[R] Number of digits of a value for problem 7.31 RFAQ SOLVED not really

[Matthieu Stigler]

Matthieu Stigler a e'crit :
> Richard.Cotton at hsl.gov.uk a e'crit :
>>>> What I mean is if R shows 2.3456 I want to obtain the info that 
>> digits=4
>>>> even if in facts the value has more (internal) digits.
>>> Try:
>>> x = 1.23456789
>>> format(x, nsmall=20)
>>> # [1] "1.23456788999999989009"
>>
>> I've just re-read the question. I suspect what you really wanted was 
>> something like:
>>
>> getndp <- function(x, tol=2*.Machine$double.eps)
>> {
>> ndp <- 0
>> while(!isTRUE(all.equal(x, round(x, ndp), tol=tol))) ndp <- ndp+1 
>> if(ndp > -log10(tol)) warning("Tolerance reached, ndp possibly 
>> underestimated.")
>> ndp }
>>
>> x = 0.123456
>> getndp(x)
>> # [1] 6
>>
>> x2 <- 3
>> getndp(x2)
>> # [1] 0
>> x3 <- 0.1234567890123456789
>> getndp(x3)
>> # [1] 16
>> # Warning message:
>> # In getndp(x3) : Tolerance reached, ndp possibly underestimated.
>>
>> Regards,
>> Richie.
>>
>> Mathematical Sciences Unit
>> HSL
>>
>
> Nice! Thanks a lot!!
> Now I can solve my problem that:
>
> target<-0.223423874568437
> target<=target+0.1-0.1 #TRUE
> target>=target+0.1-0.1 #FALSE!!
> target==target+0.1-0.1 #FALSE!!
>
> by:
> target<=round(target+0.1-0.1, getndp(target)) #TRUE
> target>=round(target+0.1-0.1, getndp(target)) #TRUE!!
> target==round(target+0.1-0.1, getndp(target)) #TRUE!!
>
>
Actually (after some trials) there is a little problem when faced with 
zeros...

 >getndp(1.0)
[1] 0

Note that it is only a partial problem because this happens only when 
the value is taken with other and so 
max(apply(matrix(allvalues,1,getndp))) will give the expected answer.

But when faced with only one value...

Note that I thought on a very different way which was starting from the 
point that as.character seems to work:

a<-as.character(1.23134)
b<-strsplit(a,"\\.")

nchar(b)
But I don't see how to do so that I can extract only the second element 
(maybe otherwise with sub or grep?)... But the method you gave is really 
more reliable than my "makeshift repair"

Thanks again!