为什么在R中用list [n]而不是list [[n]]索引到列表中，却不能达到您的期望? [英] Why does indexing into a list with list[n] instead of list[[n]] in R not do what you would expect?

问题描述

在R列表中，为什么用[n]而不是[[n]]进行索引时不会像非R程序员所期望的那样返回第n个元素?

In an R list, why does indexing with [n] instead of [[n]] not return the nth element as non-R-programmers would expect?

lfile <- list('fileA.xls', 'file2.xls', 'fileY.xls')
ll <- list(list(1,2), list(3), list(4,5,6), list(7,8))
lv <- list(c(1,2), c(3), c(4,5,6), c(7,8))

> lfile[2]
[[1]]
[1] "file2.xls" # returns SUBLIST, not n'th ELEMENT

> lfile[[2]]
[1] "file2.xls" # returns ELEMENT

推荐答案

因为通常l[n]返回一个子列表(长度可能为一个)，而l[[n]]返回一个元素:

Because in general l[n] returns a sublist (possibly of length one), whereas l[[n]] returns an element:

> lfile[2]
[[1]]
[1] "file2.xls" # returns SUBLIST of length one, not n'th ELEMENT

> lfile[[2]]
[1] "file2.xls" # returns ELEMENT

来自 R入门手册:6.1列表 :

将Lst [[1]]与Lst [1] 区分开来非常重要.

It is very important to distinguish Lst[[1]] from Lst[1].

"[[…]]"是用于选择单个元素的运算符，

"[…]"是一般的下标运算符.

因此，前者是列表Lst中的第一个对象，如果它是命名列表，则不包括名称.后者是列表Lst的子列表，仅由第一个条目组成.如果它是一个命名列表，则名称将转移到子列表中.

Thus the former is the first object in the list Lst, and if it is a named list the name is not included. The latter is a sublist of the list Lst consisting of the first entry only. If it is a named list, the names are transferred to the sublist.

这是一个R陷阱(R与其他语言的区别是显而易见的，且文献记载不足)，尽管一些R用户一如既往地坚持说，它确实按照它说的去做(在某些地方是深层的且未编制索引). doc.但是， 的帮助页面list 仅显示您如何创建列表，但不显示如何对其进行索引(！).仅

This is an R gotcha (R being different to other languages in a non-obvious and under-documented way), although as ever some R users will insist it's doing exactly what it says, (somewhere deep and unindexed) in the doc. However, the help-page for list only shows you how to create a list but does not show how to index into it(!) Only ?[ or ?Extract manpages actually tell you how to index into a list(!)

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