如何用 R 中另一个变量的分位数创建一个变量？

Question

我正在尝试使用“dplyr”命令 mutate 创建一个变量，它必须指示另一个变量的分位数。

例如：

# 1.  Fake data:
data <- data.frame(
  "id" = seq(1:20),
  "score" = round(rnorm(20,30,20)))

# 2. Creating varaible 'Quantile_5'
data <-data %>% 
  mutate(Quntile_5 = ????)

到目前为止，我已经创建了一个函数来识别 returns 分位数作为一个因素，并且它确实有效

# 3. Create a function:
quantile5 <- function(x){
  x = ifelse(
    x < quantile(x,0.2),1,
    ifelse(x >= quantile(x,0.2) & x < quantile(x,0.4),2,
           ifelse(x >= quantile(x,0.4) & x < quantile(x,0.6),3,
                  ifelse(x >= quantile(x,0.6) & x < quantile(x,0.8),4,5
                         ))))
  return(as.factor(x))
}

# 4. Running the code:
data <-data %>% 
  mutate(Quntile_5 = quantile5(score))

# 5. Result:
data

   id score Quntile_5
1   1    55         5
2   2    56         5
3   3    26         3
4   4    42         3
5   5    41         3
6   6    26         3
7   7    57         5
8   8    12         1
9   9    21         2
10 10    25         2
11 11    37         3
12 12    18         2
13 13    54         5
14 14    47         4
15 15    52         4
16 16    -4         1
17 17    53         4
18 18    51         4
19 19    -7         1
20 20    -2         1

但是，如果我想创建一个变量“Quantile_100”作为一个因子，指示每个观察值在 1 到 100 之间的哪个位置（在较大数据集的上下文中），这不是一个很好的解决方案。有没有更简单的方法来创建这些五分位数变量？

Answer 1

希望这就是您要找的：

library(dplyr)

data <- data.frame(
  "id" = seq(1:20),
  "score" = round(rnorm(20,30,20)))


data %>%
  mutate(quantile100 = findInterval(score, quantile(score, probs = seq(0, 1, 0.01)), 
                                    rightmost.closed = TRUE)) %>%
  slice_head(n = 10)

   id score quantile100
1   1    59          95
2   2    47          90
3   3    83         100
4   4    33          53
5   5     7          11
6   6    26          43
7   7    16          16
8   8    18          27
9   9    33          53
10 10    47          90

我选择关闭最右边的垃圾箱，使最大类别不超过 100。我们也可以用你自己的例子来验证，结果是一样的：


df %>%
  mutate(quantile100 = findInterval(score, quantile(score, probs = seq(0, 1, 0.2)), 
                                    rightmost.closed = TRUE)) %>%
  slice_head(n = 10)


   id score quantile5
1   1    55         5
2   2    56         5
3   3    26         3
4   4    42         3
5   5    41         3
6   6    26         3
7   7    57         5
8   8    12         1
9   9    21         2
10 10    25         2

数据

structure(list(id = 1:20, score = c(55L, 56L, 26L, 42L, 41L, 
26L, 57L, 12L, 21L, 25L, 37L, 18L, 54L, 47L, 52L, -4L, 53L, 51L, 
-7L, -2L)), class = "data.frame", row.names = c("1", "2", "3", 
"4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", 
"16", "17", "18", "19", "20"))

Answer 2

这里有两个选项 cut :

1.

library(dplyr)

data %>% mutate(quantile100 = cut(score, 100, label = FALSE))

#This is similar to @Anoushiravan R `findInterval` function.
data %>% 
    mutate(quantile100 = cut(score, unique(quantile(score, seq(0, 1, 0.01))), labels = FALSE))

如何用 R 中另一个变量的分位数创建一个变量？

How to create a variable with the quantiles of another one in R?

r

dataframe

dplyr

quantile