R 数据帧列表

我有一个叫做“mylist”的列表。它包含两个项目。每个项目都是数据帧列表。列表的第一项是1个数据帧的列表，第二项是2个数据帧的列表，如下所示：-

str(mylist1)
List of 1
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  ..$ salary   : num [1:3] 21000 23400 26800
  ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
> str(mylist2)
List of 2
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  ..$ salary   : num [1:3] 21000 23400 26800
  ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe1" "Peter Gynn1" "Jolie Hope1"
  ..$ salary   : num [1:3] 20000 25000 30000
  ..$ startdate: Date[1:3], format: "2011-11-01" "2009-03-25" "2008-03-14"
> str(mylist)
List of 2
 $ :List of 1
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  .. ..$ salary   : num [1:3] 21000 23400 26800
  .. ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
 $ :List of 2
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  .. ..$ salary   : num [1:3] 21000 23400 26800
  .. ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe1" "Peter Gynn1" "Jolie Hope1"
  .. ..$ salary   : num [1:3] 20000 25000 30000
  .. ..$ startdate: Date[1:3], format: "2011-11-01" "2009-03-25" "2008-03-14"

mylist1
[[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

> mylist2
[[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

[[2]]
     employee salary  startdate
1   John Doe1  20000 2011-11-01
2 Peter Gynn1  25000 2009-03-25
3 Jolie Hope1  30000 2008-03-14

> mylist
[[1]]
[[1]][[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14


[[2]]
[[2]][[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

[[2]][[2]]
     employee salary  startdate
1   John Doe1  20000 2011-11-01
2 Peter Gynn1  25000 2009-03-25
3 Jolie Hope1  30000 2008-03-14

testvar <- mylist

列表本身看起来是这样的：-

str(mylist1)
List of 1
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  ..$ salary   : num [1:3] 21000 23400 26800
  ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
> str(mylist2)
List of 2
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  ..$ salary   : num [1:3] 21000 23400 26800
  ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe1" "Peter Gynn1" "Jolie Hope1"
  ..$ salary   : num [1:3] 20000 25000 30000
  ..$ startdate: Date[1:3], format: "2011-11-01" "2009-03-25" "2008-03-14"
> str(mylist)
List of 2
 $ :List of 1
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  .. ..$ salary   : num [1:3] 21000 23400 26800
  .. ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
 $ :List of 2
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  .. ..$ salary   : num [1:3] 21000 23400 26800
  .. ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe1" "Peter Gynn1" "Jolie Hope1"
  .. ..$ salary   : num [1:3] 20000 25000 30000
  .. ..$ startdate: Date[1:3], format: "2011-11-01" "2009-03-25" "2008-03-14"

mylist1
[[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

> mylist2
[[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

[[2]]
     employee salary  startdate
1   John Doe1  20000 2011-11-01
2 Peter Gynn1  25000 2009-03-25
3 Jolie Hope1  30000 2008-03-14

> mylist
[[1]]
[[1]][[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14


[[2]]
[[2]][[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

[[2]][[2]]
     employee salary  startdate
1   John Doe1  20000 2011-11-01
2 Peter Gynn1  25000 2009-03-25
3 Jolie Hope1  30000 2008-03-14

testvar <- mylist

如果我将列表“mylist”分配给如下变量：-

str(mylist1)
List of 1
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  ..$ salary   : num [1:3] 21000 23400 26800
  ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
> str(mylist2)
List of 2
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  ..$ salary   : num [1:3] 21000 23400 26800
  ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
 $ :'data.frame':   3 obs. of  3 variables:
  ..$ employee : chr [1:3] "John Doe1" "Peter Gynn1" "Jolie Hope1"
  ..$ salary   : num [1:3] 20000 25000 30000
  ..$ startdate: Date[1:3], format: "2011-11-01" "2009-03-25" "2008-03-14"
> str(mylist)
List of 2
 $ :List of 1
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  .. ..$ salary   : num [1:3] 21000 23400 26800
  .. ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
 $ :List of 2
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe" "Peter Gynn" "Jolie Hope"
  .. ..$ salary   : num [1:3] 21000 23400 26800
  .. ..$ startdate: Date[1:3], format: "2010-11-01" "2008-03-25" "2007-03-14"
  ..$ :'data.frame':    3 obs. of  3 variables:
  .. ..$ employee : chr [1:3] "John Doe1" "Peter Gynn1" "Jolie Hope1"
  .. ..$ salary   : num [1:3] 20000 25000 30000
  .. ..$ startdate: Date[1:3], format: "2011-11-01" "2009-03-25" "2008-03-14"

mylist1
[[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

> mylist2
[[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

[[2]]
     employee salary  startdate
1   John Doe1  20000 2011-11-01
2 Peter Gynn1  25000 2009-03-25
3 Jolie Hope1  30000 2008-03-14

> mylist
[[1]]
[[1]][[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14


[[2]]
[[2]][[1]]
    employee salary  startdate
1   John Doe  21000 2010-11-01
2 Peter Gynn  23400 2008-03-25
3 Jolie Hope  26800 2007-03-14

[[2]][[2]]
     employee salary  startdate
1   John Doe1  20000 2011-11-01
2 Peter Gynn1  25000 2009-03-25
3 Jolie Hope1  30000 2008-03-14

testvar <- mylist

但以下命令给出了一个错误：-

str(get(paste0("testvar", "[[1]]")))

错误

为什么上面的命令找不到testvar对象，它实际上是列表“mylist”。我希望能够获得列表“mylist”第一项的结构（甚至类）。我需要以编程的方式来做，不能硬编码

有什么建议吗

致意

---

Deepak

get

或

mget

仅返回在全局环境中创建的对象<代码>“testvar”是一个使用值创建的对象，而“testvar[[1]]”不是对象标识符，它只是

列表的元素之一。因此，我们使用
[

get("testvar")[[1]]


这类似于获取data.frame的列

data(mtcars)
get("mtcars") # // => works
get("mtcars[[1]]") # // => returns error

get（“mtcars[[1]]”中出错：未找到对象“mtcars[[1]]”


我们不清楚为什么需要使用
get
。如果目的是在
mylist
上循环，可以使用
lappy

lapply(mylist, function(innerlst) yourfun(innerlst))

akrun@DeepakAgarwal它正在检查在全局环境中创建的标识符名称，即未创建的
testvar[[1]]
，但您只有
testvar
，这就是它没有创建的原因work@DeepakAgarwal.当你做了一个akrunarkun