获取向量的最后n个元素。有没有比使用length（）函数更好的方法？ [英] Getting the last n elements of a vector. Is there a better way than using the length() function?

问题描述

如果为了论证的缘故，我想在Python中使用10长度向量的最后五个元素，我可以在范围索引中使用 - 运算符，所以：

If, for argument's sake, I want the last five elements of a 10-length vector in Python, I can use the "-" operator in the range index so:

>>> x = range(10)
>>> x
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> x[-5:]
[5, 6, 7, 8, 9]
>>>

R中最好的方法是什么？有没有比我目前使用length（）函数更简洁的方法？

What is the best way to do this in R? Is there a cleaner way than my current technique which is to use the length() function?

> x <- 0:9
> x
 [1] 0 1 2 3 4 5 6 7 8 9
> x[(length(x) - 4):length(x)]
[1] 5 6 7 8 9
> 

这个问题与btw的时间序列分析有关，它通常只适用于最近的数据。

The question is related to time series analysis btw where it is often useful to work only on recent data.

推荐答案

参见？tail 和？head 用于一些方便的功能：

see ?tail and ?head for some convenient functions:

> x <- 1:10
> tail(x,5)
[1]  6  7  8  9 10

对于参数清酒：除了最后五个要素之外的所有内容都是：

For the argument's sake : everything but the last five elements would be :

> head(x,n=-5)
[1] 1 2 3 4 5

正如@Martin Morgan在评论中所说，还有另外两种可能性比尾部解决方案更快，以防你必须在1亿个值的向量上进行一百万次。为了便于阅读，我会选择尾巴。

As @Martin Morgan says in the comments, there are two other possibilities which are faster than the tail solution, in case you have to carry this out a million times on a vector of 100 million values. For readibility, I'd go with tail.

test                                        elapsed    relative 
tail(x, 5)                                    38.70     5.724852     
x[length(x) - (4:0)]                           6.76     1.000000     
x[seq.int(to = length(x), length.out = 5)]     7.53     1.113905     

基准代码：

require(rbenchmark)
x <- 1:1e8
do.call(
  benchmark,
  c(list(
    expression(tail(x,5)),
    expression(x[seq.int(to=length(x), length.out=5)]),
    expression(x[length(x)-(4:0)])
  ),  replications=1e6)
)

这篇关于获取向量的最后n个元素。有没有比使用length（）函数更好的方法？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！