我会这样做:

library(data.table)
setDT(DF1); setDT(DF2)

DF1[DF2, x := ifelse(is.na(x), i.x, x), on=c("y","z")]

给出

     x y z
1: 153 a 1
2: 163 b 1
3: 184 d 1
4: 123 a 2
5: 145 e 2
6: 176 c 2
7: 124 b 1
8: 199 a 2

评论.这种方法不是很好，因为它合并了DF1的 whole ，而我们只需要合并is.na(x)的子集.在这里，改进看起来像(感谢@Arun):

Comments. This approach isn't so great, since it merges the whole of DF1, while we only need to merge the subset where is.na(x). Here, the improvement looks like (thanks, @Arun):

DF1[is.na(x), x := DF2[.SD, x, on=c("y", "z")]]

这种方式类似于@RHertel的答案.

This way is analogous to @RHertel's answer.