CODEWITHSUNDEEP
rdata.table
Pete900
Pete900

Reputation: 2166

Renaming multiple factors simltaneously with data table

I love data table and use it to conditionally rename or add factors. However, I can't seem to do more than one factor at a time. Here is an example:

a <- rep(c("A", "B", "C", "D"), each=3)
b <- 1:12
df <- data.frame(a,b)
DT <- data.table(df)

Now add new column "New" which for all "A"s in column "a" is equal to "z"

DT[a=="A", New:="z"]

This works nicely. Now if I want to change say "A" and "C" to be equal to "z":

DT[a==c("A", "C"), New:="z"]

Gives me funny answers:

dput(DT)
structure(list(a = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 
3L, 4L, 4L, 4L), .Label = c("A", "B", "C", "D"), class = "factor"), 
    b = 1:12, New = c("z", NA, "z", NA, NA, NA, NA, "z", NA, 
    NA, NA, NA)), .Names = c("a", "b", "New"), row.names = c(NA, 
-12L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000000140788>, index = structure(integer(0), a = integer(0)))

I'm sure it's something simple, I can't seem to find it on SO (queue the dupe!). Thanks

To confirm, my desired output is:

dput(DT)
structure(list(a = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 
3L, 4L, 4L, 4L), .Label = c("A", "B", "C", "D"), class = "factor"), 
    b = 1:12, New = c("z", "z", "z", NA, NA, NA, "z", "z", "z", 
    NA, NA, NA)), .Names = c("a", "b", "New"), row.names = c(NA, 
-12L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x0000000000140788>, index = structure(integer(0), a = integer(0)))

Upvotes: 1

Views: 95

Answers (1)

Jaap
Jaap

Reputation: 83215

You should use %in% instead of ==, thus you will need:

DT[a %in% c("A", "C"), New:="z"]

which gives:

> DT
    a  b New
 1: A  1   z
 2: A  2   z
 3: A  3   z
 4: B  4  NA
 5: B  5  NA
 6: B  6  NA
 7: C  7   z
 8: C  8   z
 9: C  9   z
10: D 10  NA
11: D 11  NA
12: D 12  NA

Used data:

a <- rep(c("A", "B", "C", "D"), each=3)
b <- 1:12
DT <- data.table(a,b)

In a dataframe you could do:

df <- data.frame(a,b)
df$New <- NA
df[df$a %in% c("A", "C"), "New"] <- "z"

to achieve the same result.

Upvotes: 5

Related Questions