在R中索引嵌套列表结构 [英] Indexing nested list structure in R

问题描述

我已经从STATA过渡到R，并且我正在尝试不同的数据类型，以使R的数据结构在我的脑海中清晰可见.

I have transitioned from STATA to R, and I was experimenting with different data types so that R's data structures are clear in my mind.

这是我设置数据结构的方式:

Here's how I set up my data structure:

b<-list(u=5,v=12)
c<-list(u=7)
j<-list(name="Joe",salary=55000,union=T)
bcj<-list(b,c,j)

现在，我正在尝试找出访问u = 5的不同方法.我相信可以通过三种方式:

Now, I was trying to figure out different ways to access u=5. I believe there are three ways:

Try1:

bcj[[1]][[1]]

我有5个.正确！

Try2:

bcj[[1]][["u"]]

我有5个.正确！

Try3:

bcj[[1]]$u

我有5个.正确！

Try4

bcj[[1]][1][1]

这就是我得到的:

bcj[[1]][1][1]
$u
[1] 5

class(bcj[[1]][1][1])
[1] "list"

问题1:为什么会发生这种情况?

我还尝试了以下方法:

bcj[[1]][1][1][1][1][1]
$u
[1] 5

class(bcj[[1]][1][1][1][1][1])
[1] "list"

问题2:我本来希望看到一个错误，因为我认为bcj中不存在太多列表，但是R给了我一个列表.为什么会发生这种情况?

PS:我确实在SO上查看了此线程，但这是在谈论一个不同的问题.

PS: I did look at this thread on SO, but it's talking about a different issue.

推荐答案

我认为这足以回答您的问题.考虑一个 length-1列表:

I think this is sufficient to answer your question. Consider a length-1 list:

x <- list(u = 5)
#$u
#[1] 5
length(x)
#[1] 1

x[1]
x[1][1]
x[1][1][1]
...

总是给您相同的东西

#$u
#[1] 5

换句话说，x[1]将与x相同，并且您将陷入无限递归.不管您写多少[1]，您自己都会得到x.

In other words, x[1] will be identical to x, and you fall into infinite recursion. No matter how many [1] you write, you just get x itself.

如果我创建t1<-list(u=5,v=7)，然后再执行t1[2][1][1][1] ...这也可以.但是，t1[[2]][2]给出NA

If I create t1<-list(u=5,v=7), and then do t1[2][1][1][1]...this works as well. However, t1[[2]][2] gives NA

---

这是索引列表时[[和[之间的区别.使用[总是以列表结尾，而[[会取出内容.比较:

---

That is the difference between [[ and [ when indexing a list. Using [ will always end up with a list, while [[ will take out the content. Compare:

z1 <- t1[2]
## this is a length-1 list
#$v
#[1] 7
class(z1)
# "list"

z2 <- t1[[2]]
## this takes out the content; in this case, a vector
#[1] 7
class(z2)
#[1] "numeric"

如上所述，当您执行z1[1][1]...时，总是会遇到z1本身.如果您执行z2[2]，则肯定会得到一个NA，因为z2仅具有一个元素，而您要求的是第二个元素.

When you do z1[1][1]..., as discussed above, you always end up with z1 itself. While if you do z2[2], you surely get an NA, because z2 has only one element, and you are asking for the 2nd element.

也许这篇文章和我的回答对您有用:使用方括号和数字提取嵌套列表元素?

Perhaps this post and my answer there is useful for you: Extract nested list elements using bracketed numbers and names?

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