Question

我在R中有以下数据框：

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

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

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

我想通过取消分割列表unsplit(z, test$c2)来合并这些因素。这会产生警告：

[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

我想采用所有因子级别的联合然后解压缩，以便不会发生此错误：

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，所以我需要迭代所有列表元素（不只是两个）。这样做的最佳方式是什么？

Answer 1

你不能只是unlist() z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]

或没有结果因素上的名称：

> 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为一个不需要附加软件包的简单单行程序：

> (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]

Answer 2

如果我理解你的问题，我认为你使这个问题比需要的要复杂一些。这是使用plyr的一个解决方案。我们将按c2变量进行分组：

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

未分割列表，合并因子

2 个答案: