在lapply函数中访问和保留列表名称 [英] Access and preserve list names in lapply function

问题描述

我需要访问lapply函数中的列表名称.我在网上找到了一些线程，据说我应该遍历列表的名称，以便能够获取函数中的每个列表元素名称:

I need to access list names inside the lapply function. I've found some threads online where it's said I should iterate through the names of the list to be able to fetch each list element name in my function:

> n = names(mylist)
> mynewlist = lapply(n, function(nameindex, mylist) { return(mylist[[nameindex]]) }, mylist)
> names(mynewlist)
NULL
> names(mynewlist) = n

问题是mynewlist丢失了原始的mylist索引，我必须添加lastname()分配来恢复它们.

The problem is that mynewlist loses the original mylist indexes and I have to add that last names() assignment to restore them.

是否可以为lapply函数返回的每个元素赋予明确的索引名称?或者以其他方式确保mynewlist元素设置了正确的索引名称?我感觉如果lapply不按与mylist相同的顺序返回列表元素，则mynewlist索引名称可能是错误的.

Is there a way to give an explicit index name to each element returned by the lapply function? Or a different way to make sure mynewlist elements have the correct index names set? I feel mynewlist index names could be wrong if lapply does not return the list elements in the same order than mylist.

推荐答案

我相信默认情况下，lapply保留要迭代的名称属性.当您将myList的名称存储在n中时，该向量不再具有任何名称".因此，如果您通过将其添加回去，

I believe that lapply by default keeps the names attribute of whatever you are iterating over. When you store the names of myList in n, that vector no longer has any "names". So if you add that back in via,

names(n) <- names(myList)

并像以前一样使用lapply，您应该会得到所需的结果.

and the use lapply as before, you should get the desired result.

修改

今天早上我的大脑有点迷雾.这是另一个也许更方便的选择:

My brains a bit foggy this morning. Here's another, perhaps more convenient, option:

sapply(n,FUN = ...,simplify = FALSE,USE.NAMES = TRUE)

我一直在摸索，困惑于lapply没有USE.NAMES参数，然后我实际上查看了sapply的代码，意识到我很傻，这可能是一种更好的方法去.

I was groping about, confused that lapply didn't have a USE.NAMES argument, and then I actually looked at the code for sapply and realized I was being silly, and this was probably a better way to go.

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