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以相同的标题跟进我之前的问题，我有一个长期的分小时数据，我想以各种方式汇总数据。我想根据一天中的时间进行聚合，但也要根据聚合的组合进行聚合，例如，day-type-hourly（即周日凌晨 1 点、周日凌晨 2 点等）。另一个例子是：周末或工作日每小时。

下面的示例显示了我所做的两种聚合。我已经做到了。所以我最终得到了两个动物园对象。我接下来要做的是将聚合合并到原始数据中，这样我就可以比较聚合的错误。这就是我现在卡住的地方。

请注意，我没有使用上一个问题中的解决方案，因为我想要聚合的灵活性。

这是显示到目前为止我尝试过的代码片段。任何帮助将不胜感激。

library(zoo)
Lines <- "Index,light.kw
2013-06-14 13:00:00,3.436
2013-06-14 13:15:00,3.327
2013-06-14 13:30:00,3.319
2013-06-14 13:45:00,3.308
2013-06-14 14:00:00,3.458
2013-06-14 14:15:00,3.452
2013-06-14 14:30:00,3.445
2013-06-14 14:45:00,3.469
2013-06-14 15:00:00,3.468
2013-06-14 15:15:00,3.427
2013-06-14 15:30:00,3.168
2013-06-14 15:45:00,2.383
2013-06-15 13:00:00,0.555
2013-06-15 13:15:00,0.555
2013-06-15 13:30:00,0.555
2013-06-15 13:45:00,0.555
2013-06-15 14:00:00,0.555
2013-06-15 14:15:00,0.555
2013-06-15 14:30:00,0.555
2013-06-15 14:45:00,0.719
2013-06-15 15:00:00,0.976
2013-06-15 15:15:00,0.981
2013-06-15 15:30:00,1.116
2013-06-15 15:45:00,0.59"
con <- textConnection(Lines)
z <- read.zoo(con, header=TRUE, sep=",",
     format="%Y-%m-%d %H:%M:%S", FUN=as.POSIXct)
close(con)

index.hourly = format(index(z), "%H")
z.hourly = aggregate(z, index.hourly, mean)
z.hourly
merge(z,z.hourly)

index.dayhour = format(index(z), "%w %H")
z.dayhour = aggregate(z, index.dayhour, mean)
z.dayhour
merge(z,z.dayhour)
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1 回答 1

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根据上面DWin 的建议，这是我找到的解决方案。请注意，DWin建议的与中间列合并在 zoo 中不起作用，因此此解决方案涉及将 zoo 对象转换回数据框并将合并作为数据框。这里是：

library(zoo)
Lines <- "Index,light.kw
2013-06-14 13:00:00,3.436
2013-06-14 13:15:00,3.327
2013-06-14 13:30:00,3.319
2013-06-14 13:45:00,3.308
2013-06-14 14:00:00,3.458
2013-06-14 14:15:00,3.452
2013-06-14 14:30:00,3.445
2013-06-14 14:45:00,3.469
2013-06-14 15:00:00,3.468
2013-06-14 15:15:00,3.427
2013-06-14 15:30:00,3.168
2013-06-14 15:45:00,2.383
2013-06-15 13:00:00,0.555
2013-06-15 13:15:00,0.555
2013-06-15 13:30:00,0.555
2013-06-15 13:45:00,0.555
2013-06-15 14:00:00,0.555
2013-06-15 14:15:00,0.555
2013-06-15 14:30:00,0.555
2013-06-15 14:45:00,0.719
2013-06-15 15:00:00,0.976
2013-06-15 15:15:00,0.981
2013-06-15 15:30:00,1.116
2013-06-15 15:45:00,0.59"
con <- textConnection(Lines)
z <- read.zoo(con, header=TRUE, sep=",",
     format="%Y-%m-%d %H:%M:%S", FUN=as.POSIXct)
close(con)

# make the index for aggregation
index.hourly <- format(index(z), "%H")
# make the aggregate
z.hourly = aggregate(z, index.hourly, mean, na.rm=T)

# make a data frame from the original zoo,
# but the data frame must include the index.hourly
# so that later we can merge the data frame based
# on this index.
# First, make a zoo object of the index and then
# merge this with the original zoo.
z.index.hourly = zoo(index.hourly,index(z))
z.with.index = merge(z,z.index.hourly)
# make a dataframe of the last zoo
df1 = as.data.frame(z.with.index)
# add the index of the df1 (which is the timestamp) as a column
# as we will need the timestamp to rebuild the zoo object.
df1$Index = row.names(df1)

# make a dataframe of the aggregate zoo
df2 = as.data.frame(z.hourly)
df2$Index = row.names(df2)

# merge the two data frame
df3 = merge(df1,df2,by.x="z.index.hourly",by.y="Index",all.x=T)
df3 = df3[order(df3$Index),]
summary(df3)

# make a zoo object containing the original data and the aggregate
z.merged.agg = zoo(df3[,c(2,4)],as.POSIXct(df3$Index, tz="GMT"))
z.merged.agg
于 2013-07-25T16:26:06.953 回答