Hi, just a quick question:
Should weighted.mean be able to cope with the specific case where one
weight is Inf? I came across this when trying to implement a simple
weighted moving average algorithm for surface smoothing: these
algorithms often result in a single infinite weight when
Yan Wong wrote:
Hi, just a quick question:
Should weighted.mean be able to cope with the specific case where one
weight is Inf? I came across this when trying to implement a simple
weighted moving average algorithm for surface smoothing: these
algorithms often result in a single
On 11 Aug 2006, at 12:49, Duncan Murdoch wrote:
It makes sense in this case, but in the case where there is more
than one infinite weight, the result has to be NaN.
... it would be a lot more complicated if it were to handle this
very special case.
Yes - I see that it may not be worth
