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

[Simone Gabbriellini]

Dear Frede,

the initial question was to find probability proportional to the
values, so Rui, Dan and Boris are right...

I think I will go for Boris' solution...

best regards,
Simone

2014-04-11 7:29 GMT+02:00 Frede Aakmann Tøgersen <frtog at vestas.com>:
> 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
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> Plant Performance & Modeling
>
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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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>> and provide commented, minimal, self-contained, reproducible code.



-- 
-----------------------------------------------------------------

Simone Gabbriellini, PhD

Post-doctoral Researcher
ANR founded research project "DIFFCERAM"
GEMASS, CNRS & Paris-Sorbonne.

mobile: +39 340 39 75 626
email: simone.gabbriellini at cnrs.fr