在计算两个数据表的总和时,NA+n=NA.
> dt1 <- data.table(Name=c("Joe","Ann"), "1"=c(0,NA), "2"=c(3,NA))
> dt1
Name 1 2
1: Joe 0 3
2: Ann NA NA
> dt2 <- data.table(Name=c("Joe","Ann"), "1"=c(0,NA), "2"=c(2,3))
> dt2
Name 1 2
1: Joe 0 2
2: Ann NA 3
> dtsum <- rbind(dt1, dt2)[, lapply(.SD, sum), by=Name]
> dtsum
Name 1 2
1: Joe 0 5
2: Ann NA NA
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我不想取代所有NA与0我想是NA+NA=NA和NA+n=n得到以下结果:
Name 1 2
1: Joe 0 5
2: Ann NA 3
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这是如何在R中完成的?
更新:删除dt1中的拼写错误
mne*_*nel 10
您可以定义自己的功能以根据需要进行操作
plus <- function(x) {
if(all(is.na(x))){
c(x[0],NA)} else {
sum(x,na.rm = TRUE)}
}
rbind(dt1, dt2)[,lapply(.SD, plus), by = Name]
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