Question

我引用了一个recent, well-answered question，关于时间戳与data.table的匹配。

给出一组等间距的十分钟间隔：

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= "10 mins")

以最近的时间间隔匹配time的数据：

> head(test)
                         time id   amount
312   2018-01-20 00:02:14 PST  1 54.95083
8652  2018-01-20 00:54:41 PST  2 30.55580
13809 2018-01-20 01:19:27 PST  3 90.54592
586   2018-01-20 00:03:35 PST  1 79.76360
9077  2018-01-20 00:56:37 PST  2 75.53564
21546 2018-01-20 02:25:05 PST  3 36.60177

如何仅在test$time中包含距离最近的给定区间和的5分钟范围内的匹配项，以确保每个区间记录只有一个匹配项（通过{ {1}}）？

id

例如，上面的代码会产生意外结果：

setDT(test)[, time := as.POSIXct(time)][]
test[, .SD[.(time = intervals), on = .(time), roll = 'nearest'], by = id]

预期输出将是：

> head(test[, .SD[.(time = intervals), on = .(time), roll = 'nearest'], by = id], n = 10)
    id                time    amount
 1:  1 2018-01-20 00:00:00 0.8615881
 2:  1 2018-01-20 00:10:00 0.8615881
 3:  1 2018-01-20 00:20:00 0.8615881
 4:  1 2018-01-20 00:30:00 0.8615881
 5:  1 2018-01-20 00:40:00 0.8615881
 6:  1 2018-01-20 00:50:00 0.8615881
 7:  1 2018-01-20 01:00:00 0.8615881
 8:  1 2018-01-20 01:10:00 0.8615881
 9:  1 2018-01-20 01:20:00 0.8615881
10:  1 2018-01-20 01:30:00 0.8615881

请注意，如果id time amount 1: 1 2018-01-20 00:00:00 54.9508346 2: 1 2018-01-20 00:50:00 12.7618139 3: 1 2018-01-20 01:20:00 34.5093891 4: 1 2018-01-20 03:00:00 0.8615881 5: 2 2018-01-20 00:50:00 30.5557992 6: 2 2018-01-20 01:00:00 75.5356406 7: 2 2018-01-20 01:20:00 72.4465838 8: 2 2018-01-20 01:30:00 49.8718743 9: 2 2018-01-20 02:30:00 69.0175725 10: 3 2018-01-20 00:10:00 81.0468155 11: 3 2018-01-20 01:20:00 90.5459248 12: 3 2018-01-20 01:30:00 85.0054113 13: 3 2018-01-20 02:30:00 36.60177053中最接近的匹配距离超过5分钟（间隔和测试$时间之间的差异时间> 5分钟），则应在输出中完全排除记录。

如何在intervals，data.table或基数R中添加这些条件以匹配预期输出？

有关如何在输出中获得dplyr与最近匹配区间之间差异的建议也会有所帮助。希望这是有道理的。

test$time数据如下：

test

Answer 1

一种可能的解决方案（改编为my previous answer）：

ref <- CJ(id = test$id, time = intervals, unique = TRUE)

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

给出了所需的输出：

    id                time     amount
 1:  1 2018-01-20 00:00:00 54.9508346
 2:  1 2018-01-20 00:50:00 12.7618139
 3:  1 2018-01-20 01:20:00 34.5093891
 4:  1 2018-01-20 03:00:00  0.8615881
 5:  2 2018-01-20 00:50:00 30.5557992
 6:  2 2018-01-20 01:00:00 75.5356406
 7:  2 2018-01-20 01:20:00 72.4465838
 8:  2 2018-01-20 01:30:00 49.8718743
 9:  2 2018-01-20 02:30:00 69.0175725
10:  3 2018-01-20 00:10:00 81.0468155
11:  3 2018-01-20 01:20:00 90.5459248
12:  3 2018-01-20 01:30:00 85.0054113
13:  3 2018-01-20 02:30:00 36.6017705

使用过的数据：

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")

intervals <- seq(as.POSIXct("2018-01-20 00:00:00"), as.POSIXct("2018-01-20 03:00:00"), by = "10 mins")

setDT(test)[, time := as.POSIXct(time)][]

注意：我在创建intervals向量时没有使用时区，因为这为我提供了与test数据集相同的时区（time := as.POSIXct(time)将时区设置为CET我）。

按条件滚动连接

1 个答案: