我对R非常陌生,我有一项工作,我正在努力,我遇到了很多麻烦。我已经定义了离散概率分布:
s P(s)
0 1/9
1 4/9
2 1/9
3 0/9
4 1/9
5 0/9
6 0/9
7 1/9
8 0/9
9 1/9
现在我必须处理这个问题:
与R中可用的其他发行版一致,创建一个 概率分布的支持函数族:
f = dsidp(d) # pmf - the height of the curve/bar for digit d
p = psidp(d) # cdf - the probability of a value being d or less
d = qsidp(p) # icdf - the digit corresponding to the given
# cumulative probability p
d[] = rsidp(n) # generate n random digits based on your probability distribution.
如果有人可以帮助我开始编写这些功能,我们将不胜感激!
答案 0 :(得分:1)
首先,阅读数据:
dat <- read.table(text = "s P(s)
0 1/9
1 4/9
2 1/9
3 0/9
4 1/9
5 0/9
6 0/9
7 1/9
8 0/9
9 1/9", header = TRUE, stringsAsFactors = FALSE)
names(dat) <- c("s", "P")
将分数(表示为字符串)转换为数值:
dat$P <- sapply(strsplit(dat$P, "/"), function(x) as.numeric(x[1]) / as.numeric(x[2]))
功能:
# pmf - the height of the curve/bar for digit d
dsidp <- function(d) {
with(dat, P[s == d])
}
# cdf - the probability of a value being d or less
psidp <- function(d) {
with(dat, cumsum(P)[s == d])
}
# icdf - the digit corresponding to the given cumulative probability p
qsidp <- function(p) {
with(dat, s[sapply(cumsum(P), all.equal, p) == "TRUE"][1])
}
请注意。由于某些概率为零,因此某些数字具有相同的累积概率。在这些情况下,函数qsidp
返回最低位。
# generate n random digits based on your probability distribution.
rsidp <- function(n) {
with(dat, sample(s, n, TRUE, P))
}