6

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
Rofellos
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2 Answers2

2

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"
akrun
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0

To use this type of map/dictionary to create a new data frame column, you can also use the match method:

snape_gradebook_df<-data_frame(students=c("Harry", "Hermione", "Ron", "Neville", "Ginny", "Luna", "Draco", "Cho"),
           numeric_grade=c(4,2,2,3,4,1,4,4))
grade_map<-data_frame(numbers=c(1,2,3,4), letters=c("A", "B", "C", "D"),
                    stringsAsFactors = FALSE)
snape_gradebook_df['letter_grade']<-grade_map$letters[match(old_list, grade_map$numbers)]

I'm not sure the advantages/disadvantages of setNames vs match - just thought I would share another possible solution.

nanselm2
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