根据另一列创建具有分组值的列

Create column with grouped values based on another column

我确定以前有人问过这个问题,但我不知道要搜索什么,所以我提前道歉。

假设我有以下数据框:

grades <- data.frame(a = 1:40, b = sample(45:100, 40))

我想使用 deplyr 创建一个新变量,根据以下标准指示学生获得的成绩:90-100 = 优秀,80-90 = 非常好,等等。

我想我可以使用以下方法在 mutate() 内部嵌套 ifelse() 来获得该结果:

grades %>%
mutate(ifelse(b >= 90, "excellent"), 
       ifelse(b >= 80 & b < 90, "very_good"),
       ifelse(b >= 70 & b < 80, "fair"),
       ifelse(b >= 60 & b < 70, "poor", "fail"))

这不起作用,因为我收到错误消息 "argument no is missing, with no default")。我认为 "no" 最后会是 "fail",但显然我的语法有误。

如果我先单独过滤原始数据,然后调用ifelse,我就可以得到这个,如下:

a <- grades %>%
     filter( b >= 90) %>%
     mutate(final = ifelse(b >= 90, "excellent"))

和 rbind a、b、c 等。显然,这不是我想要的,但我想了解 ifelse() 的语法。我猜后者是有效的,因为没有任何值不符合标准,但我仍然无法弄清楚当有多个 ifelse 时如何让它工作。

所有 ifelse 都需要在彼此之内。试试这个:

mutate(ifelse(b >= 90, "excellent", 
       ifelse(b >= 80 & b < 90, "very_good",
       ifelse(b >= 70 & b < 80, "fair",
       ifelse(b >= 60 & b < 70, "poor", "fail")))))

用级别和标签定义向量,然后在 b 列上使用 cut

levels <- c(-Inf, 60, 70, 80, 90, Inf)
labels <- c("Fail", "Poor", "fair", "very good", "excellent")
grades %>% mutate(x = cut(b, levels, labels = labels))
    a   b         x
1   1  66      Poor
2   2  78      fair
3   3  97 excellent
4   4  46      Fail
5   5  89 very good
6   6  57      Fail
7   7  80      fair
8   8  98 excellent
9   9 100 excellent
10 10  93 excellent
11 11  59      Fail
12 12  51      Fail
13 13  69      Poor
14 14  75      fair
15 15  72      fair
16 16  48      Fail
17 17  74      fair
18 18  54      Fail
19 19  62      Poor
20 20  64      Poor
21 21  88 very good
22 22  70      Poor
23 23  85 very good
24 24  58      Fail
25 25  95 excellent
26 26  56      Fail
27 27  65      Poor
28 28  68      Poor
29 29  91 excellent
30 30  76      fair
31 31  82 very good
32 32  55      Fail
33 33  96 excellent
34 34  83 very good
35 35  61      Poor
36 36  60      Fail
37 37  77      fair
38 38  47      Fail
39 39  73      fair
40 40  71      fair

或使用data.table:

library(data.table)
setDT(grades)[, x := cut(b, levels, labels)]

或者只是在基数 R 中:

grades$x <- cut(grades$b, levels, labels)

备注

再次仔细查看您的初始方法后,我注意到您需要在 cut 调用中包含 right = FALSE,因为例如,90 分应该是 "excellent" ,不仅仅是 "very good"。所以它用来定义区间应该在哪里关闭(左或右),默认在右边,这与OP最初的做法略有不同。所以在 dplyr 中,它将是:

grades %>% mutate(x = cut(b, levels, labels, right = FALSE))

相应地在其他选项中。

grades$c = grades$b # creating a new column 
#and filling in the grades
grades$c[grades$c >= 90] = "exellent"
grades$c[grades$c <= 90 &  grades$c >= 80] = "very good"
grades$c[grades$c <= 80 &  grades$c >= 70] = "fair"
grades$c[grades$c <= 70 &  grades$c >= 60] = "poor"
grades$c[grades$c <= 60] = "fail"