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，我想要 Python 中 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，因此:

<预><代码>>>>x = 范围(10)>>>X[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]>>>x[-5:][5, 6, 7, 8, 9]>>>

在 R 中执行此操作的最佳方法是什么?有没有比我目前使用 length() 函数更简洁的方法?

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

顺便说一下，这个问题与时间序列分析有关，其中仅处理最近的数据通常很有用.

解决方案

查看 ?tail 和 ?head 一些方便的功能:

>x <- 1:10>尾巴(x,5)[1] 6 7 8 9 10

为了论证:除了最后五个元素之外的所有元素都是:

>头(x，n = -5)[1] 1 2 3 4 5

正如@Martin Morgan 在评论中所说的那样，还有两种比尾部解决方案更快的可能性，以防您必须在 1 亿个值的向量上执行一百万次.为了可读性，我会用尾巴.

测试已用相对尾(x，5)38.70 5.724852x[长度(x) - (4:0)] 6.76 1.000000x[seq.int(to = length(x), length.out = 5)] 7.53 1.113905

基准代码:

require(rbenchmark)x <- 1:1e8打电话(基准，c(列表(表达式(尾(x，5))，表达式(x[seq.int(to=length(x), length.out=5)]),表达式(x[长度(x)-(4:0)])), 复制=1e6))

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]
>>>

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
> 

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

解决方案

see ?tail and ?head for some convenient functions:

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

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

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

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     

benchmarking code :

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)
)

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