对不规则时间序列的定期分析

发布于 2024-12-07 08:51:55 字数 905 浏览 0 评论 0原文

我有一个不规则的时间序列（R 中的 xts），我想对其应用一些时间窗口。例如，给定如下所示的时间序列，我想计算从 2009-09-22 00:00:00 开始的每个离散 3 小时窗口中有多少个观测值：

library(lubridate)
s <- xts(c("OK", "Fail", "Service", "OK", "Service", "OK"),
         ymd_hms(c("2009-09-22 07:43:30", "2009-10-01 03:50:30",
                   "2009-10-01 08:45:00", "2009-10-01 09:48:15",
                   "2009-11-11 10:30:30", "2009-11-11 11:12:45")))

我显然不能使用 period.apply() 或 split() 来执行此操作，因为它们会忽略没有观察的周期，而且我不能给它一个开始时间。

如果我一次汇总 3 天，我对简单计数问题的期望输出（当然，我的实际任务在每个部分都更加复杂！）将是这样的：

2009-09-22    1
2009-09-25    0
2009-09-28    0
2009-10-01    3
2009-10-04    0
2009-10-07    0
2009-10-10    0
2009-10-13    0
2009-10-16    0
2009-10-19    0
2009-10-22    0
2009-10-25    0
2009-10-28    0
2009-10-31    0
2009-11-03    0
2009-11-06    0
2009-11-09    2

感谢您的指导。

原文

I have an irregular time series (xts in R) that I want to apply some time-windowing to. For example, given a time series like the following, I want to compute things like how many observations there are in each discrete 3-hour window, starting from 2009-09-22 00:00:00:

library(lubridate)
s <- xts(c("OK", "Fail", "Service", "OK", "Service", "OK"),
         ymd_hms(c("2009-09-22 07:43:30", "2009-10-01 03:50:30",
                   "2009-10-01 08:45:00", "2009-10-01 09:48:15",
                   "2009-11-11 10:30:30", "2009-11-11 11:12:45")))

I apparently can't use period.apply() or split() to do it, because those will omit periods with no observations, and I can't give it a starting time.

My desired output for the simple counting problem (though, of course, my real tasks are more complicated with each segment!) would be something like this if I aggregated 3 days at a time:

2009-09-22    1
2009-09-25    0
2009-09-28    0
2009-10-01    3
2009-10-04    0
2009-10-07    0
2009-10-10    0
2009-10-13    0
2009-10-16    0
2009-10-19    0
2009-10-22    0
2009-10-25    0
2009-10-28    0
2009-10-31    0
2009-11-03    0
2009-11-06    0
2009-11-09    2

Thanks for any guidance.

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柒夜笙歌凉 2024-12-14 08:51:55

使用 align.time 将 s 的索引放入您感兴趣的周期中。然后使用 period.apply 查找的长度每个 3 小时窗口。然后将其与具有所需所有索引值的空 xts 对象合并。

# align index into 3-hour blocks
a <- align.time(s, n=60*60*3)
# find the number of obs in each block
count <- period.apply(a, endpoints(a, "hours", 3), length)
# create an empty xts object with the desired index
e <- xts(,seq(start(a),end(a),by="3 hours"))
# merge the counts with the empty object and fill with zeros
out <- merge(e,count,fill=0)

Use align.time to put the index of s into the periods you're interested in. Then use period.apply to find the length of each 3-hour window. Then merge it with an empty xts object that has all the index values you want.

# align index into 3-hour blocks
a <- align.time(s, n=60*60*3)
# find the number of obs in each block
count <- period.apply(a, endpoints(a, "hours", 3), length)
# create an empty xts object with the desired index
e <- xts(,seq(start(a),end(a),by="3 hours"))
# merge the counts with the empty object and fill with zeros
out <- merge(e,count,fill=0)

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