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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