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

Thu Apr 10 22:39:21 CEST 2014

```Sorry, there's a bug. merge() sorts the data.frame so we need to sort x
also.

sample(sort(x), 1, prob = prob)

Em 10-04-2014 21:34, Rui Barradas escreveu:
> Hello,
>
> Inline.
>
> Em 10-04-2014 21:04, Nordlund, Dan (DSHS/RDA) escreveu:
>>> -----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
>>> 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
>>>
>>>> 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.
>
> You're right, I didn't notice that. One way of avoiding that problem is
> the following.
>
> prob <- merge(x, data.frame(x=unique(x),
> prob=unique(x)/sum(unique(x))))\$prob
> sample(x, 1, prob = prob)
>
>
>>
>>
>>
>>
>>>> Hope this helps,
>>>>
>>>>
>>>> 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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