R：如何在重叠的时间段平均(R: how to average within overlapping

我最近发布了类似的问题在这里，这是一个有点过于简单，虽然。因此，在这里，我们又来了：

假设我有一个数据帧（下面dput输出）（在本实施例5，在真实数据的许多更多）与许多不同的变量的时间序列数据：

          date          a  b  c  d  e
1  2009-10-01 00:00:00 10 20 30 40 50
2  2009-10-01 01:00:00 11 21 31 41 51
3  2009-10-01 02:00:00 12 22 32 42 52
4  2009-10-01 03:00:00 13 23 33 43 53
5  2009-10-01 04:00:00 14 24 34 44 54
6  2009-10-01 05:00:00 15 25 35 45 55
7  2009-10-01 06:00:00 16 26 36 46 56
8  2009-10-01 07:00:00 17 27 37 47 57
9  2009-10-01 08:00:00 18 28 38 48 58
10 2009-10-01 09:00:00 19 29 39 49 59
11 2009-10-01 10:00:00 20 30 40 50 60
12 2009-10-01 11:00:00 21 31 41 51 61
13 2009-10-01 12:00:00 22 32 42 52 62
14 2009-10-01 13:00:00 23 33 43 53 63
15 2009-10-01 14:00:00 24 34 44 54 64
16 2009-10-01 15:00:00 25 35 45 55 65
17 2009-10-01 16:00:00 26 36 46 56 66
18 2009-10-01 17:00:00 27 37 47 57 67
19 2009-10-01 18:00:00 28 38 48 58 68
20 2009-10-01 19:00:00 29 39 49 59 69
21 2009-10-01 20:00:00 30 40 50 60 70
22 2009-10-01 21:00:00 31 41 51 61 71
23 2009-10-01 22:00:00 32 42 52 62 72
24 2009-10-01 23:00:00 33 43 53 63 73
25 2009-10-02 00:00:00 34 44 54 64 74

而另一个数据帧“事件”与开始和停止日期定义不同的时间段（3这里，有更多的真实数据）：

   id       start                stop
1 AGH 2009-10-01 02:00:00 2009-10-01 04:00:00
2 TRG 2009-10-01 03:00:00 2009-10-01 10:00:00
3 ZUH 2009-10-01 03:00:00 2009-10-01 20:00:00

我想获得不同的这样的活动中变量的平均值的表：

   id avg(y.a) avg(y.b) avg(y.c) avg(y.d) avg(y.e)
1 AGH     13.0     23.0     33.0     43.0     53.0
2 TRG     16.5     26.5     36.5     46.5     56.5
3 ZUH     21.5     31.5     41.5     51.5     61.5

我从我以前的帖子了解到，我可以用sqldf包和一个相当简单的SQL语句做到这一点：

means <- sqldf("
+     SELECT x.id, avg(y.a), avg(y.b), avg(y.c), avg(y.d), avg(y.e) 
+     FROM events as x, data as y 
+     WHERE y.date between x.start and x.stop 
+     GROUP BY x.id 
+ ")

然而，作为真正的数据包含更多的列来平均水平，这是在我不得不过程中的各种文件名称不同，输入的所有列名到SQL语句就显得有点乏味。

因此，我宁愿在R，其中I可以简单地通过它们的数量（数据[2：100]）是指列的溶液的困难是，虽然，所述的时间段是不连续的和重叠和ID是字符串。

任何想法如何做到这一点，将不胜感激！

dput（数据）

structure(list(date = structure(c(1254348000, 1254351600, 1254355200, 
1254358800, 1254362400, 1254366000, 1254369600, 1254373200, 1254376800, 
1254380400, 1254384000, 1254387600, 1254391200, 1254394800, 1254398400, 
1254402000, 1254405600, 1254409200, 1254412800, 1254416400, 1254420000, 
1254423600, 1254427200, 1254430800, 1254434400), class = c("POSIXct", 
"POSIXt"), tzone = "Europe/Berlin"), a = 10:34, b = 20:44, c = 30:54, 
    d = 40:64, e = 50:74), .Names = c("date", "a", "b", "c", 
"d", "e"), row.names = c(NA, -25L), class = "data.frame")

dput（事件）

structure(list(id = structure(1:3, .Label = c("AGH", "TRG", "ZUH"
), class = "factor"), start = structure(c(1254355200, 1254358800, 
1254358800), class = c("POSIXct", "POSIXt"), tzone = "Europe/Berlin"), 
    stop = structure(c(1254362400, 1254384000, 1254420000), class = c("POSIXct", 
    "POSIXt"), tzone = "Europe/Berlin")), .Names = c("id", "start", 
"stop"), row.names = c(NA, -3L), class = "data.frame")

Answer 1:

最根本的问题是由于数据不归的事实; 然而，短把它变成长篇的，我们可以动态生成的SQL语句：

 library(sqldf) sql <- paste("select id, ", toString(sprintf("avg(y.%s)", names(data)[-1])), "from events as x, data as y where y.date between x.start and x.stop group by x.id") sqldf(sql)

作为替代方案，我们展示使用melt在reshape2包来将数据转换到长形式， data_long ，处理它，得到means.long和使用其转换回宽形式dcast ：

 library(reshape2) data_long <- melt(data, id.vars = "date") means_long <- sqldf(" SELECT x.id, y.variable, avg(value) FROM events as x, data_long as y WHERE y.date between x.start and x.stop GROUP BY x.id, y.variable ") means <- dcast(id ~ variable, data = means_long, value.var = "avg(value)")

Answer 2:

>  t( sapply(events$id , function(id) lapply (
            data[ data[["date"]] >= events[ events[['id']]==id, 'start'] & 
                  data[["date"]] <= events[ events[['id']]==id, 'stop' ] ,  -1 ], 
            mean) ) )
     a    b    c    d    e   
[1,] 13   23   33   43   53  
[2,] 16.5 26.5 36.5 46.5 56.5
[3,] 21.5 31.5 41.5 51.5 61.5
#  Or if you prefer:
t( sapply(events$id , function(id) data.frame( 
                       id=as.character(id), 
                       lapply (data[ data[["date"]] >= events[events[['id']]==id, 'start'] &  
                                     data[["date"]] <= events[ events[['id']]==id, 'stop' ] , -1 ],
                               mean) ,stringsAsFactors=FALSE) ) )
     id    a    b    c    d    e   
[1,] "AGH" 13   23   33   43   53  
[2,] "TRG" 16.5 26.5 36.5 46.5 56.5
[3,] "ZUH" 21.5 31.5 41.5 51.5 61.5