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r - 匹配时间范围之前和之后的记录

转载 作者:行者123 更新时间:2023-12-04 11:44:48 24 4
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我的目标是将基于时间的两个数据表与 dplyrdata.table 连接起来,特别是在事件发生之前和之后立即获取记录。

在示例数据中,这种情况下的事件是踏板车旅行。以下是四次旅行 - 两次由踏板车 1 进行,两次由踏板车 2 进行。

> testScooter
start end id
1: 2018-01-18 22:19:13 2018-01-18 22:26:31 1
2: 2018-01-18 23:29:22 2018-01-18 23:37:53 1
3: 2018-01-18 00:22:02 2018-01-18 00:29:21 2
4: 2018-01-18 00:37:52 2018-01-18 01:06:53 2

在一个单独的表中,记录的间隔几乎相等。 id s 匹配,当旅行正在进行时,踏板车被标记为 no
> intervals
id time available charge
1 1 2018-01-18 21:31:07 yes 83
2 1 2018-01-18 21:41:07 yes 83
3 1 2018-01-18 21:51:07 yes 83
4 1 2018-01-18 22:01:07 yes 83
5 1 2018-01-18 22:11:07 yes 83
6 1 2018-01-18 22:21:07 no 83
7 1 2018-01-18 22:31:07 yes 81
8 1 2018-01-18 22:41:08 yes 81
9 1 2018-01-18 22:51:08 yes 81
10 1 2018-01-18 23:01:08 yes 81
11 1 2018-01-18 23:11:08 yes 81
12 1 2018-01-18 23:21:11 yes 81
13 1 2018-01-18 23:31:07 no 81
14 1 2018-01-18 23:41:09 yes 79
15 1 2018-01-18 23:51:07 yes 79
16 2 2018-01-18 00:01:06 yes 84
17 2 2018-01-18 00:11:06 yes 84
18 2 2018-01-18 00:21:06 yes 84
19 2 2018-01-18 00:31:05 yes 80
20 2 2018-01-18 00:41:06 no 80
21 2 2018-01-18 00:51:06 no 80
22 2 2018-01-18 01:01:06 no 80
23 2 2018-01-18 01:11:05 yes 80
24 2 2018-01-18 01:21:05 yes 80
25 2 2018-01-18 01:31:05 yes 80

我试图产生的输出如下。
> output
start end id startCharge endCharge
1: 2018-01-18 22:19:13 2018-01-18 22:26:31 1 83 81
2: 2018-01-18 23:29:22 2018-01-18 23:37:53 1 81 79
3: 2018-01-18 00:22:02 2018-01-18 00:29:21 2 84 80
4: 2018-01-18 00:37:52 2018-01-18 01:06:53 2 80 80

关于如何在时间范围之前和之后匹配最近时间的任何建议都会有所帮助,也许可以使用 lubridate::new_interval() 包中的 roll='nearest'data.table 但我不确定从哪里开始。
# Here is the sample data

library(data.table)

testScooter <- setDT(
structure(list(start = structure(c(1516313953, 1516318162, 1516234922,
1516235872), tzone = "", class = c("POSIXct", "POSIXt")), end = structure(c(1516314391,
1516318673, 1516235361, 1516237613), tzone = "", class = c("POSIXct",
"POSIXt")), id = c(1, 1, 2, 2)), .Names = c("start", "end", "id"
), row.names = c(NA, -4L), class = "data.frame"))

intervals <-
structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L),
time = structure(c(1516311067, 1516311667, 1516312267, 1516312867,
1516313467, 1516314067, 1516314667, 1516315268, 1516315868,
1516316468, 1516317068, 1516317671, 1516318267, 1516318869,
1516319467, 1516233666, 1516234266, 1516234866, 1516235465,
1516236066, 1516236666, 1516237266, 1516237865, 1516238465,
1516239065), class = c("POSIXct", "POSIXt"), tzone = "UTC"),
available = c("yes", "yes", "yes", "yes", "yes", "no", "yes",
"yes", "yes", "yes", "yes", "yes", "no", "yes", "yes", "yes",
"yes", "yes", "yes", "no", "no", "no", "yes", "yes", "yes"
), charge = c(83L, 83L, 83L, 83L, 83L, 83L, 81L, 81L, 81L,
81L, 81L, 81L, 81L, 79L, 79L, 84L, 84L, 84L, 80L, 80L, 80L,
80L, 80L, 80L, 80L)), .Names = c("id", "time", "available",
"charge"), row.names = c(NA, -25L), class = "data.frame")

