Re: [R] Breaking out of multiple loops

2012-12-19 Thread MacQueen, Don
There is a break() function. Does it not do the job? -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 12/19/12 7:27 AM, "McCloskey, Bryan" wrote: >Something like this? > >n<-1000 >pythag<-function(n){ > for(i in 3:((n-

Re: [R] Breaking out of multiple loops

2012-12-19 Thread Jeff Newmiller
There is. Use while loops. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playi

Re: [R] Breaking out of multiple loops

2012-12-19 Thread McCloskey, Bryan
Something like this? n<-1000 pythag<-function(n){ for(i in 3:((n-3)/3)) for(j in (i+1):((n-i)/2)) if(i^2+j^2==(n-i-j)^2) return(i*j*(n-i-j)) } system.time(print(pythag(n))) Interesting -- seems to improve speed by ~12%. Not sure if that's because stop() has more overhead, or if functi

Re: [R] Breaking out of multiple loops

2012-12-18 Thread Duncan Murdoch
On 12-12-18 1:02 PM, McCloskey, Bryan wrote: Hey all, I'm currently working through the problems at Project Euler -- this question came up while working on Problem 9 (http://projecteuler.net/problem=9): "A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a^2 + b^2 =

[R] Breaking out of multiple loops

2012-12-18 Thread McCloskey, Bryan
Hey all, I'm currently working through the problems at Project Euler -- this question came up while working on Problem 9 (http://projecteuler.net/problem=9): "A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a^2 + b^2 = c^2. For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2