我的目标是根据时间dplyr或data.table加入两个数据表，专门用于在事件发生之前和之后立即获取记录。
在示例数据中，此案例中的事件是踏板车行程。下面是四次旅行 - 两次由踏板车1拍摄，两次由踏板车2拍摄。
> 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
在单独的表中，是间隔几乎相等的间隔的记录。当旅行正在进行时，id的匹配和踏板车被标记为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
我想要产生的输出如下。
> 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
有关如何在时间范围之前和之后的最近时间匹配的任何建议都会有所帮助，可能是使用lubridate::new_interval()包中的roll='nearest'或data.table，但我不确定开始。
# 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")
答案 0 :(得分：9)
新答案：
您可以使用双滚动连接执行此操作：
testScooter[, startCharge := intervals[testScooter, on = .(id, time = start), roll = Inf, x.charge]
][, endCharge := intervals[testScooter, on = .(id, time = end), roll = -Inf, x.charge]][]
给出了期望的结果：
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 = Inf在intervals start查找最后一次观察
roll = -Inf在intervals end查找第一个观察结果
另请参阅注意，了解新答案为何更好。
旧回答：
testScooter[intervals, on = .(id, start = time), roll = -Inf, startCharge := i.charge
][intervals, on = .(id, end = time), roll = Inf, endCharge := i.charge][]
注意：
由于@Frank注意到here on Github，data.table会在有多个匹配时返回i中的最后一个匹配，这是旧答案的情况。使用verbose = TRUE：
> 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
虽然此行为在此示例中不会导致任何问题，但效率较低，可能会导致意外结果。看这个例子（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，新答案是更好的选择。
使用过的数据：
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)
答案 1 :(得分：1)
您可以使用data.table non-equi查找最近的startCharge和endCharge，如下所示：
setDT(testScooter)
setDT(intervals)
testScooter[, startCharge := intervals[testScooter, .SD[, charge[.N], by=.(id, start)], on=.(id, time < start)]$V1]
testScooter[, endCharge := intervals[testScooter, .SD[, charge[1L], by=.(id, end)], on=.(id, time > end)]$V1]
startCharge的解释：
对于内方括号：
intervals[testScooter, .SD[, charge[.N], by=.(id, start)], on=.(id, time < start)]
您正在进行非等联接，intervals＆＃39; id匹配testScooter＆＃39; id和time intervals位于start的{{1}}之前。
testScooter分组.SD[, charge[.N], by=.(id, start)]和id并返回最新start＆＃39; intervals在每个小组time之前{/ 1}}。
同样适用于start。
答案 2 :(得分：1)
这是非R（跛脚）解决方案：
#Convert to data table
testScooter <- data.table(testScooter)
intervals <- data.table(intervals)
#Dummy data frame to store the results which we will finally
chargeDF <- data.frame(startCharge = numeric(),endCharge = numeric())
#Loop for each Unique ID
for( i in unique(intervals$id)){
newScooter <- testScooter[id == i,]
newintervals <- intervals[id == i,]
#Check if start time in intervals DF less than time in testScooter
tempStartList <- lapply(newScooter[,start], function (x) { newintervals[,time] < x})
#Check if end time in intervals DF greater than time in testScooter
tempEndList <- lapply(newScooter[,end], function (x) { newintervals[,time] > x})
#Loop through each row for a particular ID
for( j in 1:nrow(newScooter)){
#Find the value just before the condition becomes false
scharge <- tail(newintervals$charge[tempStartList[[j]]],1)
#Find the value just after the condition becomes true
echarge <- head(newintervals$charge[tempEndList[[j]]],1)
#Bind the results to the dummy df created earlier
chargeDF <- rbind(chargeDF,data.frame(startCharge = scharge,endCharge = echarge))
}
}
output <- cbind(testScooter, chargeDF)