data.table subsetting by NaN doesn't work - r

Question

I have a column in a data table with NaN values. Something like:
my.dt <- data.table(x = c(NaN, NaN, NaN, .1, .2, .2, .3), y = c(2, 4, 6, 8, 10, 12, 14))
setkey(my.dt, x)
I can use the J() function to find all instances where the x column is equal to .2
> my.dt[J(.2)]
x y
1: 0.2 10
2: 0.2 12
But if I try to do the same thing with NaN it doesn't work.
> my.dt[J(NaN)]
x y
1: NaN NA
I would expect:
x y
1: NaN 2
2: NaN 4
3: NaN 6
What gives? I can't find anything in the data.table documentation to explain why this is happening (although it may just be that I don't know what to look for). Is there any way to get what I want? Ultimately, I'd like to replace all of the NaN values with zero, using something like my.dt[J(NaN), x := 0]

Answer 1

Update: This has been fixed a while back, in v1.9.2. From NEWS:
NA, NaN, +Inf and -Inf are now considered distinct values, may be in keys, can be joined to and can be grouped. data.table defines: NA < NaN < -Inf. Thanks to Martin Liberts for the suggestions, #4684, #4815 and #4883.
require(data.table) ## 1.9.2+
my.dt[J(NaN)]
# x y
# 1: NaN 2
# 2: NaN 4
# 3: NaN 6
This issue is part design choice, part bug. There are several questions on SO and a few emails on the listserv exploring NA's in data.table key.
The main idea is outlined in the FAQ in that NA's are treated as FALSE
Please feel free chime in on the conversation in the mailing list. There was a conversation started by #Arun,
http://r.789695.n4.nabble.com/Follow-up-on-subsetting-data-table-with-NAs-td4669097.html
Also, you can read more in the answers and comments to any of the following questions on SO:
subsetting a data.table using !=<some non-NA> excludes NA too
NA in `i` expression of data.table (possible bug)
DT[!(x == .)] and DT[x != .] treat NA in x inconsistently
In the meantime, your best bet is to use is.na.
While it is slower than a radix search, it is still faster than most vector searches in R, and certainly much, much faster than any fancy workarounds
library(microbenchmark)
microbenchmark(my.dt[.(1)], my.dt[is.na(ID)], my.dt[ID==1], my.dt[!!!(ID)])
# Unit: milliseconds
expr median
my.dt[.(1)] 1.309948
my.dt[is.na(ID)] 3.444689 <~~ Not bad
my.dt[ID == 1] 4.005093
my.dt[!(!(!(ID)))] 10.038134
### using the following for my.dt
my.dt <- as.data.table(replicate(20, sample(100, 1e5, TRUE)))
setnames(my.dt, 1, "ID")
my.dt[sample(1e5, 1e3), ID := NA]
setkey(my.dt, ID)

Answer 2

Here's a fast workaround that relies a lot on what's actually happening internally (making the code a bit fragile imo). Because internally NaN is just a very very negative number, it will always be at the front of your data.table when you setkey. We can use that property to isolate those entries like so:
# this will give the index of the first element that is *not* NaN
my.dt[J(-.Machine$double.xmax), roll = -Inf, which = T]
# this is equivalent to my.dt[!is.nan(x)], but much faster
my.dt[seq_len(my.dt[J(-.Machine$double.xmax), roll = -Inf, which = T] - 1)]
Here's a benchmark for Ricardo's sample data:
my.dt <- as.data.table(replicate(20, sample(100, 1e5, TRUE)))
setnames(my.dt, 1, "ID")
my.dt[sample(1e5, 1e3), ID := NA]
setkey(my.dt, ID)
# NOTE: I have to use integer max here - because this example has integers
# instead of doubles, so I'll just add simple helper function (that would
# likely need to be extended for other cases, but I'm just dealing with the ones here)
minN = function(x) if (is.integer(x)) -.Machine$integer.max else -.Machine$double.xmax
library(microbenchmark)
microbenchmark(normalJ = my.dt[J(1)],
naJ = my.dt[seq_len(my.dt[J(minN(ID)), roll = -Inf, which = T] - 1)])
#Unit: milliseconds
# expr min lq median uq max neval
# normalJ 1.645442 1.864812 2.120577 2.863497 5.431828 100
# naJ 1.465806 1.689350 2.030425 2.600720 10.436934 100
In my tests the following minN function also covers character and logical vectors:
minN = function(x) {
if (is.integer(x)) {
-.Machine$integer.max
} else if (is.numeric(x)) {
-.Machine$double.xmax
} else if (is.character(x)) {
""
} else if (is.logical(x)) {
FALSE
} else {
NA
}
}
And you will want to add mult = 'first', e.g.:
my.dt[seq_len(my.dt[J(minN(colname)), roll = -Inf, which = T, mult = 'first'] - 1)]

Answer 3

See if this is helpful.
my.dt[!is.finite(x),]
x y
1: NaN 2
2: NaN 4
3: NaN 6