Did you read the help page?
x: numeric vector whose sample quantiles are wanted. ‘NA’ and
‘NaN’ values are not allowed unless ‘na.rm’ is ‘TRUE’.
so only 'numeric' vectors are really supported, although it does say
The default method does not allow factors, but works with objects
sufficiently like numeric vectors that ‘sort’, addition and
multiplication work correctly. In principle only sorts and
weighted means are needed, so datatimes could have quantiles - but
this is not implemented.
There is no claim that it works (let alone works well) for class
"difftime". If you follow the link to 'sort' it says
The default ‘sort’ method makes use of ‘order’ for objects with
classes, which in turn makes use of the generic function ‘xtfrm’.
and from ?xtfrm
The default method will make use of ‘==’ and ‘>’ methods for the
class of ‘x[i]’ (for integers ‘i’), and the ‘is.na’ method for the
class of ‘x’, but might be rather slow when doing so.
So, if you want this to be fast, you need to write an xtfrm method.
There is one in R-devel
xtfrm.difftime
function (x)
as.numeric(x)
and you can use that in your workspace (and your example is fast in
R-devel, because of that function I think, there being other
development work in progress in the version I tried).
On Fri, 27 Nov 2009, [email protected] wrote:
Full_Name: Hong Ooi
Version: 2.10.0
OS: Windows XP
Submission from: (NULL) (203.110.235.1)
While trying to get summary statistics on a duration variable (the difference
between a start and end date), I ran into the following issue. Using summary or
quantile (which summary calls) on a difftime object takes an extremely long time
if the object is even moderately large.
A reproducible example:
x <- as.Date(1:10000, origin="1900-01-01")
x[1:10]
[1] "1900-01-02" "1900-01-03" "1900-01-04" "1900-01-05" "1900-01-06"
[6] "1900-01-07" "1900-01-08" "1900-01-09" "1900-01-10" "1900-01-11"
d <- x - as.Date("1900-01-01")
d[1:10]
Time differences in days
[1] 1 2 3 4 5 6 7 8 9 10
system.time(summary(d[1:10]))
user system elapsed
0.01 0.00 0.01
system.time(summary(d[1:100]))
user system elapsed
0.21 0.00 0.20
system.time(summary(d[1:1000]))
user system elapsed
3.02 0.00 3.02
system.time(summary(d[1:10000]))
user system elapsed
43.56 0.04 43.66
If I unclass d, there is no problem:
system.time(summary(unclass(d[1:10000])))
user system elapsed
0 0 0
Testing with Rprof() indicates that the problem lies in [.difftime, although the
code for that function seems innocuous enough.
sessionInfo()
R version 2.10.0 (2009-10-26)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_Australia.1252 LC_CTYPE=English_Australia.1252
[3] LC_MONETARY=English_Australia.1252 LC_NUMERIC=C
[5] LC_TIME=English_Australia.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
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--
Brian D. Ripley, [email protected]
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 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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