On Mar 25, 2012, at 7:14 AM, Duncan Murdoch wrote:
> On 12-03-24 10:47 PM, J Toll wrote:
>> On Sat, Mar 24, 2012 at 7:30 PM, Duncan Murdoch
>> <murdoch.dun...@gmail.com> wrote:
>>> Do we have a format that always includes a decimal point and a given number
>>> of significant digits, but otherwise drops unnecessary characters? For
>>> example, if I wanted 5 digits, I'd want the following:
>>>
>>> Round to 5 digits:
>>> 1.234567 -> "1.2346"
>>>
>>> Drop unnecessary zeros:
>>> 1.23 -> "1.23"
>>>
>>> Force inclusion of a decimal point:
>>> 1 -> "1."
>>>
>>
>> Duncan,
>>
>> Maybe sprintf() will work for you. As it's a wrapper for C sprintf,
>> it should have its functionality.
>
> Maybe, but with which format string?
>
> Duncan Murdoch
I don't believe (though could be wrong), that you can do it all with one format
string, but can do it conditionally based upon the input. According to the C
printf documentation, the use of "#" forces a decimal point to be present, even
if there are no trailing digits. Thus:
> sprintf("%#.f", 1)
[1] "1."
The other two values seem to be handled by signif() when applied to each value
individually:
> signif(1.234567, 5)
[1] 1.2346
> signif(1.23, 5)
[1] 1.23
But, not when a vector:
> signif(c(1.234567, 1.23), 5)
[1] 1.2346 1.2300
So, wrapping that inside a function, using ifelse() to test for an integer
value:
signif.d <- function(x, digits)
{
ifelse(x == round(x),
sprintf("%.#f", x),
signif(x, digits))
}
x <- c(1.234567, 1.23, 1)
> signif.d(x, 5)
[1] "1.2346" "1.23" "1."
> signif.d(x, 6)
[1] "1.23457" "1.23" "1."
> signif.d(x, 7)
[1] "1.234567" "1.23" "1."
Not extensively tested of course, but hopefully that might work for your needs
Duncan.
Regards,
Marc Schwartz
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