从均匀分布的混合物中抽取随机数 [英] Draw random numbers from the mixture of uniform distributions

问题描述

目标

我正在尝试构建一个函数，该函数从不完全均匀分布中抽取特定数量的随机数。

I am trying to build a function that draws a specific number of random numbers from an an "incomplete uniform distribution".

我怎么称呼不完整的均匀分布？

我称不完整的均匀分布均匀分布一种概率分布，其中在一系列边界内的每个 X 值具有相同的概率。换句话说，它是一个带有孔（概率为零）的均匀分布，如下所示

I call an incomplete uniform distribution a probability distribution where each value of X within a series of boundaries have an equal probability to be picked. In other words it is a uniform distribution with holes (where the probability is zero) as represented below

x = list(12:25, 34:54, 67:90, 93:115)
y = 1/sum(25-12, 54-34, 90-67, 115-93)
plot(y=rep(y, length(unlist(x))), x=unlist(x), type="n", ylab="Probability", xlab="X")
for (xi in x)
{
    points(xi,rep(y, length(xi)), type="l", lwd=4)
}

丑陋的解决方案

这是一个缓慢而丑陋的解决方案

Here is a slow and ugly solution

IncompleteUnif = function(n,b)
{
    #################
    # "n" is the desired number of random numbers
    # "b" is a list describing the boundaries within which a random number can possibly be drawn.
    #################
    r = c() # Series of random numbers to return
    for (ni in n)
    {
        while (length(r) < n) # loop will continue until we have the "n" random numbers we need
        {
            ub = unlist(b)
            x = runif(1,min(ub), max(ub)) # one random number taken over the whole range
            for (bi in b) # The following loop test if the random number is withinn the ranges specified by "b"
            {
                if (min(bi) < x & max(bi) > x) # if found in one range then just add "x" to "r" and break
                {
                    r = append(r,x)
                    break
                }
            }
        }
    }
    return (r)
}

b = list(c(5,94),c(100,198),c(220,292), c(300,350))
set.seed(12)
IncompleteUnif(10,b)
[1]  28.929516 287.132444 330.204498  63.425103  16.693990  66.680826 226.374551  12.892821   7.872065 140.480533

推荐答案

@Gregor解决方案略有复杂的版本。

Slightly convoluted version of the @Gregor's solution.

mix_unif <- function(n, b){
  x <- c()
  ns <- rmultinom(1, n, sapply(b, diff))
  for (i in seq_along(ns)) {
    x <- c(x, runif(ns[i], b[[i]][1], b[[i]][2]))
  }
  x
}

microbenchmark(mix_unif(1e5, b),
               rmixunif(1e5, b),
               IncompleteUnif(1e5, b), 
               unit="relative")
Unit: relative
                     expr      min       lq     mean   median       uq      max neval
       mix_unif(1e+05, b) 1.000000 1.000000 1.000000 1.000000 1.000000  1.00000   100
       rmixunif(1e+05, b) 3.123515 3.235961 3.750369 3.496843 3.462529 15.73449   100
 IncompleteUnif(1e+05, b) 6.806916 7.247425 6.926282 7.188556 7.093928 18.20041   100

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