在data.table中使用`：=`来对R中两列的值求和，忽略NA

Question

I have what I think is a very simple question related to the use of data.table and the := function. I don't think I quite understand the behaviour of := and often I run into similar problems.

Here is some example data

 mat <- structure(list(
              col1 = c(NA, 0, -0.015038, 0.003817, -0.011407), 
              col2 = c(0.003745, 0.007463, -0.007407, -0.003731, -0.007491)), 
              .Names = c("col1", "col2"), 
              row.names = c(NA, 10L), 
              class = c("data.table", "data.frame"))

which gives

> mat
         col1      col2
 1:        NA  0.003745
 2:  0.000000  0.007463
 3: -0.015038 -0.007407
 4:  0.003817 -0.003731
 5: -0.011407 -0.007491

I want to create a column called col3 which gives the sum of col1 and col2. If I use

mat[,col3 := col1 + col2]

#        col1      col2      col3
#1:        NA  0.003745        NA
#2:  0.000000  0.007463  0.007463
#3: -0.015038 -0.007407 -0.022445
#4:  0.003817 -0.003731  0.000086
#5: -0.011407 -0.007491 -0.018898

then I get an NA for the first row, but I want NAs to be ignored. So I tried instead

mat[,col3 := sum(col1,col2,na.rm=TRUE)]

#        col1      col2      col3
#1:        NA  0.003745 -0.030049
#2:  0.000000  0.007463 -0.030049
#3: -0.015038 -0.007407 -0.030049
#4:  0.003817 -0.003731 -0.030049
#5: -0.011407 -0.007491 -0.030049

which is not what I am after, since it is giving me the sum of all elements of col1 and col2. I think I don't quite get := ... How can I get the sum of the element of col1 and col2 ignoring NA values?

Not sure this is relevant, but here is my sessionInfo

> sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

locale:
[1] en_AU.UTF-8/en_AU.UTF-8/en_AU.UTF-8/C/en_AU.UTF-8/en_AU.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] data.table_1.8.3

Answer 1

This is standard R behaviour, nothing really to do with data.table

Adding anything to NA will return NA

NA + 1
## NA

sum will return a single number

If you want 1 + NA to return 1

then you will have to run something like

mat[,col3 := col1 + col2]
mat[is.na(col1), col3 := col2]
mat[is.na(col2), col3 := col1]

To deal with when col1 or col2 are NA

---

EDIT - an easier solution

You could also use rowSums, which has a na.rm argument

mat[ , col3 :=rowSums(.SD, na.rm = TRUE), .SDcols = c("col1", "col2")]

rowSums is what you want (by definition, the rowSums of a matrix containing col1 and col2 , removing NA values

(@JoshuaUlrich suggested this as a comment )

Answer 2

It's not a lack of understanding of data.table but rather one regarding vectorized functions in R. You can define a dyadic operator that will behave differently than the "+" operator with regard to missing values:

 `%+na%` <- function(x,y) {ifelse( is.na(x), y, ifelse( is.na(y), x, x+y) )}

 mat[ , col3:= col1 %+na% col2]
#-------------------------------
        col1      col2      col3
1:        NA  0.003745  0.003745
2:  0.000000  0.007463  0.007463
3: -0.015038 -0.007407 -0.022445
4:  0.003817 -0.003731  0.000086
5: -0.011407 -0.007491 -0.018898

You can use mrdwad's comment to do it with sum(... , na.rm=TRUE ):

mat[ , col4 := sum(col1, col2, na.rm=TRUE), by=1:NROW(mat)]