# identical vs all.equal

https://stackoverflow.com/questions/9508518/why-are-these-numbers-not-equal

숫자 비교시, 아주아주 정확할 필요가 없다면 identical() 보다는 all.equal()을 사용하는 편이 정신건강에 좋다.

> i <- 0.1 > i <- i + 0.05 > identical(i, 0.15) [1] FALSE > identical(round(i,1), 0.15) [1] FALSE > identical(round(i,2), 0.15) [1] TRUE > all.equal(i, 0.15) [1] TRUE > i <- 0.1 > i <- i + 0.06 > identical(i, 0.16) [1] TRUE > all.equal(i, 0.16) [1] TRUE

https://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f

## 7.31 Why doesn’t R think these numbers are equal?The only numbers that can be represented exactly in R’s numeric type are integers and fractions whose denominator is a power of 2. All other numbers are internally rounded to (typically) 53 binary digits accuracy. As a result, two floating point numbers will not reliably be equal unless they have been computed by the same algorithm, and not always even then. For example R> a <- sqrt(2) R> a * a == 2 [1] FALSE R> a * a - 2 [1] 4.440892e-16 R> print(a * a, digits = 18) [1] 2.00000000000000044 The function A discussion with many easily followed examples is in Appendix G “Computational Precision and Floating Point Arithmetic”, pages 753–771 of For more information, see e.g. David Goldberg (1991), “What Every Computer Scientist Should Know About Floating-Point Arithmetic”, Here is another example, this time using addition: R> .3 + .6 == .9 [1] FALSE R> .3 + .6 - .9 [1] -1.110223e-16 R> print(matrix(c(.3, .6, .9, .3 + .6)), digits = 18) [,1] [1,] 0.299999999999999989 [2,] 0.599999999999999978 [3,] 0.900000000000000022 [4,] 0.899999999999999911 |