计算列中每个值的经验分位数

Question

我想为 data.frame 的多个列中的每个值计算经验分位数。

df <- structure(list(id = 1:20, number_1 = c(32, 220, -176, 113, 4, 
317, -55, 41, -97, 43, -8, -129, 125, 30, 44, 88, -42, 7, 73, 
-5), number_2 = c(15, -105, 77, 41, -183, 120, -216, -46, 40, 
-9, -46, 318, -92, 223, -66, 88, -45, -48, -1, -208), number_3 = c(-16, 
223, 34, 46, -62, 18, -158, 68, 124, 130, 54, 89, 89, -93, 46, 
-126, 21, -20, 118, 20)), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -20L))

# head(df)
# A tibble: 6 x 4
     id number_1 number_2 number_3
  <int>    <dbl>    <dbl>    <dbl>
1     1       32       15      -16
2     2      220     -105      223
3     3     -176       77       34
4     4      113       41       46
5     5        4     -183      -62
6     6      317      120       18

我在 SO 上找到了，但是函数 ecdf() 似乎不适用于计算 mutate().

中的列

我使用这种方法让它工作：

# generate a temporary df
quantiles_temp <- map(df, ~quantile(.x, probs = seq(0, 1, 0.2)))

# Use quantiles_temp list to check each value in df with a case_when
df_quantiles <- df %>%
  mutate(across(number_1:number_3, 
                ~case_when(
                  .x <= quantiles_temp[[cur_column()]]["20%"] ~ "0% - 20%",
                  .x >= quantiles_temp[[cur_column()]]["20%"] & .x <= quantiles_temp[[cur_column()]]["40%"] ~ "20% - 40%",
                  .x >= quantiles_temp[[cur_column()]]["40%"] & .x <= quantiles_temp[[cur_column()]]["60%"] ~ "40% - 60%",
                  .x >= quantiles_temp[[cur_column()]]["60%"] & .x <= quantiles_temp[[cur_column()]]["80%"] ~ "60% - 80%",
                  .x >= quantiles_temp[[cur_column()]]["80%"] ~ "80% - 100%"))
  )

> head(df_quantiles)
# A tibble: 6 x 4
     id number_1   number_2   number_3  
  <int> <chr>      <chr>      <chr>     
1     1 40% - 60%  60% - 80%  20% - 40% 
2     2 80% - 100% 0% - 20%   80% - 100%
3     3 0% - 20%   60% - 80%  40% - 60% 
4     4 80% - 100% 60% - 80%  40% - 60% 
5     5 20% - 40%  0% - 20%   0% - 20%  
6     6 80% - 100% 80% - 100% 20% - 40%

有没有更好的方法？

我的方法要求我生成一个临时列表 quantiles_temp 以查找每列中每个值的经验分位数。当我试图在可读性和性能方面优化我的代码时，我想知道是否有更好的方法来计算每列的分位数？

Answer 1

试试这个：

library(dplyr)
df %>%
  mutate(
    across(number_1:number_3,
           ~ cut(., quantile(., seq(0, 1, by = 0.2)),
                 labels = c("0-20%", "20-40%", "40-60%", "60-80%", "80-100%"),
                 include.lowest = TRUE))
  )
# # A tibble: 20 x 4
#       id number_1 number_2 number_3
#    <int> <fct>    <fct>    <fct>   
#  1     1 40-60%   60-80%   20-40%  
#  2     2 80-100%  0-20%    80-100% 
#  3     3 0-20%    60-80%   40-60%  
#  4     4 80-100%  60-80%   40-60%  
#  5     5 20-40%   0-20%    0-20%   
#  6     6 80-100%  80-100%  20-40%  
#  7     7 0-20%    0-20%    0-20%   
#  8     8 40-60%   20-40%   60-80%  
#  9     9 0-20%    60-80%   80-100% 
# 10    10 60-80%   40-60%   80-100% 
# 11    11 20-40%   20-40%   60-80%  
# 12    12 0-20%    80-100%  60-80%  
# 13    13 80-100%  20-40%   60-80%  
# 14    14 40-60%   80-100%  0-20%   
# 15    15 60-80%   20-40%   40-60%  
# 16    16 60-80%   80-100%  0-20%   
# 17    17 20-40%   40-60%   40-60%  
# 18    18 40-60%   20-40%   20-40%  
# 19    19 60-80%   40-60%   80-100% 
# 20    20 20-40%   0-20%    20-40%

这会产生 factors；如果您选择，如果您更喜欢字符串，请将 cut(.) 包装在 as.character 中（尽管我建议因子的序数性质为 sorting/ranking 之类的东西提供了良好的基础）。

计算列中每个值的经验分位数

Calculate empirical quantile for each value in a column

r

dplyr

quantile