元素作爲列表名稱和列表名稱作爲列表中的元素？

Question

我有一個向量列表（如下所示）。我想知道向量的每個元素在哪個列表元素中。換句話說，我想反轉列表以創建一個新列表，其中names取自矢量。

這樣做的最佳方法是什麼？

lst <- list(a=c(2, 3, 6, 10, 15, 17), b=c(4, 6, 9, 7, 6, 4, 3, 10), 
      c=c(9, 2, 1, 4, 3), d=c(3, 6, 17)) 
lst 
$a 
[1] 2 3 6 10 15 17 

$b 
[1] 4 6 9 7 6 4 3 10 

$c 
[1] 9 2 1 4 3 

$d 
[1] 3 6 17 

我想獲得以下答案。

$`1` 
[1] "c" 

$`10` 
[1] "a" "b" 

$`15` 
[1] "a" 

$`17` 
[1] "a" "d" 

$`2` 
[1] "a" "c" 

$`3` 
[1] "a" "b" "c" "d" 

$`4` 
[1] "b" "b" "c" 

$`6` 
[1] "a" "b" "b" "d" 

$`7` 
[1] "b" 

$`9` 
[1] "b" "c" 

Answer 1

這裏有一個基地R的方式與stack和unstack：

unstack(stack(lst), ind ~ values) 
# $`1` 
# [1] "c" 
# 
# $`2` 
# [1] "a" "c" 
# 
# $`3` 
# [1] "a" "b" "c" "d" 
# 
# $`4` 
# [1] "b" "b" "c" 
# 
# $`6` 
# [1] "a" "b" "b" "d" 
# 
# $`7` 
# [1] "b" 
# 
# $`9` 
# [1] "b" "c" 
# 
# $`10` 
# [1] "a" "b" 
# 
# $`15` 
# [1] "a" 
# 
# $`17` 
# [1] "a" "d" 

Answer 2

下面是使用split從基礎R的方法使用melt從 「reshape2」 之後：

library(reshape2) 
x <- melt(lst) 
split(x$L1, x$value) 
# $`1` 
# [1] "c" 
# 
# $`2` 
# [1] "a" "c" 
# 
# $`3` 
# [1] "a" "b" "c" "d" 
# 
# $`4` 
# [1] "b" "b" "c" 
# 
# $`6` 
# [1] "a" "b" "b" "d" 
# 
# $`7` 
# [1] "b" 
# 
# $`9` 
# [1] "b" "c" 
# 
# $`10` 
# [1] "a" "b" 
# 
# $`15` 
# [1] "a" 
# 
# $`17` 
# [1] "a" "d" 

---

同樣，在基礎R與stack：

x <- stack(lapply(lst, c)) 
split(as.character(x$ind), x$values) 

甚至如果您使用「lst」而不是「lst」，則更直接：

x <- stack(lst) 
split(as.character(x$ind), x$values) 

---

要闡述我的意見，我被介紹給更有效的方式是：

split(rep(names(lst), lapply(lst, nrow)), unlist(lst, use.names = FALSE)) 

應用到一個更大的lst，我們得到如下：

fun1 <- function() split(rep(names(lst), lapply(lst, nrow)), unlist(lst, use.names = FALSE)) 
fun2 <- function() { x <- stack(lapply(lst, c)) ; split(as.character(x$ind), x$values) } 
fun3 <- function() { x <- melt(lst) ; split(x$L1, x$value) } 
fun4 <- function() unstack(stack(lapply(lst, as.vector)), ind ~ values) 

## Make lst much bigger 
lst <- unlist(replicate(10000, lst, simplify = FALSE), recursive=FALSE) 
names(lst) <- make.unique(names(lst)) 

library(microbenchmark) 

system.time(fun3()) 
# user system elapsed 
# 48.338 0.000 47.643 

microbenchmark(fun1(), fun2(), fun4(), times = 5) 
# Unit: milliseconds 
# expr  min  lq median  uq  max neval 
# fun1() 454.5913 456.6793 473.901 555.8954 574.4394  5 
# fun2() 922.1282 1028.4972 1034.872 1068.4761 1150.8072  5 
# fun4() 1222.5296 1300.0643 1323.253 1339.2037 1421.1546  5 

Answer 3

unlistlist獲得矢量中的所有數字。然後，使用這些數字拆分list元素的names的向量。

split(rep(names(lst),times=sapply(lst,length)), 
     unlist(lst)) 
$`1` 
[1] "c" 

$`2` 
[1] "a" "c" 

$`3` 
[1] "a" "b" "c" "d" 

$`4` 
[1] "b" "b" "c" 

$`6` 
[1] "a" "b" "b" "d" 

$`7` 
[1] "b" 

$`9` 
[1] "b" "c" 

$`10` 
[1] "a" "b" 

$`15` 
[1] "a" 

$`17` 
[1] "a" "d"