Thanks a bundle everyone, I could not have hoped for a better set of
answers. The Darts, Dice, and Coins page is excellent, the original
papers are also very accessible, something that I have experienced
with a lot of early Computer Science papers.
In case anyone is interested in where Vose's
Hi Jason,
great answer, I learned something useful today :-)
Sebastian
On Wed, May 27, 2015 at 4:07 PM, Jason Merrill jwmerr...@gmail.com wrote:
You might want to take a look at
http://en.m.wikipedia.org/wiki/Alias_method
There's a nice, detailed exposition at
You might want to take a look at http://en.m.wikipedia.org/wiki/Alias_method
There's a nice, detailed exposition at
http://www.keithschwarz.com/darts-dice-coins/
In fact, it's already implemented in Distributions.jl:
julia c = Categorical([0.1,0.2,0.3,0.4])
Categorical(K=4, p=[0.1,0.2,0.3,0.4])
julia a = sampler(c)
AliasTable with 4 entries
julia rand(a)
4
On Wednesday, 27 May 2015 17:24:55 UTC+1, Sebastian Nowozin wrote:
Hi Jason,
great answer,
El miércoles, 27 de mayo de 2015, 18:24:55 (UTC+2), Sebastian Nowozin
escribió:
Hi Jason,
great answer, I learned something useful today :-)
I second that!
But I'm also very happy to see that the Julian experts already knew all
about it!
David.
Sebastian
On Wed, May 27, 2015 at
El miércoles, 27 de mayo de 2015, 0:27:41 (UTC+2), Sebastian Nowozin
escribió:
Hi Pontus,
I have not seen this approach before, but it is certainly interesting and
potentially useful.
I hadn't seen it either, and it is nice, yes.
As far as I am aware, the standard method for
since this looks a lot like recreating a distribution from a frequency
measurement, one of the other tools that comes to my mind (courtesy of an
old question I asked of julia-users in the development of the hist method
in base), is to compute the (continuous) Kernel Density Estimator first,
and
Hi Pontus,
I have not seen this approach before, but it is certainly interesting and
potentially useful.
Without giving an answer to your questions, some things to consider:
1. Small probabilities will be the hardest to represent.
2. Maybe there is an exact method possible by storing in each
