如何从R中的列表列表中提取元素? [英] How can I extract elements from lists of lists in R?

问题描述

我有一堆列表，其中包含列表(广义线性模型输出).我想编写一个函数，该函数将从每个列表中提取几个元素，然后将结果合并到一个数据帧中.

I have a bunch of lists containing lists within them (generalised linear model output). I want to write a function which will extract several elements from each list and then combine the results into a data frame.

我要提取modelset[[1]]$likelihood& modelset[[1]]$fixef，modelset[[2]]$likelihood& modelset[[2]]$fixef等，然后将结果合并到一个数据框中.

I want to extract modelset[[1]]$likelihood & modelset[[1]]$fixef, modelset[[2]]$likelihood & modelset[[2]]$fixef, etc, and combine the results into a data frame.

有人可以告诉我如何执行此操作吗?

Can someone give me an idea of how to do this?

很抱歉，我的问题令人困惑:我想做的事超出了我有限的编程理解.

Apologies if my question is confusing: what I am trying to do is beyond my limited programming understanding.

有关我的列表的其他信息:

Further information about my list:

modelset: Large list (16 elements, 7.3Mb)
    :List of 29
    ..$ fixef           : Named num [1:2] -1.236 -0.611
    .. ..- attr(*, "names")= chr [1:2] "(Intercept)" "SMIstd"
    ..$ likelihood      :List of 4
    .. ..$ hlik: num 238
    .. ..$ pvh : num 256
    .. ..$ pbvh: num 260
    .. ..$ cAIC: num 567

    ...etc  

推荐答案

为了优雅地解决此问题，您需要了解可以使用['…']而不是$…来访问列表元素(但您将获得一个列表)返回而不是单个元素.

In order to solve this elegantly you need to understand that you can use ['…'] instead of $… to access list elements (but you will get a list back instead of an individual element).

因此，如果要获取元素likelihood和fixef，则可以编写:

So if you want to get the elements likelihood and fixef, you can write:

modelset[[1]][c('likelihood', 'fixef')]

现在，您要为modelset中的每个元素执行此操作.这就是 lapply 的作用:

Now you want to do that for each element in modelset. That’s what lapply does:

lapply(modelset, function (x) x[c('likelihood', 'fixef')])

这行得通，但不是很像R.

This works, but it’s not very R-like.

在R中，您几乎可以看到所有内容是一个函数. […]正在调用名为[的函数(但由于[是R的特殊符号，因此需要在反引号中引用:`[`).因此，您可以这样写:

You see, in R, almost everything is a function. […] is calling a function named [ (but since [ is a special symbol for R, in needs to be quoted in backticks: `[`). So you can instead write this:

lapply(modelset, function (x) `[`(c('likelihood', 'fixef')])

哇，那根本不是很可读.但是，我们现在可以删除包装的匿名function (x)，因为在内部我们只是在调用另一个函数，然后将多余的参数移到lapply的最后一个参数:

Wow, that’s not very readable at all. However, we can now remove the wrapping anonymous function (x), since inside we’re just calling another function, and move the extra arguments to the last parameter of lapply:

lapply(modelset, `[`, c('likelihood', 'fixef'))

这行之有效并且是优雅的R代码.

This works and is elegant R code.

让我们退后一步，重新检查我们在这里所做的事情.实际上，我们有一个看起来像这样的表达式:

Let’s step back and re-examine what we did here. In effect, we had an expression which looked like this:

lapply(some_list, function (x) f(x, y))

此呼叫可以改写为

lapply(some_list, f, y)

我们正是通过somelist = modelset，f = `[`和y = c('likelihood', 'fixef')做到了这一点.

We did exactly that, with somelist = modelset, f = `[` and y = c('likelihood', 'fixef').

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