Question

我有30秒的CPU数据，如下所示。我想做的是将这些数据合并为5分钟和10分钟的平均值。

 dput(head(res,50))
structure(list(DATE = structure(c(1362114023, 1362114053, 1362114083, 
1362114113, 1362114143, 1362114150, 1362114173, 1362114180, 1362114203, 
1362114210, 1362114233, 1362114240, 1362114263, 1362114270, 1362114293, 
1362114300, 1362114330, 1362114360, 1362114390, 1362114420, 1362114450, 
1362114480, 1362114510, 1362114540, 1362114570, 1362114600, 1362114630, 
1362114660, 1362114690, 1362114720, 1362114750, 1362114780, 1362114810, 
1362114840, 1362114870, 1362114900, 1362114930, 1362114960, 1362114990, 
1362115020, 1362115050, 1362115080, 1362115111, 1362115141, 1362115171, 
1362115201, 1362115231, 1362115261, 1362115291, 1362115321), class = c("POSIXct", 
"POSIXt"), tzone = ""), CPU = c(30L, 29L, 28L, 29L, 27L, 10L, 
25L, 11L, 23L, 9L, 22L, 8L, 22L, 7L, 19L, 7L, 7L, 8L, 6L, 7L, 
6L, 7L, 8L, 8L, 7L, 6L, 8L, 8L, 9L, 8L, 9L, 10L, 9L, 8L, 8L, 
6L, 8L, 7L, 9L, 10L, 11L, 11L, 9L, 9L, 8L, 9L, 11L, 8L, 6L, 8L
)), .Names = c("DATE", "CPU"), row.names = c(132611L, 132612L, 
132613L, 132614L, 132615L, 131428L, 132616L, 131429L, 132617L, 
131430L, 132618L, 131431L, 132619L, 131432L, 132620L, 131433L, 
131434L, 131435L, 131436L, 131437L, 131438L, 131439L, 131440L, 
131441L, 131442L, 131443L, 131444L, 131445L, 131446L, 131447L, 
131448L, 131449L, 131450L, 131451L, 131452L, 131453L, 131454L, 
131455L, 131456L, 131457L, 131458L, 131459L, 131460L, 131461L, 
131462L, 131463L, 131464L, 131465L, 131466L, 131467L), class = "data.frame")

任何想法，我如何处理我的grunular数据？

Answer 1

Versions of this question have been asked and answered a bunch of times on stackoverflow.然而它一直被问到。希望这是一个满足大多数人需求的答案：

首先，使用处理不规则时间序列的包。它使它更容易。我喜欢xts。

library(xts)

mydata <- structure(list(DATE = structure(c(1362114023, 1362114053, 1362114083, 
1362114113, 1362114143, 1362114150, 1362114173, 1362114180, 1362114203, 
1362114210, 1362114233, 1362114240, 1362114263, 1362114270, 1362114293, 
1362114300, 1362114330, 1362114360, 1362114390, 1362114420, 1362114450, 
1362114480, 1362114510, 1362114540, 1362114570, 1362114600, 1362114630, 
1362114660, 1362114690, 1362114720, 1362114750, 1362114780, 1362114810, 
1362114840, 1362114870, 1362114900, 1362114930, 1362114960, 1362114990, 
1362115020, 1362115050, 1362115080, 1362115111, 1362115141, 1362115171, 
1362115201, 1362115231, 1362115261, 1362115291, 1362115321), class = c("POSIXct", 
"POSIXt"), tzone = ""), CPU = c(30L, 29L, 28L, 29L, 27L, 10L, 
25L, 11L, 23L, 9L, 22L, 8L, 22L, 7L, 19L, 7L, 7L, 8L, 6L, 7L, 
6L, 7L, 8L, 8L, 7L, 6L, 8L, 8L, 9L, 8L, 9L, 10L, 9L, 8L, 8L, 
6L, 8L, 7L, 9L, 10L, 11L, 11L, 9L, 9L, 8L, 9L, 11L, 8L, 6L, 8L
)), .Names = c("DATE", "CPU"), row.names = c(132611L, 132612L, 
132613L, 132614L, 132615L, 131428L, 132616L, 131429L, 132617L, 
131430L, 132618L, 131431L, 132619L, 131432L, 132620L, 131433L, 
131434L, 131435L, 131436L, 131437L, 131438L, 131439L, 131440L, 
131441L, 131442L, 131443L, 131444L, 131445L, 131446L, 131447L, 
131448L, 131449L, 131450L, 131451L, 131452L, 131453L, 131454L, 
131455L, 131456L, 131457L, 131458L, 131459L, 131460L, 131461L, 
131462L, 131463L, 131464L, 131465L, 131466L, 131467L), class = "data.frame")

