In stepping through 'cut.default' that is called, I get the following interval:
-0.008 0.994 1.996 2.998 4.000 5.002 6.004 7.006 8.008 In printing out to three significant digits, you will have "(0.994, 2]" or "(2,3]" as you see in the factors. If instead you used: cut(x, b = 8, dig.lab = 7) you will get: cut(x, b= 8, dig.lab=7) [1] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] [7] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (0.994,1.996] (0.994,1.996] (0.994,1.996] [13] (0.994,1.996] (1.996,2.998] (1.996,2.998] (1.996,2.998] (1.996,2.998] (1.996,2.998] [19] (1.996,2.998] (2.998,4] (2.998,4] (2.998,4] (2.998,4] (2.998,4] [25] (4,5.002] (4,5.002] (4,5.002] (4,5.002] (4,5.002] (4,5.002] [31] (4,5.002] (4,5.002] (4,5.002] (4,5.002] (4,5.002] (4,5.002] [37] (4,5.002] (5.002,6.004] (5.002,6.004] (5.002,6.004] (5.002,6.004] (5.002,6.004] [43] (6.004,7.006] (6.004,7.006] (6.004,7.006] (7.006,8.008] (7.006,8.008] (7.006,8.008] [49] (7.006,8.008] (7.006,8.008] 8 Levels: (-0.008,0.994] (0.994,1.996] (1.996,2.998] (2.998,4] (4,5.002] ... (7.006,8.008] aren't floating point numbers fun; read FAQ 7.31 On Mon, Nov 7, 2011 at 7:48 PM, JulioSergio <julioser...@gmail.com> wrote: > When I was studying the function cut I found this example: > >> x <- rep(0:8, tx0) >> x > [1] 0 0 0 0 0 0 0 0 0 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 5 5 5 5 5 5 > 5 5 5 5 6 > [39] 6 6 6 6 7 7 7 8 8 8 8 8 > >> cut(x, b = 8) > [1] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] > (-0.008,0.994] > [6] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (-0.008,0.994] (0.994,2] > [11] (0.994,2] (0.994,2] (0.994,2] (2,3] (2,3] > [16] (2,3] (2,3] (2,3] (2,3] (3,4] > [21] (3,4] (3,4] (3,4] (3,4] (4,5] > [26] (4,5] (4,5] (4,5] (4,5] (4,5] > [31] (4,5] (4,5] (4,5] (4,5] (4,5] > [36] (4,5] (4,5] (5,6] (5,6] (5,6] > [41] (5,6] (5,6] (6,7.01] (6,7.01] (6,7.01] > [46] (7.01,8.01] (7.01,8.01] (7.01,8.01] (7.01,8.01] (7.01,8.01] > 8 Levels: (-0.008,0.994] (0.994,2] (2,3] (3,4] (4,5] (5,6] ... (7.01,8.01] > > I undestand that the resulting factor yields as its first component the > corresponding "intervals" that the original vector (x) elements belong to. > So, clearly, the first element of x, 0, belongs to (-0.008,0.994] because > -0.008 < 0 <= 0.994. However, when I come to the 14th element of x, > that is, 2, I don't see that it belongs to (2,3], because 2 < 2 <= 3, > is strictly false according to the standard inverval notation > (http://en.wikipedia.org/wiki/Interval_notation). Maybe, there is something > I have not understood of either the cut function or the interval notation. > Do you have > any comments? > > Thanks, > > Sergio. > > ______________________________________________ > 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. > -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. ______________________________________________ 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.
