R:计算多组百分位数

Question

我正在使用R编程语言。

我有以下数据集:

set.seed(123)
library(dplyr)
var1 = rnorm(10000, 100,100)
var2 = rnorm(10000, 100,100)
var3 = rnorm(10000, 100,100)
var4 = rnorm(10000, 100,100)
id = 1:10000
final = data.frame(id, var1, var2, var3, var4)
final = final %>%
mutate(class1 = case_when(var1 < mean(var1) ~ "A",
TRUE ~ "B")) %>% 
mutate(class2 = case_when(var2 < mean(var2) ~ "C",
TRUE ~ "D"))

我想根据class1和class2的每一个唯一组合计算var3和var4的十分位数。

据我理解，这意味着:

• 对于所有WHERE class1 = A AND class2 = C的行，计算/分配var3和var4的十分位数
• 对于所有class1 = A AND class2 = D的行，计算/分配var3和var4的十分位数
• 对于class1 = B AND class2 = C的所有行，计算/分配var3和var4的十分位数
• 对于所有class1 = B AND class2 = D的行，计算/分配var3和var4的十分位数

这是我为此写的R代码:

final = final %>%
group_by(class1, class2) %>%
mutate(class3 = case_when(ntile(var3, 10) == 1 ~ "one",
ntile(var3, 10) == 2 ~ "two",
ntile(var3, 10) == 3 ~ "three",
ntile(var3, 10) == 4 ~ "four",
ntile(var3, 10) == 5 ~ "five",
ntile(var3, 10) == 6 ~ "six",
ntile(var3, 10) == 7 ~ "seven",
ntile(var3, 10) == 8 ~ "eight",
ntile(var3, 10) == 9 ~ "nine",
ntile(var3, 10) == 10 ~ "ten")) %>%
mutate(class4 = case_when(ntile(var4, 10) == 1 ~ "one",
ntile(var4, 10) == 2 ~ "two",
ntile(var4, 10) == 3 ~ "three",
ntile(var4, 10) == 4 ~ "four",
ntile(var4, 10) == 5 ~ "five",
ntile(var4, 10) == 6 ~ "six",
ntile(var4, 10) == 7 ~ "seven",
ntile(var4, 10) == 8 ~ "eight",
ntile(var4, 10) == 9 ~ "nine",
ntile(var4, 10) == 10 ~ "ten"))

有人能告诉我，如果我做得正确吗?

谢谢!

Answer 1

用english代替case_when可以很容易地完成

library(dplyr)
library(stringr)
final %>%
group_by(class1, class2) %>% 
mutate(across(var3:var4, 
~ as.character(english::english(ntile(.x, 10))),
.names = "{str_replace(.col, 'var', 'class')}")) %>% 
ungroup

与产出

# A tibble: 10,000 × 9
id  var1  var2    var3  var4 class1 class2 class3 class4
<int> <dbl> <dbl>   <dbl> <dbl> <chr>  <chr>  <chr>  <chr> 
1     1  44.0 337.    16.4   80.6 A      D      three  five  
2     2  77.0  83.3   77.9  126.  A      C      five   six   
3     3 256.  193.  -110.    46.2 B      D      one    four  
4     4 107.   43.2  -66.8  -17.9 B      C      one    two   
5     5 113.  123.    -9.80 190.  B      D      two    nine  
6     6 272.  213.   -66.6   98.4 B      D      one    six   
7     7 146.  238.    95.0  118.  B      D      five   six   
8     8 -26.5  76.7  256.   160.  A      C      ten    eight 
9     9  31.3 -60.1   59.5  126.  A      C      four   six   
10    10  55.4  70.2  179.   130.  A      C      eight  seven 
# … with 9,990 more rows
# ℹ Use `print(n = ...)` to see more rows