在这里，我们又来了：一个元素追加到R A名单 [英] Here we go again: append an element to a list in R

问题描述

我不开心的接受的答案，<一个href=\"http://stackoverflow.com/questions/2436688/append-an-object-to-a-list-in-r-in-amortized-constant-time\">Append一个对象在固定的时间里R中的列表？

I am not happy with the accepted answer to Append an object to a list in R in amortized constant time?

> list1 <- list("foo", pi)
> bar <- list("A", "B")

我如何可以将新的元素栏到 list1的？显然， C（）不起作用，它变平栏：

How can I append new element bar to list1? Clearly, c() does not work, it flattens bar:

> c(list1, bar)
[[1]]
[1] "foo"

[[2]]
[1] 3.141593

[[3]]
[1] "A"

[[4]]
[1] "B"

分配索引的作品：

Assignment to index works:

> list1[[length(list1)+1]] <- bar
> list1
[[1]]
[1] "foo"

[[2]]
[1] 3.141593

[[3]]
[[3]][[1]]
[1] "A"

[[3]][[2]]
[1] "B"

什么是这种方法的效率？有没有更优雅的方式？

What is the efficiency of this method? Is there a more elegant way?

推荐答案

在一个时间做一个元素时将元素添加到列表是很慢的。看到这两个例子：

Adding elements to a list is very slow when doing it one element at a time. See these two examples:

我保持结果变量在全球环境中，以避免副本评估框架，并告诉R其中寻找它与 .GlobalEnv $ ，以避免盲目搜索＆LT;＆LT; - ：

I'm keeping the Result variable in the global environment to avoid copies to evaluation frames and telling R where to look for it with .GlobalEnv$, to avoid a blind search with <<-:

Result <- list()

AddItemNaive <- function(item)
{
    .GlobalEnv$Result[[length(.GlobalEnv$Result)+1]] <- item
}

system.time(for(i in seq_len(2e4)) AddItemNaive(i))
#   user  system elapsed 
#  15.60    0.00   15.61 

慢。现在让我们尝试第二种方法：

Slow. Now let's try the second approach:

Result <- list()

AddItemNaive2 <- function(item)
{
    .GlobalEnv$Result <- c(.GlobalEnv$Result, item)
}

system.time(for(i in seq_len(2e4)) AddItemNaive2(i))
#   user  system elapsed 
#  13.85    0.00   13.89

仍然缓慢。

现在，让我们尝试使用环境，并创造这种环境，而不是将元素添加到列表中的新变量。这里的问题是，必须对变量进行命名的，所以我会用计数器作为一个字符串，命名每个项目老虎：

Now let's try using an environment, and creating new variables within this environment instead of adding elements to a list. The issue here is that variables must be named, so I'll use the counter as a string to name each item "slot":

Counter <- 0
Result <- new.env()

AddItemEnvir <- function(item)
{
    .GlobalEnv$Counter <- .GlobalEnv$Counter + 1

    .GlobalEnv$Result[[as.character(.GlobalEnv$Counter)]] <- item
}

system.time(for(i in seq_len(2e4)) AddItemEnvir(i))
#   user  system elapsed 
#   0.36    0.00    0.38 

哇快得多。 :-)它可能是一个有点尴尬的工作，但它的作品。

Whoa much faster. :-) It may be a little awkward to work with, but it works.

最后一个方法是使用一个列表，但不是一次一个元素充实它的规模，它的双打的每次列表已满的大小。该列表的大小也保持在一个专用的变量，以避免任何放缓使用长度：

A final approach uses a list, but instead of augmenting its size one element at a time, it doubles the size each time the list is full. The list size is also kept in a dedicated variable, to avoid any slowdown using length:

Counter <- 0
Result <- list(NULL)
Size <- 1

AddItemDoubling <- function(item)
{
    if( .GlobalEnv$Counter == .GlobalEnv$Size )
    {
        length(.GlobalEnv$Result) <- .GlobalEnv$Size <- .GlobalEnv$Size * 2
    }

    .GlobalEnv$Counter <- .GlobalEnv$Counter + 1

    .GlobalEnv$Result[[.GlobalEnv$Counter]] <- item
}

system.time(for(i in seq_len(2e4)) AddItemDoubling(i))
#   user  system elapsed 
#   0.22    0.00    0.22

它甚至更快。而且易于工作的任何名单。

It's even faster. And as easy to a work as any list.

让我们尝试用更多的迭代最后这两种解决方案：

Let's try these last two solutions with more iterations:

Counter <- 0
Result <- new.env()

system.time(for(i in seq_len(1e5)) AddItemEnvir(i))
#   user  system elapsed 
#  27.72    0.06   27.83 


Counter <- 0
Result <- list(NULL)
Size <- 1

system.time(for(i in seq_len(1e5)) AddItemDoubling(i))
#   user  system elapsed 
#   9.26    0.00    9.32

好了，最后一个是definetely的路要走。

Well, the last one is definetely the way to go.

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