不规则时间序列上的滚动窗口

问题描述 投票:11回答:3

我使用xts具有不规则的时间序列的事件(帖子),我想计算每周滚动的窗口(或每两周或3天等)发生的事件数。数据如下:

                    postid
2010-08-04 22:28:07    867
2010-08-04 23:31:12    891
2010-08-04 23:58:05    901
2010-08-05 08:35:50    991
2010-08-05 13:28:02   1085
2010-08-05 14:14:47   1114
2010-08-05 14:21:46   1117
2010-08-05 15:46:24   1151
2010-08-05 16:25:29   1174
2010-08-05 23:19:29   1268
2010-08-06 12:15:42   1384
2010-08-06 15:22:06   1403
2010-08-07 10:25:49   1550
2010-08-07 18:58:16   1596
2010-08-07 21:15:44   1608

应该会产生类似的结果

                    nposts
2010-08-05 00:00:00     10
2010-08-06 00:00:00      9
2010-08-07 00:00:00      5

对于2天的窗口。我研究了rollapplyapply.rolling中的PerformanceAnalytics等,它们都假设有常规的时间序列数据。我尝试将所有时间都更改为刚发布帖子的那一天,并每天使用ddply之类的内容进行分组,这使我很亲近。但是,用户可能不会每天发布信息,因此时间序列仍然不规则。我可以用0填补空白,但这可能会使我的数据膨胀很多,而且已经很大。

我该怎么办?

r time-series xts zoo
3个回答
5
投票

这里是使用xts的解决方案:

x <- structure(c(867L, 891L, 901L, 991L, 1085L, 1114L, 1117L, 1151L, 
  1174L, 1268L, 1384L, 1403L, 1550L, 1596L, 1608L), .Dim = c(15L, 1L),
  index = structure(c(1280960887, 1280964672, 1280966285, 
  1280997350, 1281014882, 1281017687, 1281018106, 1281023184, 1281025529, 
  1281050369, 1281096942, 1281108126, 1281176749, 1281207496, 1281215744),
  tzone = "", tclass = c("POSIXct", "POSIXt")), class = c("xts", "zoo"),
  .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct", "POSIXt"),
  .indexTZ = "", tzone = "")
# first count the number of observations each day
xd <- apply.daily(x, length)
# now sum the counts over a 2-day rolling window
x2d <- rollapply(xd, 2, sum)
# align times at the end of the period (if you want)
y <- align.time(x2d, n=60*60*24)  # n is in seconds

4
投票

这似乎起作用:

# n = number of days
n <- 30
# w = window width. In this example, w = 7 days
w <- 7

# I will simulate some data to illustrate the procedure
data <- rep(1:n, rpois(n, 2))

# Tabulate the number of occurences per day:
# (use factor() to be sure to have the days with zero observations included)
date.table <- table(factor(data, levels=1:n))  

mat <- diag(n)
for (i in 2:w){
  dim <- n+i-1
  mat <- mat + diag(dim)[-((n+1):dim),-(1:(i-1))]
  }

# And the answer is.... 
roll.mean.7days <- date.table %*% mat

似乎不太慢(尽管mat矩阵的尺寸为n * n)。我试图将n = 30替换为n = 3000(这将创建一个9百万个元素的矩阵= 72 MB),但在我的计算机上仍然可以很快实现。对于非常大的数据集,请首先尝试一个子集。...使用Matrix包(bandSparse)中的某些函数来创建mat矩阵,也会更快。


0
投票

带有滚轮的人可以在滚动窗口上apply any R function。 OP要求的是仅在指定时间点在滚动窗口上计算函数(长度)。使用runner的用户需要指定at参数来指示应该在哪个时间点上计算输出。我们只需将时间点的矢量传递到我们在一侧创建的runner序列作为POSIXt序列即可。要使runner与时间相关,必须使用与idx对象相对应的日期指定x。窗口长度可以设置为k = "2 days"

at <- seq(as.POSIXct("2010-08-05 00:00:00"), 
          by = "1 days",
          length.out = 4)

# [1] "2010-08-05 CEST" "2010-08-06 CEST" "2010-08-07 CEST" "2010-08-08 CEST"


runner::runner(
  x = x$postid,
  k = "2 days",
  idx = x$datetime,
  at = at,
  f = length
)
# [1]  3 10  9  5

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