[R] Testing if all elements are equal in a vector/matrix
Petr PIKAL
petr.pikal at precheza.cz
Fri Jun 19 14:18:51 CEST 2009
Hi
utkarshsinghal <utkarsh.singhal at global-analytics.com> napsal dne
17.06.2009 15:29:34:
> I will wait for the next version-2.9.1 and presently using Petr's
suggestion, i.e.,
> (x[1]*length(x))==sum(x)
> which significantly reduced the run time.
>
> The problem is now there might be only small differences ,say, of the
order of
> 10^-10 which I want to ignore.
>
> So I used:
> isTRUE(all.equal((x[1]*length(x)),sum(x)))
> as suggested in the documentation of all.equal.
>
> But this again increased the run time to five times.
>
> 1) Is there any faster way of doing the same?
Maybe (not tested)
(x[1]*length(x))==round(sum(x),10)
Petr
> 2) Will the function "anyDuplicated" treat almost equal values as
duplicated
> or not? Actually I need both the options.
>
>
> Regards
> Utkarsh
>
>
>
> Prof Brian Ripley wrote:
> On Tue, 16 Jun 2009, Prof Brian Ripley wrote:
> On Tue, 16 Jun 2009, jim holtman wrote:
> I think the only way that you are going to get it to stop on the first
> mismatch is to write your own function in C if you are concerned about
the
> time. Matching on character vectors will be even more costly since it
is
> having to loop to check the equality of each character in each element.
> This is one of the places it might pay to convert to factors and then
the
> comparison only uses the integer values assigned to the factors.
>
> Not so in a recent R: comparison of character vectors is now done by
comparing
> pointers in the first instance so (at least on a 32-bit platform) is as
fast
> as comparing integers. And on x86_64 Linux:
> x <- as.character(c(1,2,rep(1,10000000)))
> system.time(print(all(x[1] == x)))
> [1] FALSE
> user system elapsed
> 0.123 0.019 0.142
> system.time(xx <- as.factor(x))
> user system elapsed
> 9.874 0.284 10.159
> system.time(print(all(xx[1] == xx)))
> [1] FALSE
> user system elapsed
> 0.511 0.145 0.656
>
> Recent pre-release versions of R (e.g. 2.9.1 beta) allow
> system.time(anyDuplicated(x))
> user system elapsed
> 0.034 0.078 0.113
> system.time(anyDuplicated(xx))
> user system elapsed
> 0.037 0.076 0.113
>
> I'm sorry, a line got reverted here: I had edited this to say
>
> 'which is a C-level speedup of the sort the original poster seemed to be
looking for'
>
>
> On Tue, Jun 16, 2009 at 8:31 AM, utkarshsinghal <
> utkarsh.singhal at global-analytics.com> wrote:
> Hi Jim,
>
> What you are saying is correct. Although, my computer might not have
same
> speed and I am getting the following for 10M entries:
>
> user system elapsed
> 0.559 0.038 0.607
>
> Moreover, in the case of character vectors, it gets more than double.
>
> In my modeling, which is already highly time consuming, I need to do
check
> this for few thousand vectors and the entries can easily be 10M in each
> vector. So I am just looking for any possibilities of time saving. I am
> pretty sure that whenever elements are not all equal, it can be
concluded
> from any few entries (most of the times). It will be worth if I can find
a
> way which stops checking further the moment it find two distinct
elements.
>
> Regards
> Utkarsh
>
>
>
> jim holtman wrote:
>
> Just check that the first (or any other element) is equal to all the
rest:
> x = c(1,2,rep(1,10000000)) # 10,000,000
> system.time(print(all(x[1] == x)))
> [1] FALSE
> user system elapsed
> 0.18 0.00 0.19
>
> This was for 10M entries.
>
> On Tue, Jun 16, 2009 at 7:42 AM, utkarshsinghal <
> utkarsh.singhal at global-analytics.com> wrote:
>
> Hi All,
>
> There are several replies to the question below, but I think there must
> exist a better way of doing so.
> I just want to check whether all the elements of a vector are same. My
> vector has one million elements and it is highly likely that there are
> distinct elements in the first few itself. For example:
>
> > x = c(1,2,rep(1,100000))
>
> I want the answer as FALSE, which is clear from the first two
> observations itself and we don't need to check for the rest.
>
> Does anybody know the most efficient way of doing this?
>
> Regards
> Utkarsh
>
>
>
> From: Francisco J. Zagmutt <gerifalte28_at_hotmail.com
> <mailto:gerifalte28_at_hotmail.com
>
?Subject=Re:%20%5BR%5D%20Testing%20if%20all%20elements%20are%20equal%20in%20a%
> 20vector/matrix>>
>
> Date: Tue 30 Aug 2005 - 06:05:20 EST
>
>
> Hi Doran
>
> The documentation for isTRUE reads 'isTRUE(x)' is an abbreviation of
> 'identical(TRUE,x)' so actually Vincent's solutions is "cleaner" than
> using identical :)
>
> Cheers
>
> Francisco
>
> />From: "Doran, Harold" <HDoran at air.org> /
> />To: <vincent.goulet at act.ulaval.ca>, <r-help at stat.math.ethz.ch> /
> />Subject: Re: [R] Testing if all elements are equal in a vector/matrix
/
> />Date: Mon, 29 Aug 2005 15:49:20 -0400 /
> /> /
> >See ?identical
> <http://tolstoy.newcastle.edu.au/R/help/05/08/11201.html#11202qlink1>
> /> /
> />-----Original Message----- /
> />From: r-help-bounces at stat.math.ethz.ch /
> />[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Vincent Goulet
/
> />Sent: Monday, August 29, 2005 3:35 PM /
> />To: r-help at stat.math.ethz.ch /
> />Subject: [R] Testing if all elements are equal in a vector/matrix /
> /> /
> /> /
> />Is there a canonical way to check if all elements of a vector or
> matrix are /
> />the same? Solutions below work, but look hackish to me. /
> /> /
> /> > x <- rep(1, 10) /
> /> > all(x == x[1]) # == operator does not provide for small differences
/
> */>[1] TRUE /
> */> > isTRUE(all.equal(x, rep(x[1], length(x)))) # ugly /
> */>[1] TRUE /
> */> /
> />Best, /
> /> /
> />Vincent /
> />-- /
> /> Vincent Goulet, Associate Professor /
> /> ?cole d'actuariat /
> /> Universit? Laval, Qu?bec /
> /> Vincent.Goulet_at_act.ulaval.ca<
http://vincent.goulet_at_act.ulaval.ca/>
> <mailto:Vincent.Goulet_at_act.ulaval.ca
>
?Subject=Re:%20%5BR%5D%20Testing%20if%20all%20elements%20are%20equal%20in%20a%
> 20vector/matrix>
> http://vgoulet.act.ulaval.ca /
> /> /
> />______________________________________________ /
> />R-help at stat.math.ethz.ch mailing list /
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>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
> [[alternative HTML version deleted]]
>
>
> --
> Brian D. Ripley, ripley at 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 272866 (PA)
> Oxford OX1 3TG, UK Fax: +44 1865 272595
>
> ______________________________________________
> R-help at r-project.org mailing list
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> PLEASE do read the posting guide
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