使用 R 生成密度对象的随机偏差
我有一个像这样创建的密度对象 dd：
x1 <- rnorm(1000)
x2 <- rnorm(1000, 3, 2)
x <- rbind(x1, x2)
dd <- density(x)
plot(dd)
它产生了这个非常非高斯分布：
alt text http://www.cerebralmastication.com/wp-content/uploads/2009/09/nongaus.png
我最终希望获得与 rnorm 类似的分布的随机偏差偏离正态分布。
我试图解决这个问题的方法是获取内核的 CDF，然后让它告诉我如果我向它传递累积概率（逆 CDF）的变量。这样我就可以将均匀随机变量的向量转换为密度的绘图。
看来我想做的事情应该是其他人在我之前做过的基本事情。有没有简单的方法或简单的功能来做到这一点？我讨厌重新发明轮子。
FWIW我找到了这篇R帮助文章但我不能明白他们在做什么，最终的输出似乎并没有产生我想要的结果。但这可能是我不明白的一步。
我考虑过使用 Johnson来自suppdists包的分布，但Johnson不会给我我的数据所具有的漂亮的双峰驼峰。
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.
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（2）
替代方法：
Alternative approach:
这只是法线的混合。那么为什么不这样做呢：
如果您需要的只是来自该混合分布的样本，那么这应该没问题。
This is just a mixture of normals. So why not something like:
This should be fine if all you need are samples from this mixture distribution.