3

我最近在这里发布了一个类似的问题，不过这有点过于简单了。因此，我们又来了：

假设我有一个数据框（下面的 dput 输出），其中包含许多不同变量的时间序列数据（本例中为 5 个，实际数据中更多）：

          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

另一个数据框“事件”具有由开始和停止日期定义的不同时间段（此处为 3，实际数据中更多）：

   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

我想得到一张不同事件中变量平均值的表格，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，由于实际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的名称不同，因此在 SQL 语句中输入所有列名变得有点乏味。

因此，我更喜欢 R 中的解决方案，我可以简单地按列的编号 (data[2:100]) 引用列.

任何想法如何做到这一点将不胜感激！

输入（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")

输入（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")
4

2 回答 2

2

  1. 基本问题是由于数据没有归一化；但是，如果不把它变成长格式，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)

  2. 作为替代方案，我们展示了melt在 reshape2 包中使用 将数据转换为长格式，data_long处理它以给出means.long并将其转换回宽格式，使用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)")
于 2012-06-30T13:47:41.837 回答
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
于 2012-06-29T20:51:46.987 回答