,2), Col2=c('j1', 'c9', '19f', 'd18'),
stringsAsFactors=FALSE)
unname(setNames(d1[,2], d1[,1])[as.character(v1)])
#[1] "19f" "d18" "d18" "c9" "19f" "j1" "19f" "19f"
Or
d1$Col2[match(v1, d1$Col1)]
#[1] "19f" "d18" "d18" "c9" "19f" "j1" "19f" "19f"
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With