有没有一种优雅的方法可以在dplyr中将NA处理为0(NA.rm=TRUE)？
data <- data.frame(a=c(1,2,3,4), b=c(4,NA,5,6), c=c(7,8,9,NA))
data %>% mutate(sum = a + b + c)
a b c sum
1 4 7 12
2 NA 8 NA
3 5 9 17
4 6 NA NA
但我喜欢
a b c sum
1 4 7 12
2 NA 8 10
3 5 9 17
4 6 NA 10
即使我知道在许多其他情况下这不是理想的结果
你可以用这个:
library(dplyr)
data %>%
#rowwise will make sure the sum operation will occur on each row
rowwise() %>%
#then a simple sum(..., na.rm=TRUE) is enough to result in what you need
mutate(sum = sum(a,b,c, na.rm=TRUE))
输出:
Source: local data frame [4 x 4]
Groups: <by row>
a b c sum
(dbl) (dbl) (dbl) (dbl)
1 1 4 7 12
2 2 NA 8 10
3 3 5 9 17
4 4 6 NA 10