[R] Intervals in function cut

[jim holtman]

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 <juliosergio at 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.
>
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> 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.