exppois(VGAM)
exppois()所属R语言包:VGAM
The Exponential Poisson Distribution
指数的泊松分布
译者:生物统计家园网 机器人LoveR
描述----------Description----------
Density, distribution function, quantile function and random generation for the exponential poisson distribution.
密度,分布函数,指数的泊松分布的分位数函数随机生成。
用法----------Usage----------
dexppois(x, lambda, betave = 1, log = FALSE)
pexppois(q, lambda, betave = 1)
qexppois(p, lambda, betave = 1)
rexppois(n, lambda, betave = 1)
参数----------Arguments----------
参数:x, q
vector of quantiles.
向量的位数。
参数:p
vector of probabilities.
向量的概率。
参数:n
number of observations. If length(n) > 1 then the length is taken to be the number required.
若干意见。如果length(n) > 1的长度是所需的数量。
参数:lambda, betave
both positive parameters.
正面的参数。
参数:log
Logical. If log = TRUE then the logarithm of the density is returned.
逻辑。如果log = TRUE然后返回的密度的对数。
Details
详细信息----------Details----------
See exppoisson, the VGAM family function for estimating the parameters, for the formula of the probability density function and other details.
exppoisson,的VGAM的家庭功能的参数估计,公式的概率密度函数和其他细节。
值----------Value----------
dexppois gives the density, pexppois gives the distribution function, qexppois gives the quantile function, and rexppois generates random deviates.
dexppois给出了密度,pexppois给出了分布函数,qexppois给出了分位数的功能,和rexppois随机产生的偏离。
(作者)----------Author(s)----------
J. G. Lauder, jamesglauder@gmail.com
参见----------See Also----------
exppoisson.
exppoisson。
实例----------Examples----------
## Not run: [#不运行:]
lambda = 2; betave = 2; nn = 201
x = seq(-0.05, 1.05, len = nn)
plot(x, dexppois(x, lambda, betave), type = "l", las = 1, ylim = c(0, 5),
ylab = paste("[dp]exppoisson(lambda = ", lambda, ", betave = ", betave, ")"),
col = "blue", cex.main = 0.8,
main = "Blue is density, orange is cumulative distribution function",
sub = "Purple lines are the 10,20,...,90 percentiles")
lines(x, pexppois(x, lambda, betave), col = "orange")
probs = seq(0.1, 0.9, by = 0.1)
Q = qexppois(probs, lambda, betave)
lines(Q, dexppois(Q, lambda, betave), col = "purple", lty = 3, type = "h")
lines(Q, pexppois(Q, lambda, betave), col = "purple", lty = 3, type = "h")
abline(h = probs, col = "purple", lty = 3)
max(abs(pexppois(Q, lambda, betave) - probs)) # Should be 0[应为0]
## End(Not run)[#(不执行)]
转载请注明:出自 生物统计家园网(http://www.biostatistic.net)。
注:
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