How to map a column through a dictionary in R

dataframe, list, r

Solution

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"

Problem

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 ```

Original source