问题

I have a data.table that contains some groups. I operate on each group and some groups return numbers, others return NA. For some reason data.table has trouble putting everything back together. Is this a bug or am I misunderstanding? Here is an example:

dtb <- data.table(a=1:10)
f <- function(x) {if (x==9) {return(NA)} else { return(x)}}
dtb[,f(a),by=a]

Error in `[.data.table`(dtb, , f(a), by = a) : 
  columns of j don't evaluate to consistent types for each group: result for group 9 has     column 1 type 'logical' but expecting type 'integer'

My understanding was that NA is compatible with numbers in R since clearly we can have a data.table that has NA values. I realize I can return NULL and that will work fine but the issue is with NA.

回答1:

From ?NA

NA is a logical constant of length 1 which contains a missing value indicator. NA can be coerced to any other vector type except raw. There are also constants NA_integer_, NA_real_, NA_complex_ and NA_character_ of the other atomic vector types which support missing values: all of these are reserved words in the R language.

You will have to specify the correct type for your function to work -

You can coerce within the function to match the type of x (note we need any for this to work for situations with more than 1 row in a subset!

f <- function(x) {if any((x==9)) {return(as(NA, class(x)))} else { return(x)}}

More data.tableish approach

It might make more data.table sense to use set (or :=) to set / replace by reference.

set(dtb, i = which(dtb[,a]==9), j = 'a', value=NA_integer_)

Or := within [ using a vector scan for a==9

dtb[a == 9, a := NA_integer_]

Or := along with a binary search

setkeyv(dtb, 'a')
dtb[J(9), a := NA_integer_]

Useful to note

If you use the := or set approaches, you don't appear to need to specify the NA type

Both the following will work

dtb <- data.table(a=1:10)
setkeyv(dtb,'a')
dtb[a==9,a := NA]

dtb <- data.table(a=1:10)
setkeyv(dtb,'a')
set(dtb, which(dtb[,a] == 9), 'a', NA)

This gives a very useful error message that lets you know the reason and solution:

Error in [.data.table(DTc, J(9), :=(a, NA)) : Type of RHS ('logical') must match LHS ('integer'). To check and coerce would impact performance too much for the fastest cases. Either change the type of the target column, or coerce the RHS of := yourself (e.g. by using 1L instead of 1)

Which is quickest

with a reasonable large data.set where a is replaced in situ

Replace in situ

library(data.table)

set.seed(1)
n <- 1e+07
DT <- data.table(a = sample(15, n, T))
setkeyv(DT, "a")
DTa <- copy(DT)
DTb <- copy(DT)
DTc <- copy(DT)
DTd <- copy(DT)
DTe <- copy(DT)

f <- function(x) {
    if (any(x == 9)) {
        return(as(NA, class(x)))
    } else {
        return(x)
    }
}

system.time({DT[a == 9, `:=`(a, NA_integer_)]})
##    user  system elapsed 
##    0.95    0.24    1.20 
system.time({DTa[a == 9, `:=`(a, NA)]})
##    user  system elapsed 
##    0.74    0.17    1.00 
system.time({DTb[J(9), `:=`(a, NA_integer_)]})
##    user  system elapsed 
##    0.02    0.00    0.02 
system.time({set(DTc, which(DTc[, a] == 9), j = "a", value = NA)})
##    user  system elapsed 
##    0.49    0.22    0.67 
system.time({set(DTc, which(DTd[, a] == 9), j = "a", value = NA_integer_)})
##    user  system elapsed 
##    0.54    0.06    0.58 
system.time({DTe[, `:=`(a, f(a)), by = a]})
##    user  system elapsed 
##    0.53    0.12    0.66 
# The are all the same!
all(identical(DT, DTa), identical(DT, DTb), identical(DT, DTc), identical(DT, 
    DTd), identical(DT, DTe))
## [1] TRUE

Unsurprisingly the binary search approach is the fastest

回答2:

you can also do something like this :

dtb <- data.table(a=1:10)

mat <- ifelse(dtb == 9,NA,dtb$a)

The above command will give you matrix but you can change it back to data.table

new.dtb <- data.table(mat)
new.dtb
     a
 1:   1
 2:   2
 3:   3
 4:   4
 5:   5
 6:   6
 7:   7
 8:   8
 9:  NA
10:  10

Hope this helps.

回答3:

If you want to assign NAs to many variables, you could use the approach suggested here:

v_1  <- c(0,0,1,2,3,4,4,99)
v_2  <- c(1,2,2,2,3,99,1,0)
dat  <-  data.table(v_1,v_2)

for(n in 1:2) {
  chari <-  paste0(sprintf('v_%s' ,n), ' %in% c(0,99)')
  charj <- sprintf('v_%s := NA_integer_', n)
  dat[eval(parse(text=chari)), eval(parse(text=charj))]
}

来源：https://stackoverflow.com/questions/12399592/na-in-data-table

标签

data.table

NA in data.table

问题