How to recode multiple columns in R - r

Question

I tried my best to recode multiple columns, but I still struggle to do it. Here what I have done:
df<-read.table(text="ZR1 Time1 ZR2 Time2 ZR3 Time3
A 60 A 56 B 44
C 61 B 44 D 78
D 62 C 78 E 66
E 58 D 46 B 45
A 54 B 23 B 23
A 57 E 24 B 100",h=T)
What I have done
for (i in 1) {
ZRi<-paste0("ZR", i)
Zi<-paste0("Z",i)}
df[,Zi]=c(A=4,B=3,C=2,D=1,E=0)
df[,Zi]=c(A=4,B=3,C=2,D=1,E=0)[df[,ZRi]]
I got this:
ZR1 Time1 ZR2 Time2 ZR3 Time3 Z1
1 A 60 A 56 B 44 4
2 C 61 B 44 D 78 3
3 D 62 C 78 E 66 2
4 E 58 D 46 B 45 1
5 A 54 B 23 B 23 4
6 A 57 E 24 B 100 4
As you can see, I could get Z1, which is wrong.
I want to get this:
ZR1 Time1 ZR2 Time2 ZR3 Time3 Z1 Z2 Z3
A 60 A 56 B 44 4 4 3
C 61 B 44 D 78 2 3 1
D 62 C 78 E 66 1 2 0
E 58 D 46 B 45 0 1 3
A 54 B 23 B 23 4 3 3
A 57 E 24 B 100 4 0 3

Answer 1

Here's the base approach (and probably fastest). You are just using the values of the ZR columns as an index into c(A=4,B=3,C=2,D=1,E=0) which becomes a translation table and then assigning those results to new columns in df:
df[ paste0("Z", 1:3) ] <-
lapply( df[ , grepl("^ZR", names(df))] , # passes "ZR" columns one-at-a-time
function(x) {c(A=4,B=3,C=2,D=1,E=0)[as.character(x)]})
Depending on what was intended as the purpose for these new columns, #User60 should be aware that this delivers numeric vectors

Answer 2

By playing with levels and labels you can get this:
for (i in 1:3) {
df[[paste0("Z",i)]] <-
factor(df[[paste0("ZR", i)]],levels=LETTERS[1:5],labels=4:0)
}
df
# ZR1 Time1 ZR2 Time2 ZR3 Time3 Z1 Z2 Z3
# 1 A 60 A 56 B 44 4 4 3
# 2 C 61 B 44 D 78 2 3 1
# 3 D 62 C 78 E 66 1 2 0
# 4 E 58 D 46 B 45 0 1 3
# 5 A 54 B 23 B 23 4 3 3
# 6 A 57 E 24 B 100 4 0 3
The created columns with this method will be factors, to have numeric instead use the following:
for (i in 1:3) {
df[[paste0("Z",i)]] <-
as.numeric(as.character(factor(df[[paste0("ZR", i)]],levels=LETTERS[1:5],labels=4:0)))
}

Answer 3

Maybe this one one-liner with dplyr could help
df %>%
mutate_at(setNames(paste0("ZR", 1:3), paste0("Z", 1:3)),
~5-as.numeric(factor(.x, levels = LETTERS[1:5])))
The trick here is to pass named vector to mutate_at to create new columns. You can coerce factor to numeric if you pre-specified the levels.

Answer 4

An alternative solution using dplyr + magrittr packages
library(dplyr); library(magrittr)
df2 <- select(df, starts_with("ZR")) %>%
lapply(as.character) %>%
mapply(`[`, list(c(A=4,B=3,C=2,D=1,E=0)), .) %>%
data.frame(df, .)
names(df2)[ncol(df2)-2:0] <- paste0("Z", 1:3)

Answer 5

Here's a more dplyr-esque method. Useful for recoding when the output isn't an integer.
library(dplyr)
# Make lookup table
lookup <- data.frame(let = LETTERS[1:5], num = 4:0, stringsAsFactors = F)
# Join with lookup table
df %>%
left_join(lookup, by = c('ZR1' = 'let')) %>%
left_join(lookup, by = c('ZR2' = 'let')) %>%
left_join(lookup, by = c('ZR3' = 'let')) %>%
rename_at(vars(matches('num')), ~paste0('Z', 1:3))
Or, with data.table
library(data.table)
lookup <- data.frame(let = LETTERS[1:5], num = 4:0, stringsAsFactors = F)
setDT(df)
df[, paste0('Z', 1:3) := lapply(df[,paste0('ZR', 1:3)],
function(x) lookup$num[match(x, lookup$let)])]