喜欢的东西expand.grid名单

Question

我有存储为所谓的“DLIST”列出的清单三个文本文档的列表上：

dlist <- structure(list(name = c("a", "b", "c"), text = list(c("the", "quick", "brown"), c("fox", "jumps", "over", "the"), c("lazy", "dog"))), .Names = c("name", "text")) 

在我脑子里，我发现它有助于图片DLIST是这样的：

name text 
1 a  c("the", "quick", "brown") 
2 b  c("fox", "jumps", "over", "the") 
3 c  c("lazy", "dog") 

这怎么能被操纵如下？这个想法是绘制它，所以可以熔化ggplot2的东西会很好。

name text 
1 a the 
2 a quick 
3 a brown 
4 b fox 
5 b jumps 
6 b over 
7 b the 
8 c lazy 
9 c dog 

这是每个单词一行，同时提供单词和其父文档。

我曾尝试：

> expand.grid(dlist) 
    name     text 
1 a  the, quick, brown 
2 b  the, quick, brown 
3 c  the, quick, brown 
4 a fox, jumps, over, the 
5 b fox, jumps, over, the 
6 c fox, jumps, over, the 
7 a    lazy, dog 
8 b    lazy, dog 
9 c    lazy, dog 

> sapply(seq(1,3), function(x) (expand.grid(dlist$name[[x]], dlist$text[[x]]))) 
    [,1]  [,2]  [,3]  
Var1 factor,3 factor,4 factor,2 
Var2 factor,3 factor,4 factor,2 

unlist(dlist) 
    name1 name2 name3 text1 text2 text3 text4 
    "a"  "b"  "c" "the" "quick" "brown" "fox" 
    text5 text6 text7 text8 text9 
"jumps" "over" "the" "lazy" "dog" 

> sapply(seq(1,3), function(x) (cbind(dlist$name[[x]], dlist$text[[x]]))) 
[[1]] 
    [,1] [,2] 
[1,] "a" "the" 
[2,] "a" "quick" 
[3,] "a" "brown" 

[[2]] 
    [,1] [,2] 
[1,] "b" "fox" 
[2,] "b" "jumps" 
[3,] "b" "over" 
[4,] "b" "the" 

[[3]] 
    [,1] [,2] 
[1,] "c" "lazy" 
[2,] "c" "dog" 

公平地说，我通过各种迷惑申请和plyr功能，真的不知道从哪里开始。我从来没有见过类似于上面的“侥幸”尝试的结果，并且不理解它。

Answer 1

如果您将dlist转换为命名列表（我认为更适合的结构），则可以使用stack()来获取所需的两列数据。

（该rev()和setNames()在第二行调用只是众多方法来调整列的顺序和名称以匹配你的问题出所需的输出之一。）

x <- setNames(dlist$text, dlist$name) 
setNames(rev(stack(x)), c("name", "text")) 
# name text 
# 1 a the 
# 2 a quick 
# 3 a brown 
# 4 b fox 
# 5 b jumps 
# 6 b over 
# 7 b the 
# 8 c lazy 
# 9 c dog 

Answer 2

Josh的答案是更甜但我想我会把我的帽子扔在戒指里。

dlist <- structure(list(name = c("a", "b", "c"), 
    text = list(c("the", "quick", "brown"), 
    c("fox", "jumps", "over", "the"), c("lazy", "dog"))), 
    .Names = c("name", "text")) 

lens <- sapply(unlist(dlist[-1], recursive = FALSE), length) 

data.frame(name = rep(dlist[[1]], lens), text = unlist(dlist[-1]), row.names = NULL) 

## name text 
## 1 a the 
## 2 a quick 
## 3 a brown 
## 4 b fox 
## 5 b jumps 
## 6 b over 
## 7 b the 
## 8 c lazy 
## 9 c dog 

这就是说，列表清单是一种尴尬的存储方法。向量列表（特别是向量列表）将更容易处理。

Answer 3

另一种解决方案，也许更普及：

do.call(rbind, do.call(mapply, c(dlist, FUN = data.frame, SIMPLIFY = FALSE))) 

#  name text 
# a.1 a the 
# a.2 a quick 
# a.3 a brown 
# b.1 b fox 
# b.2 b jumps 
# b.3 b over 
# b.4 b the 
# c.1 c lazy 
# c.2 c dog