I have a function that I apply to a column and puts results in another column and it sometimes gives me integer(0) as output. So my output column will be something like:
45
64
integer(0)
78
How can I detect these integer(0)'s and replace them by NA? Is there something like is.na() that will detect them ?
Edit: Ok I think I have a reproducible example:
df1 <-data.frame(c("267119002","257051033",NA,"267098003","267099020","267047006"))
names(df1)[1]<-"ID"
df2 <-data.frame(c("257051033","267098003","267119002","267047006","267099020"))
names(df2)[1]<-"ID"
df2$vals <-c(11,22,33,44,55)
fetcher <-function(x){
y <- df2$vals[which(match(df2$ID,x)==TRUE)]
return(y)
}
sapply(df1$ID,function(x) fetcher(x))
The output from this sapply is the source of the problem.
> str(sapply(df1$ID,function(x) fetcher(x)))
List of 6
$ : num 33
$ : num 11
$ : num(0)
$ : num 22
$ : num 55
$ : num 44
I don't want this to be a list - I want a vector, and instead of num(0) I want NA (note in this toy data it gives num(0) - in my real data it gives (integer(0)).
Here's a way to (a) replace integer(0) with NA and (b) transform the list into a vector.
# a regular data frame
> dat <- data.frame(x = 1:4)
# add a list including integer(0) as a column
> dat$col <- list(45,
+ 64,
+ integer(0),
+ 78)
> str(dat)
'data.frame': 4 obs. of 2 variables:
$ x : int 1 2 3 4
$ col:List of 4
..$ : num 45
..$ : num 64
..$ : int
..$ : num 78
# find zero-length values
> idx <- !(sapply(dat$col, length))
# replace these values with NA
> dat$col[idx] <- NA
# transform list to vector
> dat$col <- unlist(dat$col)
# now the data frame contains vector columns only
> str(dat)
'data.frame': 4 obs. of 2 variables:
$ x : int 1 2 3 4
$ col: num 45 64 NA 78