I definitely do agree with you.
Basically, I see two different ways to proceed:
1. one could first check if there are any 0 in the vector and then
return 0 without computing the product
2. or convert prod(x1, x2, x3) in prod(c(x1, x2, x3))
Both approaches are similar except for the pathological case where one
vector x1 is really huge. An example:
prod(0, 1:1e25)
1. will give 0
2. will give an error stating that the vector c(0, 1:1e255) will be
too large - in length I mean
Consequently, my opinion will be that approach 1 will be better - and
maybe faster because it'll avoide useless computations.
Best,
Mathieu
Martin Maechler a écrit :
> I think most of us would expect prod(0:1000) to return 0, and ...
>
>
> ... it does.
>
> However, many of us also expect
> prod(x1, x2) to be equivalent to
> prod(c(x1,x2))
> the same as we can expect that for min(), max(), sum() and such
> members of the "Summary" group.
>
> Consequently, prod(0, 1:1000) should also return 0,
> but as you see, it gives NaN which may be a bit puzzling...
> The explanation is relatively simple:
>
> 1) The internal implementation uses
>
> prod(x1, x2) := prod(x1) * prod(x2)
>
> which in this case is
>
> 2) 0 * Inf and that is not 0, but NaN;
>
> not necessarily because we would want that, but I think just
> because the underlying C math library does so.
>
>
> I would personally like to change both behaviors,
> but am currently only proposing to change prod() such as to
> return 0 in such cases.
> This would be S-plus compatible, in case that matters.
>
> Opinions?
>
> Martin Maechler, ETH Zurich & R-core
>
> ______________________________________________
> R-devel@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-devel
>
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