[FORGED] Re: identical() versus sapply()
On 11/04/2016 8:25 PM, Paulson, Ariel wrote:
Hi Jeff, We are splitting hairs because R is splitting hairs, and causing us problems. Integer and numeric are different R classes with different properties, mathematical relationships notwithstanding. For instance, the counterintuitive result:
The issue here is that R has grown. The as() function is newer than the
as.numeric() function, it's part of the methods package. It is a much
more complicated thing, and there are cases where they differ.
In this case, the problem is that is(1L, "numeric") evaluates to TRUE,
and nobody has written a coerce method that specifically converts
"integer" to "numeric". So the as() function defaults to doing nothing.
It takes a while to do nothing, approximately 360 times longer than
as.numeric() takes to actually do the conversion:
> microbenchmark(as.numeric(1L), as(1L, "numeric"))
Unit: nanoseconds
expr min lq mean median uq max neval
as.numeric(1L) 133 210 516.92 273.5 409.5 9444 100
as(1L, "numeric") 51464 64501 119294.31 99768.5 138321.0 1313669 100
R performance is not always simple and easy to predict, but I think
anyone who had experience with R would never use as(x, "numeric"). So
this just isn't a problem worth fixing.
Now, you might object that the documentation claims they are equivalent,
but it certainly doesn't. The documentation aims to be accurate, not
necessarily clear.
Duncan Murdoch
identical(as.integer(1), as.numeric(1))
[1] FALSE Unfortunately the reply-to chain doesn't extend far enough -- here is the original problem:
sapply(1, identical, 1)
[1] TRUE
sapply(1:2, identical, 1)
[1] FALSE FALSE
sapply(1:2, function(i) identical(as.numeric(i),1) )
[1] TRUE FALSE
sapply(1:2, function(i) identical(as(i,"numeric"),1) )
[1] FALSE FALSE These are the results of R's hair-splitting!
Ariel
________________________________ From: Jeff Newmiller <jdnewmil at dcn.davis.ca.us> Sent: Monday, April 11, 2016 6:49 PM To: Bert Gunter; Paulson, Ariel Cc: Rolf Turner; r-help at r-project.org Subject: Re: [R] [FORGED] Re: identical() versus sapply() Hypothesis regarding the thought process: integer is a perfect subset of numeric, so why split hairs? -- Sent from my phone. Please excuse my brevity. On April 11, 2016 12:36:56 PM PDT, Bert Gunter <bgunter.4567 at gmail.com> wrote: Indeed! Slightly simplified to emphasize your point: class(as(1:2,"numeric")) [1] "integer" class(as.numeric(1:2)) [1] "numeric" whereas in ?as it says: "Methods are pre-defined for coercing any object to one of the basic datatypes. For example, as(x, "numeric") uses the existing as.numeric function. " I suspect this is related to my ignorance of S4 classes (i.e. as() ) and how they relate to S3 classes, but I certainly don't get it either. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Apr 11, 2016 at 9:30 AM, Paulson, Ariel <apa at stowers.org> wrote: Ok, I see the difference between 1 and 1:2, I'll just leave it as one of those "only in R" things. But it seems then, that as.numeric() should guarantee a FALSE outcome, yet it does not. To build on what Rolf pointed out, I would really love for someone to explain this one: str(1) num 1 str(1:2) int [1:2] 1 2 str(as.numeric(1:2)) num [1:2] 1 2 str(as(1:2,"numeric")) int [1:2] 1 2 Which doubly makes no sense. 1) Either the class is "numeric" or it isn't; I did not call as.integer() here. 2) method of recasting should not affect final class. Thanks, Ariel -----Original Message----- From: Rolf Turner [mailto:r.turner at auckland.ac.nz] Sent: Saturday, April 09, 2016 5:27 AM To: Jeff Newmiller Cc: Paulson, Ariel; 'r-help at r-project.org' Subject: Re: [FORGED] Re: [R] identical() versus sapply() On 09/04/16 16:24, Jeff Newmiller wrote: I highly recommend making friends with the str function. Try str( 1 ) str( 1:2 ) Interesting. But to me counter-intuitive. Since R makes no distinction between scalars and vectors of length 1 (or more accurately I think, since in R there is *no such thing as a scalar*, only a vector of length 1) I don't see why "1" should be treated in a manner that is categorically different from the way in which "1:2" is treated. Can you, or someone else with deep insight into R and its rationale, explain the basis for this difference in treatment? for the clue you need, and then sapply( 1:2, identical, 1L ) cheers, Rolf -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 ________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] ______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.