R向量/数据帧滞后的相反功能是什么? [英] What's the opposite function to lag for an R vector/dataframe?

问题描述

我在处理R中的时间序列时遇到问题.

I have a problem dealing with time series in R.

#--------------read data

wb = loadWorkbook("Countries_Europe_Prices.xlsx") 
df = readWorksheet(wb, sheet="Sheet2")

x <- df$Year
y <- df$Index1

y <- lag(y, 1, na.pad = TRUE)
cbind(x, y)

它给了我以下输出:

        x     y
 [1,] 1974    NA
 [2,] 1975  50.8
 [3,] 1976  51.9
 [4,] 1977  54.8
 [5,] 1978  58.8
 [6,] 1979  64.0
 [7,] 1980  68.8
 [8,] 1981  73.6
 [9,] 1982  74.3
[10,] 1983  74.5
[11,] 1984  72.9
[12,] 1985  72.1
[13,] 1986  72.3
[14,] 1987  71.7
[15,] 1988  72.9
[16,] 1989  75.3
[17,] 1990  81.2
[18,] 1991  84.3
[19,] 1992  87.2
[20,] 1993  90.1

但是我希望y中的第一个值为50.8，依此类推.换句话说，我想得到一个负的滞后.我不明白，该怎么办?

But I want the first value in y to be 50.8 and so forth. In other words, I want to get a negative lag. I don't get it, how can I do it?

我的问题与该问题非常相似，但是我无法解决.我想我还是不明白解决方案...

My problem is very similar to this problem, but however I cannot solve it. I guess I still do not understand the solution(s)...

R向量/数据框中的基本滞后

推荐答案

内置的"lead"功能如何? (来自dplyr包) 它不是完全可以完成艾哈迈德(Ahmed)函数的工作吗?

How about the built-in 'lead' function? (from the dplyr package) Doesn't it do exactly the job of Ahmed's function?

cbind(x, lead(y, 1))

如果您希望能够在同一函数中计算正或负滞后，我建议使用移位"函数的较短"版本:

If you want to be able to calculate either positive or negative lags in the same function, i suggest a 'shorter' version of his 'shift' function:

shift = function(x, lag) {
  require(dplyr)
  switch(sign(lag)/2+1.5, lead(x, abs(lag)), lag(x, abs(lag)))
}

它的作用是创建2个个案，一个有滞后，另一个有领先，并根据滞后的符号选择一个个案(+1.5是一种将{-1，+1}转换为{ 1，2}替代).

What it does is creating 2 cases, one with lag the other with lead, and chooses one case depending on the sign of your lag (the +1.5 is a trick to transform a {-1, +1} into a {1, 2} alternative).

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