Replace integer(0) by NA

Question

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)).

would data.frame$column[data.frame$column == integer(0) ] <- NA work ? — cianius, Jan 19 '14 at 14:36
@pepsimax why don't you put that as an answer (maybe provide a working example)? — Roman Luštrik, Jan 19 '14 at 14:36
A data frame cannot contain `integer(0)`. Please provide a reproducible example. — Sven Hohenstein, Jan 19 '14 at 14:37
or you can check for `length()==0` see: http://stackoverflow.com/questions/6451152/how-to-catch-integer0 — holzben, Jan 19 '14 at 14:44
I am also interested to see how `integer(0)` did get into your output. But to get you going: To test if an object is `integer(0)` you could do `identical(object, integer(0))`. — Mark Heckmann, Jan 19 '14 at 14:44
pepsimax tried that doesn't work. Sven I tried to make a reproducible example and I could not. Its not possible to post the whole project here. I'll take another look at it. Here is the output from looking at my live data: `> temp[1,tn1] $ integer(0) > is.na(temp[1,tn1]) FALSE > class(temp) [1] "data.frame" > class(temp[tn1]) [1] "data.frame" > class(temp[1,tn1]) [1] "list"` — user2498193, Jan 19 '14 at 14:56
I appear to have a list inside a data.frame inside a data.frame - which is weird..... — user2498193, Jan 19 '14 at 14:59
I'm voting to close, because without a reproducible example it's nearly impossible to help you ... it would help a little if you could edit your question to show the results of `str(temp)` (and tell us what `tn1` is ...) — Ben Bolker, Jan 19 '14 at 15:01
Yeah sorry Ben I realise it is a bad question - seemingly I don't understand how I created this weirdness in the first place. Perhaps I need to take a step back. This situation came about after applying this logic to my live dataset: http://stackoverflow.com/questions/21215705/r-passing-variable-name-into-sapply/21215819?noredirect=1#21215819 — user2498193, Jan 19 '14 at 15:04
tn1 is a column name. str(temp) output is vast - I don't think you want it on here! — user2498193, Jan 19 '14 at 15:06
Don't reinvent the wheel. Try `merge(df1, df2, all.x = TRUE)` — flodel, Jan 20 '14 at 00:14

Sven Hohenstein · Accepted Answer · 2018-02-05T12:08:54.493

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

Thanks Sven! That sorted out my problem by incorporating parts of it into my function! Many thanks everyone - apologies for the poor original question! — user2498193, Jan 19 '14 at 15:28

score 4 · Answer 2 · edited Jan 19 '14 at 14:58

4

Best to do that in your function, I'll call it myFunctionForApply but that's your current function. Before you return, check the length and if it is 0 return NA:

myFunctionForApply <- function(x, ...) {
    # Do your processing
    # Let's say it ends up in variable 'ret':
    if (length(ret) == 0)
        return(NA)
    return(ret)
}

edited Jan 19 '14 at 14:58

Ben Bolker

211,554
25
370
453

answered Jan 19 '14 at 14:48

Calimo

7,510
4
39
61

2

Correct. Quite un-idiomatic code, though: `function (x, ...) if (length(ret) == 0) NA else ret` – Konrad Rudolph Jan 19 '14 at 15:09
Thanks Ben and Konrad! Appreciate your answers! – user2498193 Jan 19 '14 at 15:32
Thanks Calimo I realise now you answered and Ben edited – user2498193 Jan 19 '14 at 15:38

Replace integer(0) by NA

2 Answers2