如何根据根据条件参数计算的特定分位数创建具有值的列？

Question

一切都在标题中。为了说明，我构建了以下示例。

我有以下数据框：

date <- c("01.02.2011","01.02.2011","01.02.2011","01.02.2011","01.02.2011","01.02.2011",
          "01.02.2011","01.02.2011","01.02.2011","01.02.2011",
          "02.02.2011","02.02.2011","02.02.2011","02.02.2011","02.02.2011","02.02.2011",
          "02.02.2011","02.02.2011","02.02.2011","02.02.2011")
date <- as.Date(date, format="%d.%m.%Y")
ID <- c("A","B","C","D","E","F","G","H","I","J",
        "A","B","C","D","E","F","G","H","I","J")
values <- as.numeric(c("1","8","2","3","5","13","2","4","1","16",
            "4","2","12","16","8","1","7","11","2","10"))

df <- data.frame(ID, date, values)

看起来像：

  ID       date    values
1   A 2011-02-01      1
2   B 2011-02-01      8
3   C 2011-02-01      2
4   D 2011-02-01      3
5   E 2011-02-01      5
6   F 2011-02-01     13
7   G 2011-02-01      2
8   H 2011-02-01      4
9   I 2011-02-01      1
10  J 2011-02-01     16
11  A 2011-02-02      4
12  B 2011-02-02      2
13  C 2011-02-02     12
14  D 2011-02-02     16
15  E 2011-02-02      8
16  F 2011-02-02      1
17  G 2011-02-02      7
18  H 2011-02-02     11
19  I 2011-02-02      2
20  J 2011-02-02     10

我想创建一个新列 "QF"，它采用以下值：

1 如果 <=按日期计算的第 40 个百分位数
2 如果按日期计算的 > 第 40 个百分位和 < 第 70 个百分位
3 如果按日期计算的 > 第 70 个百分位数

我想获得：

ID       date      values   QF
1   A 2011-02-01      1     1
2   B 2011-02-01      8     3
3   C 2011-02-01      2     1
4   D 2011-02-01      3     2
5   E 2011-02-01      5     2
6   F 2011-02-01     13     3
7   G 2011-02-01      2     1
8   H 2011-02-01      4     2
9   I 2011-02-01      1     1
10  J 2011-02-01     16     3
11  A 2011-02-02      4     1
12  B 2011-02-02      2     1
13  C 2011-02-02     12     3
14  D 2011-02-02     16     3
15  E 2011-02-02      8     2
16  F 2011-02-02      1     1
17  G 2011-02-02      7     2
18  H 2011-02-02     11     3
19  I 2011-02-02      2     1
20  J 2011-02-02     10     2

如果需要对我的问题进行任何编辑，请随时告诉我

Answer 1

一个dplyr选项可以是：

df %>%
 group_by(date) %>%
 mutate(QF = cut(values, c(0, quantile(values, probs = c(0.4, 0.7, 1))),
                 labels = 1:3))

   ID    date       values QF   
   <fct> <date>      <dbl> <fct>
 1 A     2011-02-01      1 1    
 2 B     2011-02-01      8 3    
 3 C     2011-02-01      2 1    
 4 D     2011-02-01      3 2    
 5 E     2011-02-01      5 2    
 6 F     2011-02-01     13 3    
 7 G     2011-02-01      2 1    
 8 H     2011-02-01      4 2    
 9 I     2011-02-01      1 1    
10 J     2011-02-01     16 3    
11 A     2011-02-02      4 1    
12 B     2011-02-02      2 1    
13 C     2011-02-02     12 3    
14 D     2011-02-02     16 3    
15 E     2011-02-02      8 2    
16 F     2011-02-02      1 1    
17 G     2011-02-02      7 2    
18 H     2011-02-02     11 3    
19 I     2011-02-02      2 1    
20 J     2011-02-02     10 2

Answer 2

我们可以使用fndInterval

library(dplyr)
df %>%
   group_by(date) %>%
    mutate(QF = findInterval(values, c(0, quantile(values, probs = c(0.4, 0.7, 1)))))

如何根据根据条件参数计算的特定分位数创建具有值的列？

How to create a column with values based on specific quantiles calculated from a conditional argument?

r

dataframe

quantile