使用R中的其他列值来选择data.table分配的LHS列和RHS列

Question

Here is the code to produce a sample dataset:

require(data.table)
testdata <- data.table(
  X = rep(sample(1:3),5),
  Y = rep(sample(1:3),5),
  X1 = rnorm(15),
  X2 = rnorm(15),
  X3 = rnorm(15),
  Y1 = NA_character_,
  Y2 = NA_character_,
  Y3 = NA_character_
)

Initial data table:

    X Y         X1          X2          X3 Y1 Y2 Y3
 1: 3 3 -0.7098927  0.63342935  0.94470612 NA NA NA
 2: 1 2  0.3008547 -1.40043977  1.53781754 NA NA NA
 3: 2 1  0.3423140  0.34897695 -0.38402565 NA NA NA
 4: 3 3 -0.5726456 -2.24526957 -1.10947867 NA NA NA
 5: 1 2 -1.3239474 -0.53924617 -0.04103982 NA NA NA
 6: 2 1  0.2493801  0.85806647  0.96488021 NA NA NA
 7: 3 3 -2.0653505  0.05481703  1.75161043 NA NA NA
 8: 1 2 -1.3919774  0.34282832  0.50834289 NA NA NA
 9: 2 1  0.5928025 -1.11899399  0.35967102 NA NA NA
10: 3 3 -0.4704720  0.64004313 -0.17343794 NA NA NA
11: 1 2  0.3056093  2.14544631  0.43740447 NA NA NA
12: 2 1 -0.1568971  1.05091249  1.18884487 NA NA NA
13: 3 3 -1.3078670  1.07482123 -0.65367957 NA NA NA
14: 1 2  0.4622123 -0.60308532 -1.11104235 NA NA NA
15: 2 1 -0.7894978  0.33018926 -0.04700393 NA NA NA

Here is the action I want to perform: In each row,

if X = 2 and Y = 3 then Y3 <- X2

Expected Output:

    X Y         X1          X2          X3 Y1                 Y2                 Y3
 1: 3 3 -0.7098927  0.63342935  0.94470612 NA                 NA                 0.94470612
 2: 1 2  0.3008547 -1.40043977  1.53781754 NA                 0.3008547          NA
 3: 2 1  0.3423140  0.34897695 -0.38402565 0.34897695         NA                 NA
 4: 3 3 -0.5726456 -2.24526957 -1.10947867 NA                 NA                 -1.10947867
 5: 1 2 -1.3239474 -0.53924617 -0.04103982 NA                 -1.3239474         NA
 6: 2 1  0.2493801  0.85806647  0.96488021 0.85806647         NA                 NA
 7: 3 3 -2.0653505  0.05481703  1.75161043 NA                 NA                 1.75161043
 8: 1 2 -1.3919774  0.34282832  0.50834289 NA                 -1.3919774         NA
 9: 2 1  0.5928025 -1.11899399  0.35967102 -1.11899399        NA                 NA
10: 3 3 -0.4704720  0.64004313 -0.17343794 NA                 NA                 -0.17343794
11: 1 2  0.3056093  2.14544631  0.43740447 NA                 0.3056093          NA
12: 2 1 -0.1568971  1.05091249  1.18884487 1.05091249         NA                 NA
13: 3 3 -1.3078670  1.07482123 -0.65367957 NA                 NA                 -0.65367957
14: 1 2  0.4622123 -0.60308532 -1.11104235 NA                 0.4622123          NA
15: 2 1 -0.7894978  0.33018926 -0.04700393 0.33018926         NA                 NA

How can I achieve this using simple data.table syntax? I have tried get, eval(parse) etc but running into trouble each time.

Note that my actual dataset is quite large(100 plus columns) so I require a solution that doesn't rely on column numbers. I can possible write a large number of if statements as well but it looks like a bad way to do this for about 30 odd columns that need to be assigned in a similar way.

data.table version is 1.10.4 and the R version is 3.3.2

Edit: I solved it using a function. Not sure if this is the best way though as it is very very slow.

populateY <- function(input_table) {

  for(i in 1:nrow(input_table)) {
    k <- X
    j <- Y
    tempX <- paste0("input_table$X",k,"[i]")
    tempY <- paste0("input_table$Y",j,"[i]")
    eval(parse(text = paste0(tempY," <- ",tempX)))
  }    
  return(input_table)
}

Answer 1

If you're open to using the tidyverse and tibble data frames, I would do it this way.

require(tibble)
testdata <- as_tibble(testdata)

testdata <- testdata %>%
  mutate(Y3 = ifelse(X == 2 & Y == 3, X2, NA))

You can then add all the lines you need easily and legibly in the mutate function.

Else if you're going to use data.tables for sure, then I'd go with akrun's suggestion, though you'll need change the data type of column Y3 to double, or just not have it exist when you run that code.