假设我有一系列n个独立伯努利试验成功的概率,p1到pn,使得p1 != p2 !=…! = pn。给每个试验一个唯一的名称。
p <- c(0.5, 0.12, 0.7, 0.8, .02)
a <- c("A","B","C","D","E")
我知道从搜索堆栈交换(例如,这里和这里),我可以找到cdf, pmf等使用泊松二项分布函数。
我感兴趣的是每一种可能的成功和失败组合的确切概率。(例如,如果我画了一棵概率树,我想知道每个分支末端的概率。) all <- prod(p)
all
[1] 0.000672
o1 <- (0.5 * (1-0.12) * 0.7 * 0.8 * .02)
o1
[1] 0.004928
o2 <- (0.5 * 0.12 * (1-0.7) * 0.8 * .02)
o2
[1] 0.000288
…对于所有2^5种可能的成功/失败组合
在R中有什么有效的方法来处理这个?
在我的实际数据集中,试验次数是19,所以我们讨论的是概率树上的2^19条路径。
使此计算快速的关键是在对数概率空间中进行计算,以便树的每个分支的乘积是可以作为矩阵乘法的内和计算的和。通过这种方式,所有分支可以以矢量化的方式一起计算。
首先,构造一个所有分支的枚举。为此,我们使用R.utils
包中的intToBin
函数:
library(R.utils)
enum.branches <- unlist(strsplit(intToBin(seq_len(2^n)-1),split=""))
式中n
为伯努利变量个数。对于您的示例,n=5
:
matrix(enum.branches, nrow=n)
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17]
##[1,] "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "1"
##[2,] "0" "0" "0" "0" "0" "0" "0" "0" "1" "1" "1" "1" "1" "1" "1" "1" "0"
##[3,] "0" "0" "0" "0" "1" "1" "1" "1" "0" "0" "0" "0" "1" "1" "1" "1" "0"
##[4,] "0" "0" "1" "1" "0" "0" "1" "1" "0" "0" "1" "1" "0" "0" "1" "1" "0"
##[5,] "0" "1" "0" "1" "0" "1" "0" "1" "0" "1" "0" "1" "0" "1" "0" "1" "0"
## [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31] [,32]
##[1,] "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1" "1"
##[2,] "0" "0" "0" "0" "0" "0" "0" "1" "1" "1" "1" "1" "1" "1" "1"
##[3,] "0" "0" "0" "1" "1" "1" "1" "0" "0" "0" "0" "1" "1" "1" "1"
##[4,] "0" "1" "1" "0" "0" "1" "1" "0" "0" "1" "1" "0" "0" "1" "1"
##[5,] "1" "0" "1" "0" "1" "0" "1" "0" "1" "0" "1" "0" "1" "0" "1"
得到一个矩阵,其中每一列是概率树的一个分支的结果。
现在,用它来构造一个与enum.branches
大小相同的对数概率矩阵,如果enum.branches=="1"
和log(1-p)
,则值为log(p)
。对于p <- c(0.5, 0.12, 0.7, 0.8, .02)
的数据,这是:
logp <- matrix(ifelse(enum.branches == "1", rep(log(p), 2^n), rep(log(1-p), 2^n)), nrow=n)
然后,对对数概率求和,取指数,得到概率的乘积:
result <- exp(rep(1,n) %*% logp)
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
##[1,] 0.025872 0.000528 0.103488 0.002112 0.060368 0.001232 0.241472 0.004928 0.003528 7.2e-05
[,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
##[1,] 0.014112 0.000288 0.008232 0.000168 0.032928 0.000672 0.025872 0.000528 0.103488 0.002112
[,21] [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30]
##[1,] 0.060368 0.001232 0.241472 0.004928 0.003528 7.2e-05 0.014112 0.000288 0.008232 0.000168
[,31] [,32]
##[1,] 0.032928 0.000672
result
将与enum.branches
中的分支数顺序相同。
可以将计算封装到函数中:
enum.prob.product <- function(n, p) {
enum.branches <- unlist(strsplit(intToBin(seq_len(2^n)-1),split=""))
exp(rep(1,n) %*% matrix(ifelse(enum.branches == "1", rep(log(p), 2^n), rep(log(1-p), 2^n)), nrow=n))
}
使用19
独立伯努利变量计时:
n <- 19
p <- runif(n)
system.time(enum.prob.product(n,p))
## user system elapsed
## 24.064 1.470 26.082
这是在我的2 GHz MacBook(大约2009年)上。应该注意的是,计算本身是相当快的;它是概率树分支的枚举(我猜unlist
在其中),这是花费大量时间的。如果有任何来自社区的其他方法的建议,我们将不胜感激。
在base R中试试:
p <- c(0.5, 0.12, 0.7, 0.8, .02)
a <- c("A","B","C","D","E")
n <- length(p)
apply(expand.grid(replicate(n,list(0:1)))[n:1], 1,
function(x) prod(p[which(x==1)])*prod(1-p[which(x==0)]))
#[1] 0.025872 0.000528 0.103488 0.002112 0.060368 0.001232 0.241472 0.004928 0.003528 0.000072 0.014112 0.000288 0.008232 0.000168 0.032928 0.000672 0.025872
#[18] 0.000528 0.103488 0.002112 0.060368 0.001232 0.241472 0.004928 0.003528 0.000072 0.014112 0.000288 0.008232 0.000168 0.032928 0.000672