[R] Rounding

[Tony Plate]

At 09:22 PM 9/19/2002 +1200, you wrote:

>Now my questions is: how do I work out the number of digits x has?


If you're working with decimal digits (as most people do) that would be
   ceiling(log10(abs(x)))
for most cases.  The problem cases are round powers of 10, such as 1000, 
for which the above expression returns 3.  You'll have to work out whether 
or not that is appropriate for your task.

Another more indirect method is:
   nchar(formatC(floor(abs(x)), format="f", digits=0))
which will tell you number of digits before the decimal point in an 
infinite precision decimal rendition of x.  E.g., it will return 4 for 
1000, and 3 for 999.999999.  The latter may or may not be what you want, 
since 999.999999 usually prints as 1000 (because of rounding implicit in 
printing).

-- Tony Plate

PS.  If you do need to treat round powers of 10 as special cases, be sure 
to test on many examples to make sure your arithmetic behaves correctly 
when implemented with binary floating point arithmetic; here's a table 
illustrating the challenges:

 > cbind(i=i,"=="=log10(10**i)==i,"<"=log10(10**i)<i,">"=log10(10**i)>i)
        i == < >
  [1,] -6  0 0 1
  [2,] -5  1 0 0
  [3,] -4  0 0 1
  [4,] -3  0 0 1
  [5,] -2  0 0 1
  [6,] -1  0 0 1
  [7,]  0  1 0 0
  [8,]  1  1 0 0
  [9,]  2  1 0 0
[10,]  3  0 1 0
[11,]  4  1 0 0
[12,]  5  1 0 0
[13,]  6  0 1 0
[14,]  7  1 0 0
[15,]  8  1 0 0
[16,]  9  0 1 0
[17,] 10  1 0 0
[18,] 11  1 0 0
[19,] 12  0 1 0
[20,] 13  0 1 0
[21,] 14  1 0 0
[22,] 15  0 1 0
[23,] 16  1 0 0
[24,] 17  1 0 0
[25,] 18  0 1 0
[26,] 19  1 0 0
[27,] 20  1 0 0
[28,] 21  0 1 0
[29,] 22  1 0 0
[30,] 23  0 1 0
[31,] 24  0 1 0
 >

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