sample(sort(x), 1, prob = prob)
Rui Barradas 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-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Simone Gabbriellini Sent: Thursday, April 10, 2014 11:59 AM To: Rui Barradas Cc: r-help@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 <ruipbarra...@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.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) Rui BarradasHope 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 proportionaltothe 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-richermechanism?Best regards, SimoneDan Daniel J. Nordlund, PhD Research and Data Analysis Division Services & Enterprise Support Administration Washington State Department of Social and Health Services______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
