如何获得R时间序列中下一行与上一行之间的差异？ [英] How to get the difference between next and previous row in an R time series?

问题描述

我有一个相当大的时间序列，其中包含约14k对4个变量的观察值（日期， x ， y ， z ）。

I have a quite large time series comprising around 14k observations of 4 variables (date, x, y, z).

我怎么（相反）函数 diff（df $ vector，lag = 1）到计算当前值之间的差（ t ）和前一个（ t-1 ）），为每个值计算下一个值（ t + 1 ）和上一个值（ t-1 ）？

How can I, (contrary to the function diff( df$vector, lag = 1) which computes the difference between the current value (t) and the previous one (t-1)), calculate for each value the difference between the next value (t+1) and the previous value (t-1)?

推荐答案

因此，为了理解请求...生成一些数据：

So, to understand the request... Generate some data:

set.seed(11)
a = sample(1:10, 10)

数据给出为：

3  1  5  9  7  8  6  4  2 10

需要 T + 1与T-1 ：

T = 0 => No computation
T = 1 => 5 - 3 = 2
T = 2 => 9 - 1 = 8
...
T = 9 => 10 - 4 = 6
T = 10 => No computation

随着计算的建立...

With that being established...

#' Future Difference
#' 
#' Obtain the lagged difference between X[t+1+lag] - X[t-1-lag]
#' @param x   A \code{vec}
#' @param lag A \code{integer} indicating the lag
#' @return A \code{vec} with differences taken at T+lag v. T-lag
#' @examples
#' set.seed(11)
#' a = sample(1:10, 12)
#' fdiff(a)
fdiff = function(x, lag = 1){
  # Number of obs
  n = length(x)

  # Trigger error to prevent subset
  if(n < 2+lag){stop("`x` must be greater than `2+lag`")}

  # X_(T+1) - X_(T-1)
  x[(2+lag):n] - x[1:(n-lag-1)]
}

在 a 上调用会给出：

fdiff(a)

2  8  2 -1 -1 -4 -4  6

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