如何倒序向量? [英] How to reverse order a vector?

问题描述

假设我有一个向量 v，我如何得到它的反向，即最后一个元素?

Suppose I have a vector v, how do I get its reverse, i.e. last element first?

首先想到的是v[length(v):1]，但是当v是numeric(0)<时它返回NA/code>，虽然用户通常希望排序不返回任何内容，但不排序返回不可用的内容 - 这对我来说确实有很大的不同.

The first thing that comes to me is v[length(v):1], but it returns NA when v is numeric(0), while user normally expect sorting nothing returns nothing, not sorting nothing returns the unavailable thing - it does make a big difference in my case.

推荐答案

你就快到了；rev 满足您的需求:

You are almost there; rev does what you need:

rev(1:3)
# [1] 3 2 1
rev(numeric(0))
# numeric(0)

原因如下:

rev.default
# function (x) 
# if (length(x)) x[length(x):1L] else x
# <bytecode: 0x0b5c6184>
# <environment: namespace:base>

在numeric(0)的情况下，length(x)返回0.由于if需要一个逻辑条件，它强制length(x) 到 TRUE 或 FALSE.当 x 为 0 并且 TRUE 为任何其他数字时，as.logical(x) 恰好是 FALSE.

In the case of numeric(0), length(x) returns 0. As if requires a logical condition, it coerces length(x) to TRUE or FALSE. It happens that as.logical(x) is FALSE when x is 0 and TRUE for any other number.

因此，if (length(x)) 可以精确地测试您想要的内容 - x 的长度是否为零.如果不是，length(x):1L 有一个理想的效果，否则就不需要反转任何东西，正如@floder 在评论中所解释的那样.

Thus, if (length(x)) tests precisely what you want - whether x is of length zero. If it isn't, length(x):1L has a desirable effect, and otherwise there is no need to reverse anything, as @floder has explained in the comment.

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