最佳答案

新答案:

您可以使用双滚动连接来做到这一点:

testScooter[, startCharge := intervals[testScooter, on = .(id, time = start), roll = Inf, x.charge]
][, endCharge := intervals[testScooter, on = .(id, time = end), roll = -Inf, x.charge]][]

这给出了所需的结果:

                 start                 end id startCharge endCharge
1: 2018-01-18 23:19:13 2018-01-18 23:26:31 1 83 81
2: 2018-01-19 00:29:22 2018-01-19 00:37:53 1 81 79
3: 2018-01-18 01:22:02 2018-01-18 01:29:21 2 84 80
4: 2018-01-18 01:37:52 2018-01-18 02:06:53 2 80 80


这是做什么的:
  • roll = Infintervals
  • 之前查找 start 中的最后一个观察
  • roll = -Infintervals
  • 之后寻找 end 中的第一个观察值

    另请参阅有关为什么新答案更好的注释。

    旧答案:
    testScooter[intervals, on = .(id, start = time), roll = -Inf, startCharge := i.charge
    ][intervals, on = .(id, end = time), roll = Inf, endCharge := i.charge][]

    注意:

    正如@Frank 指出的 here on Github ,当有多个匹配项时, data.table 返回 i 中的最后一个匹配项,旧答案就是这种情况。使用 verbose = TRUE 运行代码时,请查看以下输出:
    > testScooter[intervals, on = .(id, start = time), roll = -Inf, startCharge := i.charge, verbose = TRUE][]
    Calculated ad hoc index in 0 secs
    Starting bmerge ...done in 0 secs
    Detected that j uses these columns: startCharge,i.charge
    Assigning to 16 row subset of 4 rows
    start end id startCharge
    1: 2018-01-18 22:19:13 2018-01-18 22:26:31 1 83
    2: 2018-01-18 23:29:22 2018-01-18 23:37:53 1 81
    3: 2018-01-18 00:22:02 2018-01-18 00:29:21 2 84
    4: 2018-01-18 00:37:52 2018-01-18 01:06:53 2 80

    尽管此行为在本示例中不会导致任何问题,但效率较低,并且可能会导致意外结果。请参阅此示例(由@Frank 提供):
    > data.table(a = 1:2)[data.table(a = c(2L, 2L), v = 3:4), on=.(a), v := i.v, verbose = TRUE][]
    Calculated ad hoc index in 0 secs
    Starting bmerge ...done in 0 secs
    Detected that j uses these columns: v,i.v
    Assigning to 2 row subset of 2 rows
    a v
    1: 1 NA
    2: 2 4

    因此,新的答案是更好的选择。

    使用数据:
    testScooter <- structure(list(start = structure(c(1516313953, 1516318162, 1516234922, 1516235872), tzone = "UTC", class = c("POSIXct", "POSIXt")),
    end = structure(c(1516314391, 1516318673, 1516235361, 1516237613), tzone = "UTC", class = c("POSIXct", "POSIXt")),
    id = c(1L, 1L, 2L, 2L)),
    .Names = c("start", "end", "id"), row.names = c(NA, -4L), class = "data.frame")
    setDT(testScooter)

    intervals <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L),
    time = structure(c(1516311067, 1516311667, 1516312267, 1516312867, 1516313467, 1516314067, 1516314667, 1516315268, 1516315868, 1516316468, 1516317068, 1516317671, 1516318267, 1516318869, 1516319467, 1516233666, 1516234266, 1516234866, 1516235465, 1516236066, 1516236666, 1516237266, 1516237865, 1516238465, 1516239065), class = c("POSIXct", "POSIXt"), tzone = "UTC"),
    available = c("yes", "yes", "yes", "yes", "yes", "no", "yes", "yes", "yes", "yes", "yes", "yes", "no", "yes", "yes", "yes", "yes", "yes", "yes", "no", "no", "no", "yes", "yes", "yes"),
    charge = c(83L, 83L, 83L, 83L, 83L, 83L, 81L, 81L, 81L, 81L, 81L, 81L, 81L, 79L, 79L, 84L, 84L, 84L, 80L, 80L, 80L, 80L, 80L, 80L, 80L)),
    .Names = c("id", "time", "available", "charge"), row.names = c(NA, -25L), class = "data.frame")
    setDT(intervals)

    关于r - 匹配时间范围之前和之后的记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49726148/

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