如何反转向量? [英] How to reverse order a vector?

问题描述

假设我有一个向量v，如何得到它的倒数，即最后一个元素在先?

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

我遇到的第一件事是v[length(v):1]，但是当v是numeric(0)时，它返回NA，而用户通常期望不进行任何排序将不返回任何内容，不进行任何排序将返回不可用的内容-这确实使它变得很大我的情况有所不同.

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

您快到了； rev做您需要的事情:

You are almost there; rev does what you need:

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

这是为什么:

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

对于numeric(0)，length(x)返回0.由于if需要逻辑条件，因此它将length(x)强制为TRUE或FALSE.当x为0且TRUE为其他任何数字时，as.logical(x)为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))精确测试了您想要的内容-x的长度是否为零.如果不是，则length(x):1L具有令人满意的效果，否则就无需撤销任何操作，如@floder在评论中解释的那样.

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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