data.table将NA替换为多列的均值和id

要评估仅包含列名的列,我们可以使用get().我们需要lapply()在多个列上执行此操作.

## determine the column names that contain NA values
nm <- names(dat)[colSums(is.na(dat)) != 0]
## replace with the mean - by 'id'
dat[, (nm) := lapply(nm, function(x) {
    x <- get(x)
    x[is.na(x)] <- mean(x, na.rm = TRUE)
    x
}), by = id]

这给了更新 dat

   id     var1     var2 var3
1:  1 1.666667 4.000000    4
2:  1 1.000000 4.000000    4
3:  1 2.000000 4.000000    4
4:  1 2.000000 4.000000    3
5:  2 1.000000 5.000000    5
6:  2 1.000000 5.000000    5
7:  2 2.000000 4.666667    5
8:  2 2.000000 4.000000    4

更新:使用您更新的问题,为避免在包含NA的所有列上运行此问题,请不要使用nm.只需使用自己的矢量tomean.

tomean <- c("var1", "var2")
dat[, (tomean) := lapply(tomean, function(x) {
    x <- get(x)
    x[is.na(x)] <- mean(x, na.rm = TRUE)
    x
}), by = id]

这给了

   id     var1     var2 var3
1:  1 1.666667 4.000000    4
2:  1 1.000000 4.000000    4
3:  1 2.000000 4.000000    4
4:  1 2.000000 4.000000   NA
5:  2 1.000000 5.000000    5
6:  2 1.000000 5.000000    5
7:  2 2.000000 4.666667    5
8:  2 2.000000 4.000000    4