替换data.table的多个满足条件的列中的值

Question

library(data.table)
data = data.table("cat" = c(0,5,NA,0,0,0),
                  "horse" = c(0,4,2,1,1,3),
                  "fox" = c(2,2,NA,NA,7,0))

I wish to replace values of 'cat' and 'fox' that are equal to '0' or '2' with '-99'

I can do it one at a time but how to do them both?

dat[fox == 0 | fox == 2, fox := -99]

Answer 1

Another approach with data.table is using a for(...) set(...) -approach, which is in this case both fast and memory efficient:

cols <- c('fox', 'cat')

# option 1
for (j in cols) d[get(j) %in% c(0, 2), (j) := -99]

# option 2 (thx to @Cole for highlighting)
for (j in cols) set(d, which(d[[j]] %in% c(0, 2)), j, value = -99)

# option 3 (thx to @Frank for highlighting)
for (j in cols) d[.(c(0,2)), on = j, (j) := -99]

which gives:

 > d cat horse fox 1: -99 0 -99 2: 5 4 -99 3: NA 2 NA 4: -99 1 NA 5: -99 1 7 6: -99 3 -99

---

d  <- data.table("cat"   = c(0,5,NA,0,0,0),
                 "horse" = c(0,4,2,1,1,3),
                 "fox"   = c(2,2,NA,NA,7,0))

Answer 2

Here's a not-so-elegant way of doing this:

> data
   cat horse fox
1:   0     0   2
2:   5     4   2
3:  NA     2  NA
4:   0     1  NA
5:   0     1   7
6:   0     3   0

> data[, c('fox', 'cat') := list(ifelse(cat %in% c(0,2) | fox %in% c(0,2), 99, cat ), ifelse(cat %in% c(0,2) | fox %in% c(0,2), 99, cat ))]
> data
   cat horse fox
1:  99     0  99
2:  99     4  99
3:  NA     2  NA
4:  99     1  99
5:  99     1  99
6:  99     3  99

I'm calling (c('cat', 'fox')) explicitly, but you could save them as mycols and assign using := operator: data[, mycols:=...]

Similarly, I'm passing a list explicitly based on the conditions - this could be better done using a function instead.

Answer 3

If I understand, this would work as well:

cols = c("cat", "fox")
data[, (cols) := lapply(.SD, function (x) fifelse(x %in% c(0, 2), -99, x)), .SDcols = cols]