[R] converting numeric to integer

[Gabor Grothendieck]

Try this:

as.integer(x + sign(x) * .5)


On Sun, May 17, 2009 at 6:01 AM, Thomas Mang <thomas.mang at fiwi.at> wrote:
> Hi,
>
> The problem is, x might be negative. If x == -6.9999999, the result should
> be -7, not -6. That's why my original proposal had the ifelse-condition (one
> could alternatively write sign(x) * 0.5, BTW.
> I agree however that in my proposal, the round(x) is redundant, one can use
> x itself as left-hand argument for the sum operation. Is there however still
> a more 'elegant' way ?
>
> thanks,
> Thomas
>
>
>
> Gabor Grothendieck wrote:
>>
>> Try:
>>
>> as.integer(x + 0.5)
>>
>> assuming the calculation error is less than 0.5 .
>>
>> On Sat, May 16, 2009 at 2:49 PM, Thomas Mang <thomas.mang at fiwi.at> wrote:
>>>
>>> Hello,
>>>
>>> Suppose I have x, which is a variable of class numeric. The calculations
>>> performed to yield x imply that mathematically it should be an integer ,
>>> but
>>> due to round-off errors, it might not be (and so in either direction).
>>> The
>>> error is however small, so round(x) will yield the appropriate integer
>>> value. Moreover, this integer values is guaranteed to be representable by
>>> an
>>> 'integer' class, that is -2^31 < x < 2^31, and logically it is an integer
>>> anyway. So I want to convert x from class 'numeric' to 'integer'. What is
>>> the most elegant, but always correct way, to achieve this conversion ?
>>>
>>> What comes to mind is of course something along:
>>>
>>> x = as.integer(round(x))
>>>
>>> I am, however, not sure if this always works, because I do not know if
>>> the
>>> round-function is guaranteed to return a numeric value which, in finite
>>> binary representation, is always >= the underlying mathematical integer.
>>> If
>>> that is however guaranteed, that would of course be a simple + elegant
>>> one.
>>>
>>> An alternative I came up with is:
>>>
>>> x = as.integer(round(x) + ifelse(x >= 0, 0.5, -0.5))
>>> Where I explicitly add a bit to ensure the finite binary representation
>>> must
>>> be >= the underlying integer, and then truncate the decimal digits.
>>> IMO, this one is always guaranteed to work, at least within the numerical
>>> range of what integers are limited to anyway.
>>>
>>>
>>> What's your opinion on the issue ?
>>> Any other solution ?
>>>
>>> Thanks a lot in advance and cheers,
>>> Thomas
>>>
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>>
>
>