找到R中某个百分位数对应的x值

Question

我有一个密度分布。x列表示x轴上的值，而y列表示相应的密度。

我想找到第2.5个和97.5个百分位数，以及它们对应的x值。结束数据帧应该看起来像cri，并填充了x列。

library(tidyverse)
x = seq(-10,10, 0.1)
y = dnorm(x, mean = 0, sd = 2)
df = tibble(x,y) 
cri <- df %>%
summarise(lwr = quantile(y, probs = 0.025),
upr = quantile(y, probs = 0.975),
mean = mean(y)) %>%
pivot_longer(cols = everything()) %>% 
mutate(x = NA)
cri
#> # A tibble: 3 x 3
#>   name       value x    
#>   <chr>      <dbl> <lgl>
#> 1 lwr   0.00000122 NA   
#> 2 upr   0.197      NA   
#> 3 mean  0.0498     NA

^{创建于2022-09-13由reprex包(v2.0.1(}

Answer 1

假设存在完全匹配，则可以使用match:

cri <- df %>%
summarise(lwr = quantile(y, probs = 0.025),
upr = quantile(y, probs = 0.975),
mean = mean(y)) %>%
pivot_longer(cols = everything()) %>% 
mutate(exact_x = df$x[match(value, df$y)]) %>%
rowwise %>% 
mutate(closest_x = df$x[which.min(abs(value - df$y))]) %>%
ungroup()
cri
# # A tibble: 3 × 4
#   name       value exact_x closest_x
#   <chr>      <dbl>   <dbl>     <dbl>
# 1 lwr   0.00000122  -9.8      -9.8  
# 2 upr   0.197       -0.300    -0.300
# 3 mean  0.0498      NA         3.3  

Answer 2

另一种查找值的方法是使用left_join。

library(dplyr)
library(tidyr)
x = seq(-10,10, 0.1)
y = dnorm(x, mean = 0, sd = 2)
df = tibble(x,y) 
cri <- df %>%
summarise(lwr = quantile(y, probs = 0.025),
upr = quantile(y, probs = 0.975),
mean = mean(y)) %>%
pivot_longer(cols = everything()) %>% 
left_join(., df, by = c("value"="y"))
# A tibble: 4 x 3
name       value      x
<chr>      <dbl>  <dbl>
1 lwr   0.00000122 -9.8  
2 lwr   0.00000122  9.8  
3 upr   0.197      -0.300
4 mean  0.0498     NA