Question

考虑以下data.frame：

data_frame = structure(list(AGREEDTIME = structure(c(1451785888.76968, 1451785945.59156, 
1451786041.22135, 1451786089.17713, 1451786671.55922, 1451786853.52841, 
1451787231.03475, 1451787641.43011, 1451787999.77345, 1451788571.08314, 
1451788695.76539, 1451788769.29787, 1451788891.90181, 1451789206.47645, 
1451789653.27497, 1451789740.52194, 1451789875.7906, 1451789971.7024, 
1451790030.94949, 1451790681.31701), class = c("POSIXct", "POSIXt"
), tzone = "UTC"), EA = c(2383, 1064, 848, 68, 2665, 277, 175, 
2761, 773, 426, 1164, 600, 413, 371, 733, 259, 976, 297, 1973, 
1022), FW = structure(c(2L, 2L, 1L, 1L, 3L, 1L, 2L, 2L, 2L, 3L, 
1L, 3L, 3L, 1L, 1L, 3L, 2L, 3L, 3L, 1L), .Label = c("NZNR", "WZNI", 
"HSCW"), class = "factor"), CP = structure(c(1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), .Label = "OZU/NQV", class = "factor")), .Names = c("AGREEDTIME", 
"EA", "FW", "CP"), row.names = c(NA, -20L), class = "data.frame")

或以更易读的形式：

            AGREEDTIME   EA   FW      CP
1  2016-01-03 01:51:28 2383 WZNI OZU/NQV
2  2016-01-03 01:52:25 1064 WZNI OZU/NQV
3  2016-01-03 01:54:01  848 NZNR OZU/NQV
4  2016-01-03 01:54:49   68 NZNR OZU/NQV
5  2016-01-03 02:04:31 2665 HSCW OZU/NQV
6  2016-01-03 02:07:33  277 NZNR OZU/NQV
7  2016-01-03 02:13:51  175 WZNI OZU/NQV
8  2016-01-03 02:20:41 2761 WZNI OZU/NQV
9  2016-01-03 02:26:39  773 WZNI OZU/NQV
10 2016-01-03 02:36:11  426 HSCW OZU/NQV
11 2016-01-03 02:38:15 1164 NZNR OZU/NQV
12 2016-01-03 02:39:29  600 HSCW OZU/NQV
13 2016-01-03 02:41:31  413 HSCW OZU/NQV
14 2016-01-03 02:46:46  371 NZNR OZU/NQV
15 2016-01-03 02:54:13  733 NZNR OZU/NQV
16 2016-01-03 02:55:40  259 HSCW OZU/NQV
17 2016-01-03 02:57:55  976 WZNI OZU/NQV
18 2016-01-03 02:59:31  297 HSCW OZU/NQV
19 2016-01-03 03:00:30 1973 HSCW OZU/NQV
20 2016-01-03 03:11:21 1022 NZNR OZU/NQV

现在，我的目标是，对于这个矩阵的每一行，

对满足以下所有行的EA值求和：

AGREEDTIME在窗口中[AGREEDTIME（i）-3600sec，AGREEDTIME（i）-1800sec]
FW == FW [i]
CP == CP [i]

使用天真的R代码，实际结果应为：

slow_function <- function(ind, data_frame){
    index     = data_frame[ind, ]
    index_set = which(data_frame$AGREEDTIME > index$AGREEDTIME - 3600 & 
            data_frame$AGREEDTIME < index$AGREEDTIME - 1800 &
            data_frame$FW == index$FW &
            data_frame$CP == index$CP)
    sum(data_frame$EA[index_set])
}
data_frame_results = data.frame(data_frame, results = sapply(1:nrow(data_frame), slow_function, data_frame = data_frame))

现在，我正在尝试使用data.table来获得相同的结果。我的第一个尝试是：

data_table  <- data.table(data_frame)
setkey(data_table, FW, CP, AGREEDTIME)

less_slow_function <- function(ind, data_table){
    ATm = data_table$AGREEDTIME[ind] - 3600
    ATM = data_table$AGREEDTIME[ind] - 1800
    fw  = data_table$FW[ind]
    cp  = data_table$CP[ind]
    as.numeric(data_table[J(fw, cp)][AGREEDTIME > ATm & AGREEDTIME < ATM, 'EA'][,lapply(.SD, sum)])
}
data_table_results = data.table(data_table, results = sapply(1:nrow(data_table), less_slow_function, data_table = data_table))

给出了所需的数字。我的问题是：使用data.table魔法获得相同结果的方法是否比我天真的解决方案更好？

data.table中的条件滚动时间窗口总和

0 个答案: