双循环在许多列中迭代以查找R中的离群值

inkz8wg9 于 2023-04-03 发布在其他

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我有一个数据框，其中包含一个人的“id”和两个特征（“x”e“y”），如下所示：

id = c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","A12","A13","A14","A15","A16","A17","A18","A19","A20","A21","A22","A23","A24")
x = c(10,4,6,8,9,8,7,6,12,14,11,9,8,4,5,10,14,12,15,7,10,14,24,28)
y = c(1.5,1.2,5,2,0.8,4,1,1.1,1.2,1.4,1.3,1.6,0.9,0.8,1,1.1,1.3,1.5,1.2,1.1,1,1.2,1.1,1)
a = data.frame(id,x,y)

我希望有一个循环来迭代每个特质和每个个体，这样我就可以创建一个新的数据框（或a的新列），其中如果个体是离群值，则为1，如果不是，则为0。将离群值视为偏离特质均值± 3 sd的任何点。
在本例中，“x”的离群值为28，“y”的离群值为5。所需的结果可能如下所示：

id = c("A1","A2","A3","A4","A5","A6","A7","A8","A9","A10","A11","A12","A13","A14","A15","A16","A17","A18","A19","A20","A21","A22","A23","A24")
x_out = c(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1)
y_out = c(0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)
a_out = data.frame(id, x_out, y_out)

你知道如何在循环中实现吗？我的想法是，如果我包含新的特征或个体，我不需要改变循环。谢谢！

r

来源：https://stackoverflow.com/questions/75881375/double-loop-to-iterate-in-many-columns-to-find-outliers-in-r

1条答案

按热度按时间

rsl1atfo1#

不需要循环，你可以一次测试所有列的绝对z得分（abs(scale())）是否为>= 3：

a_out <- a
a_out[, -1] <- as.integer(abs(scale(a[, -1])) >= 3)

或者使用dplyr：

library(dplyr)

a_out <- a %>% 
  mutate(across(!id, \(x) as.integer(abs(scale(x)) >= 3)))
# same output as above

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双循环在许多列中迭代以查找R中的离群值

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