3 messages · george.leigh@dpi.qld.gov.au, Brian Ripley, Peter Dalgaard
Full_Name: George Leigh Version: 1.8.1 OS: Windows 2000 Submission from: (NULL) (203.25.1.208) The following example gives the correct answer when the first argument of tapply is a numeric vector, but an incorrect answer when it is a factor. If the function used by tapply is "length", the type and contents of the first argument should make no difference, provided it has the same length as the second argument.
x = c(NA, 1) y = factor(x) tapply(x, y, length)
1 1
tapply(y, y, length)
1 2
Full_Name: George Leigh Version: 1.8.1 OS: Windows 2000 Submission from: (NULL) (203.25.1.208) The following example gives the correct answer when the first argument of tapply is a numeric vector, but an incorrect answer when it is a factor. If the function used by tapply is "length", the type and contents of the first argument should make no difference, provided it has the same length as the second argument.
Not so:
split(x, y)
$"1" [1] 1
split(y, y)
$"1" [1] <NA> 1 Levels: 1 Note that as there is only one level, NA must be 1 in y, whereas it does not have to be in x. So the answer for a factor in your problem is definitely correct, if fortuitous. R does the same as S in this example. If there were more than one level in y, the issue is less clearcut. Probably y[[k]] <- x[f == k] in split.default should be x[f %in% k] Note too z <- x; class(x) <- "foo"
split(z, y)
$"1" [1] NA 1
x = c(NA, 1)
y = factor(x) tapply(x, y, length)
1 1
tapply(y, y, length)
1 2
Brian D. Ripley, ripley@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272860 (secr) Oxford OX1 3TG, UK Fax: +44 1865 272595
george.leigh@dpi.qld.gov.au writes:
Full_Name: George Leigh Version: 1.8.1 OS: Windows 2000 Submission from: (NULL) (203.25.1.208) The following example gives the correct answer when the first argument of tapply is a numeric vector, but an incorrect answer when it is a factor. If the function used by tapply is "length", the type and contents of the first argument should make no difference, provided it has the same length as the second argument.
x = c(NA, 1) y = factor(x) tapply(x, y, length)
1 1
tapply(y, y, length)
1 2
The core of this is that
split(y,y)
$"1" [1] <NA> 1 Levels: 1
split(x,y)
$"1"
[1] 1
which in turn comes from the innards of split.default:
...
if (is.null(attr(x, "class")) && is.null(names(x)))
return(.Internal(split(x, f)))
lf <- levels(f)
y <- vector("list", length(lf))
names(y) <- lf
for (k in lf) y[[k]] <- x[f == k]
y
Factors have a class attribute, so you don't use the internal code in
that case and
y[y=="1"]
[1] <NA> 1
Levels: 1
I think the line in split.default needs to read
for (k in lf) y[[k]] <- x[!is.na(f) & f == k]
O__ ---- Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~~~~~~~~~ - (p.dalgaard@biostat.ku.dk) FAX: (+45) 35327907