MASS 包的 "fitdistr"：处理被操纵的随机数据时出错
MASS packages' "fitdistr": Error when dealing with manipulated random data
背景：
下面我使用 R 生成了一些随机的 beta 数据并稍微操纵数据的形状以达到我所说的"Final" 在我的代码中。我在我的代码中直方图 "Final"。
问题：
我想知道为什么在尝试使用 MASS 包“ 将 "beta" 分布拟合到 ”Final“ 数据时fitdistr" 函数，我收到以下错误（关于如何避免此错误的任何建议）？
Error in stats::optim(x = c(0.461379379270288, 0.0694261016478062, 0.76934266883081, :
initial value in 'vmmin' is not finite
这是我的 R 代码：
require(MASS)
## Generate some data and manipulate it
set.seed(47)
Initial = rbeta(1e5, 2, 3)
d <- density(Initial)
b.5 <- dbeta(seq(0, 1, length.out = length(d$y)), 50, 50)
b.5 <- b.5 / (max(b.5) / max(d$y)) # Scale down to max of original density
b.6 <- dbeta(seq(0, 1, length.out = length(d$y)), 60, 40)
b.6 <- b.6 / (max(b.6) / max(d$y))
# Collect maximum densities at each x to use as sample probability weights
p <- pmax(d$y, b.5, b.6)
Final <- sample(d$x, 1e4, replace = TRUE, prob = p) ## THIS IS MY FINAL DATA
hist(Final, freq = F, ylim = c(0, 2)) ## HERE IS A HISTOGRAM
m <- MASS::fitdistr(Final, "beta", ## RUN THIS TO SEE HOW THE ERROR COMES UP
start = list(shape1 = 1, shape2 = 1))
这是代码。
和你的代码一样，我只是去掉了负的 beta 值。
library(MASS)
set.seed(47)
Initial = rbeta(1e5, 2, 3)
d <- density(Initial)
b.5 <- dbeta(seq(0, 1, length.out = length(d$y)), 50, 50)
b.5 <- b.5 / (max(b.5) / max(d$y)) # Scale down to max of original
density
b.6 <- dbeta(seq(0, 1, length.out = length(d$y)), 60, 40)
b.6 <- b.6 / (max(b.6) / max(d$y))
# Collect maximum densities at each x to use as sample probability weights
p <- pmax(d$y, b.5, b.6)
Final <- sample(d$x, 1e4, replace = TRUE, prob = p) ## THIS IS MY FINAL DATA
hist(Final, freq = F, ylim = c(0, 2)) ## HERE IS A HISTOGRAM
# replace negative beta values with smallest value > 0
Final[Final<= 0] <- min(Final[Final>0])
hist(Final, freq = F, ylim = c(0, 2))
m <- MASS::fitdistr(x = Final, densfun = "beta",
start = list(shape1 = 1, shape2 = 1))
形状参数如下：
> m
shape1 shape2
1.99240852 2.90219720
(0.02649853) (0.04010168)
注意它给出了一些警告。
背景：
下面我使用 R 生成了一些随机的 beta 数据并稍微操纵数据的形状以达到我所说的"Final" 在我的代码中。我在我的代码中直方图 "Final"。
问题：
我想知道为什么在尝试使用 MASS 包“ 将 "beta" 分布拟合到 ”Final“ 数据时fitdistr" 函数，我收到以下错误（关于如何避免此错误的任何建议）？
Error in stats::optim(x = c(0.461379379270288, 0.0694261016478062, 0.76934266883081, :
initial value in 'vmmin' is not finite
这是我的 R 代码：
require(MASS)
## Generate some data and manipulate it
set.seed(47)
Initial = rbeta(1e5, 2, 3)
d <- density(Initial)
b.5 <- dbeta(seq(0, 1, length.out = length(d$y)), 50, 50)
b.5 <- b.5 / (max(b.5) / max(d$y)) # Scale down to max of original density
b.6 <- dbeta(seq(0, 1, length.out = length(d$y)), 60, 40)
b.6 <- b.6 / (max(b.6) / max(d$y))
# Collect maximum densities at each x to use as sample probability weights
p <- pmax(d$y, b.5, b.6)
Final <- sample(d$x, 1e4, replace = TRUE, prob = p) ## THIS IS MY FINAL DATA
hist(Final, freq = F, ylim = c(0, 2)) ## HERE IS A HISTOGRAM
m <- MASS::fitdistr(Final, "beta", ## RUN THIS TO SEE HOW THE ERROR COMES UP
start = list(shape1 = 1, shape2 = 1))
这是代码。
和你的代码一样，我只是去掉了负的 beta 值。
library(MASS)
set.seed(47)
Initial = rbeta(1e5, 2, 3)
d <- density(Initial)
b.5 <- dbeta(seq(0, 1, length.out = length(d$y)), 50, 50)
b.5 <- b.5 / (max(b.5) / max(d$y)) # Scale down to max of original
density
b.6 <- dbeta(seq(0, 1, length.out = length(d$y)), 60, 40)
b.6 <- b.6 / (max(b.6) / max(d$y))
# Collect maximum densities at each x to use as sample probability weights
p <- pmax(d$y, b.5, b.6)
Final <- sample(d$x, 1e4, replace = TRUE, prob = p) ## THIS IS MY FINAL DATA
hist(Final, freq = F, ylim = c(0, 2)) ## HERE IS A HISTOGRAM
# replace negative beta values with smallest value > 0
Final[Final<= 0] <- min(Final[Final>0])
hist(Final, freq = F, ylim = c(0, 2))
m <- MASS::fitdistr(x = Final, densfun = "beta",
start = list(shape1 = 1, shape2 = 1))
形状参数如下：
> m
shape1 shape2
1.99240852 2.90219720
(0.02649853) (0.04010168)
注意它给出了一些警告。