使用R生成具有密度对象的随机随机偏差

Question

I have a density object dd created like this:

x1 <- rnorm(1000) 
x2 <- rnorm(1000, 3, 2) 
x <- rbind(x1, x2)
dd <- density(x) 
plot(dd)

Which produces this very non-Gaussian distribution:

alt text http://www.cerebralmastication.com/wp-content/uploads/2009/09/nongaus.png

I would ultimately like to get random deviates from this distribution similar to how rnorm gets deviates from a normal distribution.

The way I am trying to crack this is to get the CDF of my kernel and then get it to tell me the variate if I pass it a cumulative probability (inverse CDF). That way I can turn a vector of uniform random variates into draws from the density.

It seems like what I am trying to do should be something basic that others have done before me. Is there a simple way or a simple function to do this? I hate reinventing the wheel.

FWIW I found this R Help article but I can't grok what they are doing and the final output does not seem to produce what I am after. But it could be a step along the way that I just don't understand.

I've considered just going with a Johnson distribution from the suppdists package but Johnson won't give me the nice bimodal hump which my data has.

Answer 1

替代方法：

sample(x, n, replace = TRUE)

Answer 2

This is just a mixture of normals. So why not something like:

rmnorm <- function(n,mean, sd,prob) {
    nmix <- length(mean)
    if (length(sd)!=nmix) stop("lengths should be the same.")
    y <- sample(1:nmix,n,prob=prob, replace=TRUE)
    mean.mix <- mean[y]
    sd.mix <- sd[y]
    rnorm(n,mean.mix,sd.mix)
}
plot(density(rmnorm(10000,mean=c(0,3), sd=c(1,2), prob=c(.5,.5))))

This should be fine if all you need are samples from this mixture distribution.