当负索引可能是整数(0)时，如何安全地从向量中删除任何内容? [英] How to safely drop nothing from a vector when the negative index could be integer(0)?

问题描述

假设我有一个向量 x = 1:10，它是通过连接另外两个向量 a = integer(0) 和 b = 1 来构造的:10 在一起(这是一个边缘情况).我想稍后再次将组合向量拆分为 a 和 b .我原以为我可以安全地将它们分开:

Suppose I have a vector x = 1:10, and it is constructed by concatenating two other vectors a = integer(0) and b = 1:10 together (this is an edge case). I want to split up the combined vector again into a and b later on. I would have thought I could safely separate them with:

i = seq_along(a)
x[i]
x[-i]

但我发现当我使用 x[-integer(0)] 我得到 integer(0) 返回，而不是 x正如我天真地认为的那样.做这种事情的最佳方法是什么?

But I discovered that when I use x[-integer(0)] I get integer(0) returned, instead of x itself as I naively thought. What is the best way to do this sort of thing?

推荐答案

如果你想使用负索引并且索引可能退化为 integer(0)(例如，索引是从which)，将一个大的越界"值填充到索引中.删除越界"值没有副作用.

If you want to use negative indexing and the index may degenerate to integer(0) (for example, the index is computed from which), pad a large "out-of-bound" value to the index. Removing an "out-of-bound" value has no side effect.

x <- 1:10
i <- integer(0)
x[-c(i, 11)]  ## position 11 is "out-of-bound"
# [1]  1  2  3  4  5  6  7  8  9 10

如果你麻烦设置这个out-of-bound"值，这里有一个规范的选择:2 ^ 31，因为这个值已经超出了 32 位有符号整数的表示范围，但是它不是 Inf.

If you bother setting this "out-of-bound" value, here is a canonical choice: 2 ^ 31, because this value has exceeded representation range of 32-bit signed integer, yet it is not Inf.

另一种方法是对 length(i) 进行 if 测试.例如:

An alternative way is to do an if test on length(i). For example:

if (length(i)) x[-i] else x

注意:不要将函数 ifelse 用于此目的.

Caution: don't use function ifelse for this purpose.

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