如何将fitdistrplus :: fitdist摘要转换为整洁的格式? [英] How to convert fitdistrplus::fitdist summary into tidy format?
问题描述
我有以下代码:
x <- c(
0.367141764080875, 0.250037975705769, 0.167204185003365, 0.299794433447383,
0.366885973041269, 0.300453205296379, 0.333686861081341, 0.33301168850398,
0.400142004893329, 0.399433677388411, 0.366077304765104, 0.166402979455671,
0.466624230750293, 0.433499934139897, 0.300017278751768, 0.333673696762895,
0.29973685692478
)
fn <- fitdistrplus::fitdist(x,"norm")
summary(fn)
#> Fitting of the distribution ' norm ' by maximum likelihood
#> Parameters :
#> estimate Std. Error
#> mean 0.32846024 0.01918923
#> sd 0.07911922 0.01355908
#> Loglikelihood: 19.00364 AIC: -34.00727 BIC: -32.34084
#> Correlation matrix:
#> mean sd
#> mean 1 0
#> sd 0 1
基本上,它需要一个向量并尝试使用< a href = https://cran.r-project.org/web/packages/fitdistrplus/index.html rel = noreferrer> fitdistrplus包。
Basically, it takes a vector and tried to fit the distribution using fitdistrplus package.
我尝试查看扫帚包,但是没有。
I tried looking at the broom package, but it doesn't have a function that covers that.
推荐答案
调用 broom :: tidy( fn)
,您会收到一条错误消息:
When you call broom::tidy(fn)
you receive an error that says:
错误:类fitdist的对象没有整洁的方法
Error: No tidy method for objects of class fitdist
这是因为 broom
中的此函数仅包含有限个对象,这些对象是好使用,请参见方法(整洁)
以获取完整列表。 (了解更多有关R中S3方法的信息。更多此处)。
This is because this function from broom
only has a finite number objects that are "good to use", see methods(tidy)
for the complete list. (Read more about S3 methods in R. More here).
因此该函数不会适用于对象 fitdist
,但适用于 MASS $ c $中的
fitdistr
对象c>(更著名)。
So the function doesn't work for an object fitdist
but works for a fitdistr
object from MASS
(more "famous").
然后我们可以将分配给
,然后使用 fn
类扫帚
:
class(fn) <- ("fitdist", "fitdistr")
# notice that I've kept the original class and added the other
# you shouldn't overwrite classes. ie: don't to this: class(fn) <- "fitdistr"
broom::tidy(fn)
# # A tibble: 2 x 3
# term estimate std.error
# <chr> <dbl> <dbl>
# 1 mean 0.328 0.0192
# 2 sd 0.0791 0.0136
请注意只能看到参数
。如果您希望看到更多内容并将所有内容整理为整洁,则应向我们详细说明您的预期输出。
Note that you can only see the parameters
. If you wish to see more and organize everything as "tidy", you should tell us more about your expected output.
broom :: tidy( )
让您走得更远,如果您想要更多,我先定义自己的方法函数,该函数可用于类
fitdist
对象用作引用 tidy.fitdistr
方法,并对其进行调整。
broom::tidy()
gets you this far, if you want more I'd start by defining my own method function that works for a class
fitdist
object using as reference the tidy.fitdistr
method, and adapting it.
如何使用S3方法为类 fitdist <修改原始
broom :: tidy()
代码的示例/ code>。
Example of how I'd adapt from the original broom::tidy()
code, using the S3 method for the class fitdist
.
定义自己的方法(类似于定义自己的函数的方式):
Define your own method (similar to how you define your own function):
# necessary libraries
library(dplyr)
library(broom)
# method definition:
tidy.fitdist <- function(x, ...) { # notice the use of .fitdist
# you decide what you want to keep from summary(fn)
# use fn$ecc... to see what you can harvest
e1 <- tibble(
term = names(x$estimate),
estimate = unname(x$estimate),
std.error = unname(x$sd)
)
e2 <- tibble(
term = c("loglik", "aic", "bic"),
value = c(unname(x$loglik), unname(x$aic), unname(x$bic))
)
e3 <- x$cor # I prefer this to: as_tibble(x$cor)
list(e1, e2, e3) # you can name each element for a nicer result
# example: list(params = e1, scores = e2, corrMatr = e3)
}
现在,您可以通过这种方式调用此新的方法
:
This is how you can call this new method
now:
tidy(fn) # to be more clear this is calling your tidy.fitdist(fn) under the hood.
# [[1]]
# # A tibble: 2 x 3
# term estimate std.error
# <chr> <dbl> <dbl>
# 1 mean 0.328 0.0192
# 2 sd 0.0791 0.0136
#
# [[2]]
# # A tibble: 3 x 2
# term value
# <chr> <dbl>
# 1 loglik 19.0
# 2 aic -34.0
# 3 bic -32.3
#
# [[3]]
# mean sd
# mean 1 0
# sd 0 1
注意 class
是:
class(fn)
[1] "fitdist"
所以现在您实际上不需要分配 fitdistr
(来自 MASS
)类。
So now you don't actually need to assign the fitdistr
(from MASS
) class as before.
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