Data frames and is.nan()

Question

I was using sum(is.na(my.df)) to check whether my data frame contained any NAs, which worked as I expected, but sum(is.nan(my.df)) did not work as I expected.

> my.df <- data.frame(a=c(1, 2, 3), b=c(5, NA, NaN)) > my.df   a   b 1 1   5 2 2  NA 3 3 NaN > is.na(my.df)          a     b [1,] FALSE FALSE [2,] FALSE  TRUE [3,] FALSE  TRUE > is.nan(my.df)     a     b  FALSE FALSE  > sum(is.na(my.df)) [1] 2 > sum(is.nan(my.df)) [1] 0 

Oh dear. Is there a reason for the inconsistency in behaviour? Is it for a lack of implementation, or is it intentional? What does the return value of is.nan(my.df) signify? Is there a good reason not to use is.nan() on a whole data frame?

In the documentation for is.na( ) and is.nan( ), the argument types seem the same (although they don't specifically list data frames):

is.na(): x R object to be tested: the default methods handle atomic vectors, lists and pairlists. is.nan(): x R object to be tested: the default methods handle atomic vectors, lists and pairlists.

Answer 1

From ?is.nan:

All elements of logical,integer and raw vectors are considered not to be NaN, and elements of lists and pairlists are also unless the element is a length-one numeric or complex vector whose single element is NaN. 

The columns of a data frame are technically "elements of a list", so is.nan(df) returns a vector with length equal to the number of columns of the data frame, which is TRUE only if the column consists of a single NaN element:

> is.nan(data.frame(a=NaN,b=NA,c=1))     a     b     c   TRUE FALSE FALSE  

If you want behavior matching that of is.na, use apply:

sum(apply(my.df,2,is.nan)) 

The answer is 1 rather than 2 because is.nan(NA) is FALSE ...

edit: alternatively, you can just turn the data frame into a matrix:

 sum(is.nan(as.matrix(my.df))) 

update: this behaviour changed shortly (two months) after the question was asked, in R version 2.14 (October 2011): from the NEWS file,

o The default methods for is.finite(), is.infinite() and is.nan() now signal an error if their argument is not an atomic vector.

Answer 2

The is.nan function does not work with lists for some odd reason. Why it differs from is.na is beyond me and appears to be a language design issue. However, there is a simple solution:

df <- data.frame(a=c(1, 2, 3), b=c(5, NA, NaN))  df <- data.frame(sapply(df, function(x) ifelse(is.nan(x), NA, x))) df   a  b 1 1  5 2 2 NA 3 3 NA