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Intervals in function cut

5 messages · JulioSergio, Duncan Murdoch, jim holtman +1 more

JulioSergio
# 
When I was studying the function cut I found this example:
[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
[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.
Duncan Murdoch
#
On 11-11-07 7:48 PM, JulioSergio wrote:
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
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:

  
    

      
      
    

  


  


      

    

    

      
      

        
      

    

    
1 day later

    

      
      

        


  

        

      JulioSergio
    

    

      
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jim holtman <jholtman <at> gmail.com> writes:
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Thanks Jim,
very useful your information,

Sergio.