mydata.xts <- xts(mydata$CPU, order.by = mydata$DATE)

然后，调整period.apply基础架构，以便在运行中轻松聚合到不同的窗口：

apply.periodly <- function (x, FUN, period, k=1, ...) 
{
  if (!require("xts")) {
    stop("Need 'xts'")
  }
  ep <- endpoints(x, on=period, k=k)
  period.apply(x, ep, FUN, ...)
}

现在，创建您的聚合。

mydata.10m <- apply.periodly(x = mydata.xts, FUN = mean, period = "minutes", k = 10)
mydata.5m <- apply.periodly(x = mydata.xts, FUN = mean, period = "minutes", k = 5)

请注意，输出时间戳将反映每个聚合窗口中的 last 输入时间戳。

mydata.10m
                     [,1]
2013-03-01 00:09:30 14.80
2013-03-01 00:19:31  8.55
2013-03-01 00:22:01  8.40

mydata.5m
                        [,1]
2013-03-01 00:04:53 19.93333
2013-03-01 00:09:30  7.10000
2013-03-01 00:14:30  8.30000
2013-03-01 00:19:31  8.80000
2013-03-01 00:22:01  8.40000

但是，您可以向上或向下舍入时间戳：

align.time.down=function(x,n){index(x)=index(x)-n;align.time(x,n)}

mydata.10m <- align.time(mydata.10m, 10*60)
mydata.10m
#                      [,1]
# 2013-03-01 00:10:00 14.80
# 2013-03-01 00:20:00  8.55
# 2013-03-01 00:30:00  8.40

mydata.5m <- align.time.down(mydata.5m, 5*60)
mydata.5m
#                         [,1]
# 2013-03-01 00:00:00 19.93333
# 2013-03-01 00:05:00  7.10000
# 2013-03-01 00:10:00  8.30000
# 2013-03-01 00:15:00  8.80000
# 2013-03-01 00:20:00  8.40000

Answer 2

您希望聚合哪些时间，以及如何报告？例如，您想汇总00:00 - 04:59或00:01 - 05:00，并在期间开始或结束时报告？

要从x：00聚合到x + 04:59并在句点开头报告，请使用floor创建向下舍入到最接近的5分钟的时间戳：

data <- structure(...)
data$DATE.5mindown <- as.POSIXct(floor(as.numeric(data$DATE) / (5 * 60)) *
  (5 * 60), origin='1970-01-01')
aggregate(CPU ~ DATE.5mindown, data, mean)
#         DATE.5mindown      CPU
# 1 2013-03-01 00:00:00 19.93333
# 2 2013-03-01 00:05:00  7.10000
# 3 2013-03-01 00:10:00  8.30000
# 4 2013-03-01 00:15:00  8.80000
# 5 2013-03-01 00:20:00  8.40000

要从x：01聚合到x + 5:00并在句点结束时报告，请使用ceiling创建四舍五入到最接近的5分钟的时间戳：

data$DATE.5minup <- as.POSIXct(ceiling(as.numeric(data$DATE) / (5 * 60)) *
  (5 * 60), origin='1970-01-01')
aggregate(CPU ~ DATE.5minup, data, mean)
#           DATE.5minup       CPU
# 1 2013-03-01 00:05:00 19.125000
# 2 2013-03-01 00:10:00  7.000000
# 3 2013-03-01 00:15:00  8.300000
# 4 2013-03-01 00:20:00  9.111111
# 5 2013-03-01 00:25:00  8.400000

如何将30个粒度数据聚合为5分钟数据

2 个答案: