Question

关于将每小时数据汇总到每日数据的previous question后，我想继续（a）每月汇总和（b）将每月汇总合并到原始数据框中。

我的原始数据框如下所示：

Lines <- "Date,Outdoor,Indoor
01/01/2000 01:00,30,25
01/01/2000 02:00,31,26
01/01/2000 03:00,33,24
02/01/2000 01:00,29,25
02/01/2000 02:00,27,26
02/01/2000 03:00,39,24
12/01/2000 02:00,27,26
12/01/2000 03:00,39,24
12/31/2000 23:00,28,25"

我的previous question已经回复了每日汇总数据，然后我可以找到从那里生成每月汇总数据的方法，如下所示：

Lines <- "Date,Month,OutdoorAVE
01/01/2000,Jan,31.33
02/01/2000,Feb,31.67
12/01/2000,Dec,31.33"

其中OutdoorAVE是每日最低和最高室外温度的月平均值。我最终想要的是这样的：

Lines <- "Date,Outdoor,Indoor,Month,OutdoorAVE
01/01/2000 01:00,30,25,Jan,31.33
01/01/2000 02:00,31,26,Jan,31.33
01/01/2000 03:00,33,24,Jan,31.33
02/01/2000 01:00,29,25,Feb,31.67
02/01/2000 02:00,27,26,Feb,31.67
02/01/2000 03:00,39,24,Feb,31.67
12/01/2000 02:00,27,26,Dec,31.33
12/01/2000 03:00,39,24,Dec,31.33
12/31/2000 23:00,28,25,Dec,31.33"

我不知道如何做到这一点。非常感谢任何帮助。

Answer 1

尝试ave，例如POSIXlt来提取月份：

zz <- textConnection(Lines)
Data <- read.table(zz,header=T,sep=",",stringsAsFactors=F)
close(zz)

Data$Month <- strftime(
     as.POSIXlt(Data$Date,format="%m/%d/%Y %H:%M"),
     format='%b')
Data$outdoor_ave <- ave(Data$Outdoor,Data$Month,FUN=mean)

给予：

> Data
              Date Outdoor Indoor Month outdoor_ave
1 01/01/2000 01:00      30     25   Jan    31.33333
2 01/01/2000 02:00      31     26   Jan    31.33333
3 01/01/2000 03:00      33     24   Jan    31.33333
4 02/01/2000 01:00      29     25   Feb    31.66667
5 02/01/2000 02:00      27     26   Feb    31.66667
6 02/01/2000 03:00      39     24   Feb    31.66667
7 12/01/2000 02:00      27     26   Dec    31.33333
8 12/01/2000 03:00      39     24   Dec    31.33333
9 12/31/2000 23:00      28     25   Dec    31.33333

编辑：然后如上图所示，只计算数据中的月份并使用合并：

zz <- textConnection(Lines2) # Lines2 is the aggregated data
Data2 <- read.table(zz,header=T,sep=",",stringsAsFactors=F)
close(zz)

> merge(Data,Data2[-1],all=T)
  Month             Date Outdoor Indoor OutdoorAVE
1   Dec 12/01/2000 02:00      27     26      31.33
2   Dec 12/01/2000 03:00      39     24      31.33
3   Dec 12/31/2000 23:00      28     25      31.33
4   Feb 02/01/2000 01:00      29     25      31.67
5   Feb 02/01/2000 02:00      27     26      31.67
6   Feb 02/01/2000 03:00      39     24      31.67
7   Jan 01/01/2000 01:00      30     25      31.33
8   Jan 01/01/2000 02:00      31     26      31.33
9   Jan 01/01/2000 03:00      33     24      31.33

Answer 2

这与您的问题相关，但您可能希望使用RSQLite和单独的表来代替各种聚合值，并使用简单的SQL命令连接表。如果您使用多种聚合，您的数据框很容易变得庞大和丑陋。

Answer 3

这是一个zoo / xts解决方案。请注意，此处Month是数字，因为您无法在zoo / xts对象中混合类型。

require(xts) # loads zoo too
Lines1 <- "Date,Outdoor,Indoor
01/01/2000 01:00,30,25
01/01/2000 02:00,31,26
01/01/2000 03:00,33,24
02/01/2000 01:00,29,25
02/01/2000 02:00,27,26
02/01/2000 03:00,39,24
12/01/2000 02:00,27,26
12/01/2000 03:00,39,24
12/31/2000 23:00,28,25"
con <- textConnection(Lines1)
z <- read.zoo(con, header=TRUE, sep=",",
    format="%m/%d/%Y %H:%M", FUN=as.POSIXct)
close(con)

zz <- merge(z, Month=.indexmon(z),
    OutdoorAVE=ave(z[,1], .indexmon(z), FUN=mean))
zz
#                     Outdoor Indoor Month OutdoorAVE
# 2000-01-01 01:00:00      30     25     0   31.33333
# 2000-01-01 02:00:00      31     26     0   31.33333
# 2000-01-01 03:00:00      33     24     0   31.33333
# 2000-02-01 01:00:00      29     25     1   31.66667
# 2000-02-01 02:00:00      27     26     1   31.66667
# 2000-02-01 03:00:00      39     24     1   31.66667
# 2000-12-01 02:00:00      27     26    11   31.33333
# 2000-12-01 03:00:00      39     24    11   31.33333
# 2000-12-31 23:00:00      28     25    11   31.33333

更新：如何使用两个不同的数据集获得上述结果。

Lines2 <- "Date,Month,OutdoorAVE
01/01/2000,Jan,31.33
02/01/2000,Feb,31.67
12/01/2000,Dec,31.33"
con <- textConnection(Lines2)
z2 <- read.zoo(con, header=TRUE, sep=",", format="%m/%d/%Y",
    FUN=as.POSIXct, colClasses=c("character","NULL","numeric"))
close(con)

zz2 <- na.locf(merge(z1, Month=.indexmon(z1), OutdoorAVE=z2))[index(z1)]
# same output as zz (above)

合并R中的汇总数据

3 个答案: