Hi utkarshsinghal <utkarsh.sing...@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.sing...@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.sing...@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" <hdo...@air.org> / > />To: <vincent.gou...@act.ulaval.ca>, <r-h...@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-boun...@stat.math.ethz.ch / > />[mailto:r-help-boun...@stat.math.ethz.ch] On Behalf Of Vincent Goulet / > />Sent: Monday, August 29, 2005 3:35 PM / > />To: r-h...@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-h...@stat.math.ethz.ch mailing list / > />https://stat.ethz.ch/mailman/listinfo/r-help / > />PLEASE do read the posting guide! / > />http://www.R-project.org/posting-guide.html<http://www.r-project.org/ > posting-guide.html>/ > /> / > />______________________________________________ / > />r-h...@stat.math.ethz.ch mailing list / > />https://stat.ethz.ch/mailman/listinfo/r-help / > />PLEASE do read the posting guide! / > />http://www.R-project.org/posting-guide.html<http://www.r-project.org/ > posting-guide.html>/ > > [[alternative HTML version deleted]] > > > ______________________________________________ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html< http://www.r-project.org/posting-guide.html> > 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, rip...@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@r-project.org mailing list > 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. ______________________________________________ R-help@r-project.org mailing list 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.
