4

我在 R 中有以下数据框：

  c1 c2  
1 10  a  
2 20  a  
3 30  b  
4 40  b

我然后split它如下z = lapply(split(test$c1, test$c2), function(x) {cut(x,2)}) ： z然后是：

$a  
[1] (9.99,15] (15,20]  
Levels: (9.99,15] (15,20]

$b  
[1] (30,35] (35,40]
Levels: (30,35] (35,40]

然后我想通过取消拆分列表来合并这些因素unsplit(z, test$c2)。这会产生一个警告：

[1] (9.99,15] (15,20]   <NA>      <NA>     
Levels: (9.99,15] (15,20]
Warning message:
In `[<-.factor`(`*tmp*`, i, value = 1:2) :
  invalid factor level, NAs generated

我想将所有因子级别合并，然后取消拆分，以免发生此错误：

z$a = factor(z$a, levels=c(levels(z$a), levels(z$b)))
unsplit(z, test$c2)
[1] (9.99,15] (15,20]   (30,35]   (35,40]  
Levels: (9.99,15] (15,20] (30,35] (35,40]

在我的真实数据框中，我有一个非常大的列表，所以我需要遍历所有列表元素（不仅仅是两个）。做这个的最好方式是什么？

4

2 回答 2

4

你不能只是unlist() z代替吗？

> unlist(z)
       a1        a2        b1        b2 
(9.99,15]   (15,20]   (30,35]   (35,40] 
Levels: (9.99,15] (15,20] (30,35] (35,40]

或没有结果因子的名称：

> unlist(z, use.names=FALSE)
[1] (9.99,15] (15,20]   (30,35]   (35,40]  
Levels: (9.99,15] (15,20] (30,35] (35,40]

您可以将所有内容合并到一个不需要附加包的简单单行中：

> (test2 <- within(test, newvar <- unlist(lapply(split(c1, c2), cut, 2))))
  c1 c2    newvar
1 10  a (9.99,15]
2 20  a   (15,20]
3 30  b   (30,35]
4 40  b   (35,40]
于 2011-04-29T18:50:58.847 回答
4

如果我正确理解了您的问题，我认为您使这比需要的要复杂一些。这是使用plyr. 我们将按c2变量分组：

require(plyr)
ddply(test, "c2", transform, newvar = cut(c1, 2))

返回：

  c1 c2    newvar
1 10  a (9.99,15]
2 20  a   (15,20]
3 30  b   (30,35]
4 40  b   (35,40]

并具有以下结构：

'data.frame':   4 obs. of  3 variables:
 $ c1    : num  10 20 30 40
 $ c2    : Factor w/ 2 levels "a","b": 1 1 2 2
 $ newvar: Factor w/ 4 levels "(9.99,15]","(15,20]",..: 1 2 3 4
于 2011-04-29T18:53:12.643 回答