R help - Feb 2008 - Number of digits of a value for problem 7.31 RFAQ

Hello dear R users!

I did not find a function which gives information about the number of 
digits of a value shown by R.
Do you know one?

I need it to solve the problem (see RFAQ 7.31)that 0.2==0.2+0.1-0.1 FALSE
The solution suggested in RFAQ is to use isTRUE(all.equal(0.2,0.2+0.1-0.1))

But if I want to compare inequality:
0.2<=0.2 +0.1-0.1 TRUE
but 0.2<=0.2 +0.1-0.1 FALSE
bad!
but in this case all.equal does not work I think... Unless to write a 
double condition like
0.2>0.2 +0.1-0.1 | isTRUE(all.equal(0.2,0.2 +0.1-0.1))

The solution I found is to round the values, because 
0.2==round(0.2+0.1-0.1,2) TRUE
However, one has to know the number of digits of the target value. How 
can you do when it is unknown?

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.

I did not find a way to represent the number of digits but I found that 
as.numeric(as.character(x)) rounds the values to the number of digits 
shown. But is there a better way to make it?

Thanks!

> I need it to solve the problem (see RFAQ 7.31)that 0.2==0.2+0.1-0.1 
FALSE
> The solution suggested in RFAQ is to use 
isTRUE(all.equal(0.2,0.2+0.1-0.1))
> 
> But if I want to compare inequality:
> 0.2<=0.2 +0.1-0.1 TRUE
> but 0.2<=0.2 +0.1-0.1 FALSE
> bad!
> but in this case all.equal does not work I think... Unless to write a 
> double condition like
> 0.2>0.2 +0.1-0.1 | isTRUE(all.equal(0.2,0.2 +0.1-0.1))
> 
> The solution I found is to round the values, because 
> 0.2==round(0.2+0.1-0.1,2) TRUE
> However, one has to know the number of digits of the target value. How 
> can you do when it is unknown?
You can do this by using all.equal to check for cases close to the 
boundary:

if(isTRUE(all.equal(0.2, 0.2 + 0.1 - 0.1))) 
{
  message("Possible rounding problems.") 
  #investigate further
} else if(0.2 <= 0.2 + 0.1 - 0.1)
{
  #your code here
}

all.equal has a tolerance parameter that you can set to see how strict you 
want the equality to be; you may want to make this value smaller.
> 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"

Regards,
Richie.

Mathematical Sciences Unit
HSL



------------------------------------------------------------------------
ATTENTION:

This message contains privileged and confidential inform...{{dropped:20}}

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!