具有基于时间的窗口的不规则时间序列上的优化滚动函数

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问题描述

有没有办法使用 rollapply(来自 zoo 包或类似的东西)优化函数(rollmean、rollmedian 等)来计算滚动具有基于时间的窗口的函数，而不是基于多个观察的函数?我想要的很简单:对于不规则时间序列中的每个元素，我想计算一个具有 N 天窗口的滚动函数.也就是说，该窗口应包括当前观察前 N 天的所有观察.时间序列也可能包含重复项.

下面是一个例子.给定以下时间序列:

 日期值2011 年 1 月 11 日 52011 年 1 月 11 日 42011 年 1 月 11 日 22011 年 8 月 11 日 12011 年 11 月 13 日 02011 年 11 月 14 日 02011 年 15 月 11 日 02011 年 11 月 18 日 121/11/2011 42011 年 5 月 12 日 3

具有 5 天窗口的滚动中位数，向右对齐，应进行以下计算:

>C(中位数(c(5))，中位数(c(5,4))，中位数(c(5,4,2))，中位数(c(1))，中位数(c(1,0))，中位数(c(0,0))，中位数(c(0,0,0))，中位数(c(0,0,0,1))，中位数(c(1,4))，中位数(c(3)))[1] 5.0 4.5 4.0 1.0 0.5 0.0 0.0 0.0 2.5 3.0

我已经找到了一些解决方案，但它们通常很棘手，这通常意味着缓慢.我设法实现了自己的滚动函数计算.问题在于，对于很长的时间序列，中位数(rollmedian)的优化版本可能会产生巨大的时间差异，因为它考虑了窗口之间的重叠.我想避免重新实现它.我怀疑rollapply参数有一些技巧可以使它起作用，但我无法弄清楚.提前感谢您的帮助.

解决方案

截至 v1.9.8 版(CRAN 2016 年 11 月 25 日)，data.table 已经获得了执行 non-equi joins 的能力，可以在这里使用.

OP 已请求

对于不规则时间序列中的每个元素，我想计算一个具有 N 天窗口的滚动函数.也就是说，窗口应该包括当前 N 天之前的所有观测值观察.时间序列也可能包含重复项.

请注意，OP 已要求包括在当前观察之前最多 N 天的所有观察.这与请求当前 day 前 N 天的所有观察结果不同.

对于后者，我希望 1/11/2011 的 one 值，即 median(c(5, 4, 2)) = 4.

显然，OP 期望基于观察的滚动窗口限制为 N 天.因此，非等连接的连接条件也要考虑行号.

库(data.table)n_days

[1] 5.0 4.5 4.0 1.0 0.5 0.0 0.0 0.0 2.5 3.0

为了完整起见，基于天的滚动窗口可能的解决方案是:

setDT(DT)[.(ud = unique(date), ld = unique(date) - n_days), on = .(date <= ud, date >= ld),中位数(as.double(值))，按 = .EACHI]

 date date V11: 2011-11-01 2011-10-27 4.02:2011-11-08 2011-11-03 1.03: 2011-11-13 2011-11-08 0.54: 2011-11-14 2011-11-09 0.05:2011-11-15 2011-11-10 0.06: 2011-11-18 2011-11-13 0.07: 2011-11-21 2011-11-16 2.58: 2011-12-05 2011-11-30 3.0

`数据`

库(data.table)DT

Is there some way to use rollapply (from zoo package or something similar) optimized functions (rollmean, rollmedian etc) to compute rolling functions with a time-based window, instead of one based on a number of observations? What I want is simple: for each element in an irregular time series, I want to compute a rolling function with a N-days window. That is, the window should include all the observations up to N days before the current observation. Time series may also contain duplicates.

Here follows an example. Given the following time series:

      date  value
 1/11/2011      5
 1/11/2011      4
 1/11/2011      2
 8/11/2011      1
13/11/2011      0
14/11/2011      0
15/11/2011      0
18/11/2011      1
21/11/2011      4
 5/12/2011      3

A rolling median with a 5-day window, aligned to the right, should result in the following calculation:

> c(
    median(c(5)),
    median(c(5,4)),
    median(c(5,4,2)),
    median(c(1)),
    median(c(1,0)),
    median(c(0,0)),
    median(c(0,0,0)),
    median(c(0,0,0,1)),
    median(c(1,4)),
    median(c(3))
   )

 [1] 5.0 4.5 4.0 1.0 0.5 0.0 0.0 0.0 2.5 3.0

I already found some solutions out there but they are usually tricky, which usually means slow. I managed to implement my own rolling function calculation. The problem is that for very long time series the optimized version of median (rollmedian) can make a huge time difference, since it takes into account the overlap between windows. I would like to avoid reimplementing it. I suspect there are some trick with rollapply parameters that will make it work, but I cannot figure it out. Thanks in advance for the help.

解决方案

As of version v1.9.8 (on CRAN 25 Nov 2016), data.table has gained the ability to perform non-equi joins which can be used here.

The OP has requested

Note that the OP has requested to include all the observations up to N days before the current observation. This is different to request all the observations up to N days before the current day.

For the latter, I would expect one value for 1/11/2011, i.e., median(c(5, 4, 2)) = 4.

Apparently, the OP expects an observation-based rolling window which is limited to N days. Therefore, the join conditions of the non-equi join have to consider the row number as well.

library(data.table)
n_days <- 5L
setDT(DT)[, rn := .I][
  .(ur = rn, ud = date, ld = date - n_days),
  on = .(rn <= ur, date <= ud, date >= ld),
  median(as.double(value)), by = .EACHI]$V1

For the sake of completeness, a possible solution for the day-based rolling window could be:

setDT(DT)[.(ud = unique(date), ld = unique(date) - n_days), on = .(date <= ud, date >= ld),
   median(as.double(value)), by = .EACHI]

`Data`

library(data.table)
DT <- fread("      date  value
 1/11/2011      5
 1/11/2011      4
 1/11/2011      2
 8/11/2011      1
13/11/2011      0
14/11/2011      0
15/11/2011      0
18/11/2011      1
21/11/2011      4
 5/12/2011      3")[
   # coerce date from character string to integer date class
   , date := as.IDate(date, "%d/%m/%Y")]

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