如何获得一辆汽车的 mpg 比其他汽车最低的概率？

Question

我正在使用它来获取汽车 mpg 的均值和标准差

df1 <- mtcars; df1$rownames = rownames(df1)
df2 <- mtcars; df2$rownames = rownames(df2)
df2$mpg = df2$mpg + rnorm(nrow(df2),0,3)
data = rbind(df1, df2)

我使用一个函数来获取一辆汽车的 mpg 低于其他汽车的概率

df = plyr::ddply(data,~rownames,summarise,mean=mean(mpg),sd=sd(mpg))

f <- function(x, y){
  n1 = df$mean[x]; n2 = df$mean[y]; sd1 = df$sd[x]; sd2 = df$sd[y]
  pnorm(0, mean = n1 - n2, sd = sqrt(sd1^2 + sd2^2))
}

res <- outer(X = 1:nrow(df), Y = 1:nrow(df), f)
dimnames(res) <- list(df$rownames, df$rownames)
res <- data.frame(res)
res <- tibble::rownames_to_column(res, 'p1')

datalong_2 <- tidyr::gather(res, 'p2', 'value', -1) # output

现在我想要一辆汽车的 mpg 比其他汽车最低的概率。我绑这个：

cars = unique(datalong_2$p1)
win <- data.frame(sapply(1:length(cars), function(x) setNames(prod(subset(datalong_2, p1 == cars[x] & p2 != cars[x])$value),cars[x])))

colnames(win) <- "prob"
win$prob <- round(win$prob,4)

但是概率加起来不为一。我如何更改此代码以获得 table 每辆汽车的 mpg 比其他汽车最低的概率？

Answer 1

这是 dplyr/data.table 方法与 return 概率

的比较

library(dplyr)
library(data.table)
library(tidyr)
library(tibble)
# // input data
df <- mtcars[1] %>% 
         rownames_to_column("car")

-测试

# // dplyr
system.time({
        out <- df %>%
                uncount(10000, .id = "run") %>%
                rowwise() %>%
                mutate(sim_mpg = rpois(1, lambda = mpg)) %>%

                group_by(run) %>%
                arrange(sim_mpg) %>%
                mutate(lowest_mpg = row_number() == 1) %>%

                group_by(car) %>%
                summarize(chance_lowest = mean(lowest_mpg),
                orig_mpg = first(mpg))

})
# user system elapsed
# 1.715 0.074 1.787


# // data.table
system.time({

     df_expand <- setDT(df)[rep(seq_len(.N), 10000)][, run := rowid(car)]

     out2 <- df_expand[, sim_mpg := rpois(1, lambda = mpg), 1:nrow(df_expand)
        ][order(sim_mpg), lowest_mpg := seq_len(.N) == 1 ,run
        ][, .(chance_lowest = mean(lowest_mpg), orig_mpg = first(mpg)), .(car)]
})
# user system elapsed
# 0.704 0.050 0.757
sum(out$chance_lowest)
#[1] 1

如何获得一辆汽车的 mpg 比其他汽车最低的概率？

How to get the probabilities of one car having the lowest mpg than the rest of the cars?

r

normal-distribution

probability