data.table将NA替换为多列的均值和id

use*_*905 5 r missing-data data.table

如果我有以下data.table:

dat <- data.table("id"=c(1,1,1,1,2,2,2,2), "var1"=c(NA,1,2,2,1,1,2,2),
              "var2"=c(4,4,4,4,5,5,NA,4), "var3"=c(4,4,4,NA,5,5,5,4))
   id var1 var2 var3
1:  1   NA    4    4
2:  1    1    4    4
3:  1    2    4    4
4:  1    2    4   NA
5:  2    1    5    5
6:  2    1    5    5
7:  2    2   NA    5
8:  2    2    4    4
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如何用id中每列的平均值替换缺失值?在我的实际数据中,我有许多变量,对于我想要替换的变量,所以如何以一般方式完成,例如它不会替换为var3而只替换为var1和var2?:

tomean=c("var1", "var2")
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我试过这样的东西,但我还没有找到解决方案:

dat[, (tomean) := mean(tomean, na.rm=TRUE), by=id, .SDcols = tomean]
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Ric*_*ven 9

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]
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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
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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]
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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
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