使用 'R'中的数据表迭代替换NA

Question

我正在尝试用来自适当组的随机样本替换 NA。例如，在第 2 行中，NA 来自“法国”，年龄和时间为“20-30”“30-40”。因此，我想对所有其他“法国”、“20-30”、“30-40”观察结果的响应列进行随机抽样。

我有下面的代码，效果很好，但每个值都被替换为相同的随机样本。例如，如果我有多个“法国”、“20-30”、“30-40”NA，则它们对应的 R2 将相同。

我希望每个 NA 都被独立采样，但 data.table 似乎“一次”完成，因此我不能那样做。有什么想法吗？

DT <- data.table(mydf, key = "Country,Age,Time")
DT[, R2 := ifelse(is.na(Response), sample(na.omit(Response), 1), 
                  Response), by = key(DT)]
DT
#    Index Country   Age  Time Response R2
# 1:     5  France 20-30 30-40        1  1
# 2:     6  France 20-30 30-40       NA  2
# 3:     7  France 20-30 30-40        2  2
# 4:     1 Germany 20-30 15-20        1  1
# 5:     2 Germany 20-30 15-20       NA  1
# 6:     3 Germany 20-30 15-20        1  1
# 7:     4 Germany 20-30 15-20        0  0

mydf 在哪里

mydf <- structure(list(Index = 1:7, Country = c("Germany", "Germany", 
"Germany", "Germany", "France", "France", "France"), Age = c("20-30", 
"20-30", "20-30", "20-30", "20-30", "20-30", "20-30"), Time = c("15-20", 
"15-20", "15-20", "15-20", "30-40", "30-40", "30-40"), Response = c(1L, 
NA, 1L, 0L, 1L, NA, 2L)), .Names = c("Index", "Country", "Age", 
"Time", "Response"), class = "data.frame", row.names = c(NA, -7L))

Answer 1

我会这样做:

DT[, is_na := is.na(Response)]
nas <- DT[, sample(Response[!is_na], sum(is_na), TRUE) ,
             by=list(Country, Age, Time)]$V1
DT[, R2 := Response][(is_na), R2 := nas]