Find the corresponding x value for a percentile in R?-CodePudding

I have a density distribution. The x column represents the values on the x axis, while the y column represents the corresponding density.

I would like to find the 2.5th and 97.5 percentiles and their corresponding x values. The end dataframe should look like cri with the x column filled in.

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

^{Created on 2022-09-13 by the reprex package (v2.0.1)}

CodePudding user response：

Assuming there is an exact match, you can use 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

CodePudding user response：

Another way to find values is using 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