I saw a list of questions asked in stack overflow, regarding the following, but never got a satisfactory answer. I will follow up on the following question Blend of na.omit and na.pass using aggregate?
> test <- data.frame(name = rep(c("A", "B", "C"), each = 4),
var1 = rep(c(1:3, NA), 3),
var2 = 1:12,
var3 = c(rep(NA, 4), 1:8))
> test
name var1 var2 var3
1 A 1 1 NA
2 A 2 2 NA
3 A 3 3 NA
4 A NA 4 NA
5 B 1 5 1
6 B 2 6 2
7 B 3 7 3
8 B NA 8 4
9 C 1 9 5
10 C 2 10 6
11 C 3 11 7
12 C NA 12 8
When I try out the given solution, instead of mean I try to find out the sum
aggregate(. ~ name, test, FUN = sum, na.action=na.pass, na.rm=TRUE)
the solution doesn't work as usual. Accordingly, it converts NA to 0, So the sum of NAs is 0. It displays it as 0 instead of NaN.
Why doesn't the following work for FUN=sum.And how to make it work?
Create a lambda function with a condition to return NaN when all elements are NA
aggregate(. ~ name, test, FUN = function(x) if(all(is.na(x))) NaN
else sum(x, na.rm = TRUE), na.action=na.pass)
-output
name var1 var2 var3
1 A 6 10 NaN
2 B 6 26 10
3 C 6 42 26
It is an expected behavior with sum and na.rm = TRUE. According to ?sum
the sum of an empty set is zero, by definition.
> sum(c(NA, NA), na.rm = TRUE)
[1] 0