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.m...@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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