我最近发布了一个类似的问题here，但有点过于简单了。因此，我们再来一次：
假设我有一个数据帧（下面的输出输出），其中包含许多不同变量的时间序列数据（本例中为5，真实数据中有更多）：
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
和另一个数据框“事件”，具有由开始和结束日期定义的不同时间段（此处为3，实际数据中有更多）：
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
我想得到一个表格，列出不同事件中变量的平均值，如下所示：
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
我从上一篇文章中了解到，我可以使用sqldf包和一个相当简单的SQL语句来做到这一点：
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
+ ")
但是，由于真实数据包含更多要平均的列，这些列在我必须处理的各种文件中命名不同，因此在SQL语句中键入所有列名会变得有点乏味。
因此我更喜欢R中的一个解决方案，我可以简单地通过它们的数字来引用列（数据[2：100]）然而，困难在于时间段是非连续的和重叠的以及ids是字符串。
任何想法如何做到这一点将非常感谢！
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")
答案 0 :(得分：2)
基本问题是由于数据未规范化;但是，如果没有将它放入长形式，我们可以动态生成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，我们会在reshape2包中使用melt来将数据转换为长格式data_long，处理它以提供means.long并转换它使用dcast返回宽屏：
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)")
答案 1 :(得分：1)
> 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