On 30-Jan-2013 20:39:34 Berend Hasselman wrote: > > On 30-01-2013, at 21:32, Dave Mitchell <dmmtc...@gmail.com> wrote: > >> Why, in R, does (0.1 + 0.05) > 0.15 evaluate to True? What am I missing >> here? How can I ensure this (ostensibly incorrect) behavior doesn't >> introduce bugs into my code? Thanks for your time. >> > > R-FAQ 7.31: > http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-num > bers-are-equal_003f > > Berend
And, to put Dave's specific query into the context of that FAQ: (0.1 + 0.05) > 0.15 # [1] TRUE (0.1 + 0.05) - 0.15 # [1] 2.775558e-17 so that tiny 2.775558e-17 is the inexactitude (due to finite binary representation). As an interesting variant: (1.0 + 0.5) > 1.5 # [1] FALSE (1.0 + 0.5) - 1.5 # [1] 0 and that is because 1.0 and 0.5, and also 1.5, have exact finite binary representations, e.g.: 1.0 == 1.0000000000000000 0.5 == 0.1000000000000000 1.5 == 1.1000000000000000 whereas 0.1, 0.5 and 0.15 are these numbers divided by 10 = 2*5; and while you can exactly do the "/2" part (just shift right by one binary place), you can't exactly divide by 5 in finite binary arithmetic (any more than you can exactly divide by 3 in decimal), because 5 is not a factor of the base (2) of the binary representation (whereas, in decimal, both 2 and 5 are factors of 10; but 3 isn't). Whereas R has the function all.equal() to give the "right" answer for most things like (0.1 + 0.05) == 0.15 # [1] FALSE all.equal((0.1 + 0.05), 0.15) # [1] TRUE (because it tests for equality to within a very small tolerance, by default the square root of the binary precision of a double precision binary real number), R does not have a straightforward method for testing the truth of "(0.1 + 0.05) > 0.15" (and that is because the question is not clearly discernible from the expression, when imprecision underlies it). You could emulate all.equal() on the lines of (0.1 + 0.05) > (0.15 + .Machine$double.eps^0.5) # [1] FALSE (0.1 + 0.05) < (0.15 - .Machine$double.eps^0.5) # [1] FALSE (or similar). Observe that .Machine$double.eps^0.5 # [1] 1.490116e-08 .Machine$double.eps # [1] 2.220446e-16 (0.1 + 0.05) - 0.15 # [1] 2.775558e-17 Hoping this helps, Ted. ------------------------------------------------- E-Mail: (Ted Harding) <ted.hard...@wlandres.net> Date: 30-Jan-2013 Time: 23:22:53 This message was sent by XFMail ______________________________________________ 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.
