[R] Intervals in function cut

[Duncan Murdoch]

On 11-11-07 7:48 PM, JulioSergio 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?

The actual break is 1.996, but the formatting rounded that to 2 for 
printing.  You can see it more accurately if you set dig.lab to 4 or more.

Duncan Murdoch

>
> Thanks,
>
> Sergio.
>
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