通过id将数据与最接近的时间值匹配

Question

我已经生成了一系列小时时间戳:

intervals <- seq(as.POSIXct("2018-01-20 00:00:00", tz = 'America/Los_Angeles'), as.POSIXct("2018-01-20 03:00:00", tz = 'America/Los_Angeles'), by="hour")

> intervals
[1] "2018-01-20 00:00:00 PST" "2018-01-20 01:00:00 PST" "2018-01-20 02:00:00 PST"
[4] "2018-01-20 03:00:00 PST" 

给定具有杂乱且间隔不均匀的时间戳的数据集,如何将该数据集中的时间值与最接近的每小时时间戳匹配id,并删除其间的其他时间戳？例如:

> test
                         time      id     amount
312   2018-01-20 00:02:14 PST       1 54.9508346
8652  2018-01-20 00:54:41 PST       2 30.5557992
13809 2018-01-20 01:19:27 PST       3 90.5459248
586   2018-01-20 00:03:35 PST       1 79.7635973
9077  2018-01-20 00:56:37 PST       2 75.5356406
21546 2018-01-20 02:25:05 PST       3 36.6017705
7275  2018-01-20 00:47:45 PST       1 12.7618139
12768 2018-01-20 01:15:30 PST       2 72.4465838
1172  2018-01-20 00:08:01 PST       3 81.0468155
24106 2018-01-20 03:04:10 PST       1  0.8615881
14464 2018-01-20 01:25:04 PST       2 49.8718743
15344 2018-01-20 01:29:30 PST       3 85.0054113
14255 2018-01-20 01:23:22 PST       1 34.5093891
21565 2018-01-20 02:25:40 PST       2 69.0175725
15602 2018-01-20 01:31:32 PST       3 61.8602426

会产生:

> output
             interval id     amount
1 2018-01-20 01:00:00  1 12.7618139
2          2018-01-20  1 54.9508346
3 2018-01-20 03:00:00  1  0.8615881
4 2018-01-20 01:00:00  2 75.5356400
5 2018-01-20 02:00:00  2 69.0175700
6          2018-01-20  3 81.0468200
7 2018-01-20 01:00:00  3 90.5459200
8 2018-01-20 02:00:00  3 36.6017700

我知道存在可能的解决方案 data.table

setDT(reference)[data, refvalue, roll = "nearest", on = "datetime"]

有roll = nearest,但是如何在intervals每个idin中找到最接近的匹配test并保留amount属性？

任何建议,将不胜感激!以下是示例数据:

 dput(test)
structure(list(time = c("2018-01-20 00:02:14 PST", "2018-01-20 00:54:41 PST", 
"2018-01-20 01:19:27 PST", "2018-01-20 00:03:35 PST", "2018-01-20 00:56:37 PST", 
"2018-01-20 02:25:05 PST", "2018-01-20 00:47:45 PST", "2018-01-20 01:15:30 PST", 
"2018-01-20 00:08:01 PST", "2018-01-20 03:04:10 PST", "2018-01-20 01:25:04 PST", 
"2018-01-20 01:29:30 PST", "2018-01-20 01:23:22 PST", "2018-01-20 02:25:40 PST", 
"2018-01-20 01:31:32 PST"), id = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 
1, 2, 3, 1, 2, 3), amount = c(54.9508346011862, 30.5557992309332, 
90.5459248460829, 79.763597343117, 75.5356406327337, 36.6017704829574, 
12.7618139144033, 72.4465838400647, 81.0468154959381, 0.861588073894382, 
49.8718742514029, 85.0054113194346, 34.5093891490251, 69.0175724914297, 
61.8602426256984)), .Names = c("time", "id", "amount"), row.names = c(312L, 
8652L, 13809L, 586L, 9077L, 21546L, 7275L, 12768L, 1172L, 24106L, 
14464L, 15344L, 14255L, 21565L, 15602L), class = "data.frame")

Answer 1

另一种选择是加入里面j有data.table:

# convert 'test' to a 'data.table' first with 'setDT'
# and convert the 'time'-column tot a datetime format
setDT(test)[, time := as.POSIXct(time)][]

# preform the join
test[, .SD[.(time = intervals), on = .(time), roll = 'nearest'], by = id]

这使:

    id                time     amount
 1:  1 2018-01-20 00:00:00 54.9508346
 2:  1 2018-01-20 01:00:00 12.7618139
 3:  1 2018-01-20 02:00:00 34.5093891
 4:  1 2018-01-20 03:00:00  0.8615881
 5:  2 2018-01-20 00:00:00 30.5557992
 6:  2 2018-01-20 01:00:00 75.5356406
 7:  2 2018-01-20 02:00:00 69.0175725
 8:  2 2018-01-20 03:00:00 69.0175725
 9:  3 2018-01-20 00:00:00 81.0468155
10:  3 2018-01-20 01:00:00 90.5459248
11:  3 2018-01-20 02:00:00 36.6017705
12:  3 2018-01-20 03:00:00 36.6017705

另外,在上述方法的一些amount-值被分配给一个以上的time通过id.如果你不想那样,只想保留最接近a的那个,time你可以按如下方式改进方法:

test[, r := rowid(id)
     ][, .SD[.(time = intervals)
             , on = .(time)
             , roll = 'nearest'
             , .(time, amount, r, time_diff = abs(x.time - i.time))
             ][, .SD[which.min(time_diff)], by = r]
       , by = id][, c('r','time_diff') := NULL][]

这使:

    id                time     amount
 1:  1 2018-01-20 00:00:00 54.9508346
 2:  1 2018-01-20 01:00:00 12.7618139
 3:  1 2018-01-20 02:00:00 34.5093891
 4:  1 2018-01-20 03:00:00  0.8615881
 5:  2 2018-01-20 00:00:00 30.5557992
 6:  2 2018-01-20 01:00:00 75.5356406
 7:  2 2018-01-20 02:00:00 69.0175725
 8:  3 2018-01-20 00:00:00 81.0468155
 9:  3 2018-01-20 01:00:00 90.5459248
10:  3 2018-01-20 02:00:00 36.6017705