[R] how to select an element from a vector based on a probability

[Frede Aakmann Tøgersen]

I think you have calculated the wrong probabilities. Shouldn't it be

> x <- c(2,2,6,2,1,1,1,3)
> MASS::fractions(table(x)/length(x))
x
  1   2   3   6 
3/8 3/8 1/8 1/8 
>

Yours sincerely / Med venlig hilsen


Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
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> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
> On Behalf Of Nordlund, Dan (DSHS/RDA)
> Sent: 10. april 2014 22:04
> To: Simone Gabbriellini; Rui Barradas
> Cc: r-help at r-project.org
> Subject: Re: [R] how to select an element from a vector based on a
> probability
> 
> > -----Original Message-----
> > From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> > project.org] On Behalf Of Simone Gabbriellini
> > Sent: Thursday, April 10, 2014 11:59 AM
> > To: Rui Barradas
> > Cc: r-help at r-project.org
> > Subject: Re: [R] how to select an element from a vector based on a
> > probability
> >
> > Hello, Rui,
> >
> > it does, indeed!
> >
> > thanks,
> > Simone
> >
> > 2014-04-10 20:55 GMT+02:00 Rui Barradas <ruipbarradas at sapo.pt>:
> > > Hello,
> > >
> > > Use ?sample.
> > >
> > > sample(x, 1, prob = x)
> > >
> 
> Just be aware that, in using this method, the probability of selection of a
> particular value will also be a function of how frequent the value is.  For
> example,
> 
> set.seed(7632)
> x <- c(2,2,6,2,1,1,1,3)
> table(sample(x, 10000, prob=x, replace=TRUE))
> 
>    1    2    3    6
> 1664 3340 1696 3300
> 
> 
> The probability that a vector position with a value of 1 will be selected is 1/18
> (in this particular example).  However, the probability that a value of 1 will be
> selected is 1/6 since there are three 1's.  The probability of selecting the
> position with a value of 3 is 3/18.  But since there is only one position with a
> value of 3, the probability of getting the value 1 on any given sample is equal
> to the probability of getting the value 3.
> 
> 
> 
> 
> > > Hope this helps,
> > >
> > > Rui Barradas
> > >
> > > Em 10-04-2014 19:49, Simone Gabbriellini escreveu:
> > >
> > >> Hello List,
> > >>
> > >> I have an array like:
> > >>
> > >> c(4, 3, 5, 4, 2, 2, 2, 4, 2, 6, 6, 7, 5, 5, 5, 10, 10, 11, 10,
> > >> 12, 10, 11, 9, 12, 10, 36, 35, 36, 36, 36, 35, 35, 36, 37, 35,
> > >> 35, 38, 35, 38, 36, 37, 36, 36, 37, 36, 35, 35, 36, 36, 35, 35,
> > >> 36, 35, 38, 35, 35, 35, 36, 35, 35, 35, 6, 5, 8, 6, 6, 7, 1,
> > >> 7, 7, 8, 9, 7, 8, 7, 7, 13, 13, 13, 14, 13, 13, 13, 14, 14, 15,
> > >> 15, 14, 13, 14, 39, 39, 39, 39, 39, 39, 41, 40, 39, 39, 39, 39,
> > >> 40, 39, 39, 41, 41, 40, 39, 40, 41, 40, 41, 40, 40, 40, 39, 41,
> > >> 39, 39, 39, 39, 40, 39, 39, 40, 40, 39, 39, 39, 1, 4, 3, 4)
> > >>
> > >> I would like to pick up an element with a probability proportional
> > to
> > >> the element value, thus higher values should be picked up more often
> > >> than small values (i.e., picking up 38 should be more probable than
> > >> picking up 3)
> > >>
> > >> Do you have any idea on how to code such a rich-get-richer
> > mechanism?
> > >>
> > >> Best regards,
> > >> Simone
> > >>
> > >
> >
> >
> >
> 
> Dan
> 
> Daniel J. Nordlund, PhD
> Research and Data Analysis Division
> Services & Enterprise Support Administration
> Washington State Department of Social and Health Services
> 
> 
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