NA in data.table

前端 未结 3 1566
遇见更好的自我
遇见更好的自我 2021-02-20 15:24

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.ta

相关标签:
3条回答
  • 2021-02-20 15:45

    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.

    0 讨论(0)
  • 2021-02-20 15:57

    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))]
    }
    
    0 讨论(0)
  • 2021-02-20 16:11

    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.table*ish* 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

    0 讨论(0)
提交回复
热议问题