Question

我有一个清单

  mylist<-list( "[3] C(+57.02)", c("[11] C(+57.02)", "[16] C(+57.02)"), "[3] C(+57.02)",c ("[5] M(+15.99)", "[6] M(+15.99)"))

我想得到一个这样的表，其中名称是对象列表编号。

      1               2                   2                 3                   
"[3] C(+57.02)" , "[11] C(+57.02)", "[16] C(+57.02)", "[3] C(+57.02)",  
      4               4
"[5] M(+15.99)", "[6] M(+15.99)"

我试图这样做但不完全是我想要的。

table(unlist(mylist))

[11] C(+57.02) [16] C(+57.02)  [3] C(+57.02)  [5] M(+15.99)  [6] M(+15.99) 

         1              1              2              1              1

Answer 1

您可以使用rep + seq_along + lengths + setNames：

setNames(unlist(mylist), rep(seq_along(mylist), lengths(mylist)))
##                1                2                2                3                4                4 
##  "[3] C(+57.02)" "[11] C(+57.02)" "[16] C(+57.02)"  "[3] C(+57.02)"  "[5] M(+15.99)"  "[6] M(+15.99)"

更好的演示可能是使用stack：

stack(setNames(mylist, seq_along(mylist)))
##           values ind
## 1  [3] C(+57.02)   1
## 2 [11] C(+57.02)   2
## 3 [16] C(+57.02)   2
## 4  [3] C(+57.02)   3
## 5  [5] M(+15.99)   4
## 6  [6] M(+15.99)   4

Answer 2

我使用了一个命名的字符向量。不知道为什么你想要table。

数据：

mylist<-list( "[3] C(+57.02)", c("[11] C(+57.02)", "[16] C(+57.02)"), "[3] C(+57.02)",c ("[5] M(+15.99)", "[6] M(+15.99)"))

代码：

v <- unlist(mylist)
names(v) <- rapply(mylist,length) %>% rep(seq_along(mylist),.)

输出：

> v
               1                2                2                3                4                4 
 "[3] C(+57.02)" "[11] C(+57.02)" "[16] C(+57.02)"  "[3] C(+57.02)"  "[5] M(+15.99)"  "[6] M(+15.99)" 
>

从列表中创建一个表

2 个答案: