在两个时间戳之间在 R 中左连接

问题描述 投票：0回答：3

我的目标是在 

intervals
上执行左连接，其中 
bike_id
匹配，并且 
created_at
中的 
records
时间戳位于 
start
表中的 
end
和 
intervals
之间

> class(records)
[1] "data.table" "data.frame"
> class(intervals)
[1] "data.table" "data.frame"

> records
  bike_id          created_at         resolved_at
1   28780 2019-05-03 08:29:18 2019-05-03 08:35:37
2   28780 2019-05-03 21:05:28 2019-05-03 21:07:28
3   28780 2019-05-04 21:13:39 2019-05-04 21:15:40
4   28780 2019-05-07 17:24:20 2019-05-07 17:26:39
5   28780 2019-05-08 11:34:32 2019-05-08 12:16:44
6   28780 2019-05-08 23:38:39 2019-05-08 23:40:36


> intervals
   bike_id               start                 end id
1:   28780 2019-05-03 04:44:45 2019-05-03 16:58:56  1
2:   28780 2019-05-04 07:07:39 2019-05-04 14:48:29  2
3:   28780 2019-05-07 23:28:32 2019-05-08 12:56:24  3
4:   28780 2019-05-10 06:06:21 2019-05-10 13:12:08  4
5:   28780 2019-05-12 05:21:24 2019-05-12 11:35:52  5
6:   28780 2019-05-13 08:44:54 2019-05-13 12:28:31  6

在这种情况下，输出将如下所示

> output
  bike_id          created_at         resolved_at   id
1   28780 2019-05-03 08:29:18 2019-05-03 08:35:37    1
2   28780 2019-05-03 21:05:28 2019-05-03 21:07:28  NULL   
3   28780 2019-05-04 21:13:39 2019-05-04 21:15:40  NULL
4   28780 2019-05-07 17:24:20 2019-05-07 17:26:39  NULL
5   28780 2019-05-08 11:34:32 2019-05-08 12:16:44  NULL
6   28780 2019-05-08 23:38:39 2019-05-08 23:40:36  NULL

我尝试使用解决方案发布在这里使用

tidyverse
但这会导致R耗尽内存（尽管两个表中的记录量只有大约100K）

library('fuzzyjoin')

fuzzy_left_join(
 records, intervals,
  by = c(
    "bike_id" = "bike_id",
    "created_at" = "start",
    "created_at" = "end"
    ),
  match_fun = list(`==`, `>=`, `<=`)
  ) %>%
  select(id, bike_id = bike_id.x, created_at, start, end)

这会引发错误：

Error: vector memory exhausted (limit reached?)

是否有另一种方法可以在 

data.table
中甚至在基本 R 中使用 
merge()
进行滚动连接？通过 id 连接两个数据帧以及连接表中其他两个数据帧之间的时间戳的好方法是什么？

这是数据

dput(intervals)
structure(list(bike_id = c(28780L, 28780L, 28780L, 28780L, 28780L, 
28780L), start = structure(c(1556858685, 1556953659, 1557271712, 
1557468381, 1557638484, 1557737094), class = c("POSIXct", "POSIXt"
), tzone = "UTC"), end = structure(c(1556902736, 1556981309, 
1557320184, 1557493928, 1557660952, 1557750511), class = c("POSIXct", 
"POSIXt"), tzone = "UTC"), id = c(1, 2, 3, 4, 5, 6)), row.names = c(NA, 
-6L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x1030056e0>)

dput(records)
structure(list(bike_id = c(28780L, 28780L, 28780L, 28780L, 28780L, 
28780L), created_at = structure(c(1556872158.796, 1556917528.845, 
1557004419.928, 1557249860.939, 1557315272.396, 1557358719.333
), class = c("POSIXct", "POSIXt"), tzone = "UTC"), resolved_at = structure(c(1556872537.867, 
1556917648.118, 1557004540.056, 1557249999.892, 1557317804.183, 
1557358836.202), class = c("POSIXct", "POSIXt"), tzone = "UTC")), row.names = c(NA, 
6L), class = "data.frame")
r dplyr data.table tidyverse
3个回答
5
投票

我们可以使用

data.table
非等价连接

library(data.table)
setDT(records)[intervals, on = .(bike_id, created_at >= start, created_at <= end)]
3
投票

我知道 OP 要求提供 

tidyverse
或 
data.table
解决方案，但 SQL 似乎是实现此目的的完美工具：

library(sqldf)

sqldf("select a.*, b.id 
        from records as a
        left join intervals as b
          on a.bike_id = b.bike_id and
            a.created_at >= b.start and
            a.created_at <= b.end")

或使用 

between
作为替代语法：

sqldf("select a.*, b.id 
        from records as a
        left join intervals as b
          on a.bike_id = b.bike_id and
            a.created_at between b.start and b.end")

编辑：正如@G 所指出的。 Grothendieck，我们可以在读入数据之前设置环境的时区（使用 

Sys.setenv
）以匹配 OP 的时区。

输出：

  bike_id          created_at         resolved_at id
1   28780 2019-05-03 08:29:18 2019-05-03 08:35:37  1
2   28780 2019-05-03 21:05:28 2019-05-03 21:07:28 NA
3   28780 2019-05-04 21:13:39 2019-05-04 21:15:40 NA
4   28780 2019-05-07 17:24:20 2019-05-07 17:26:39 NA
5   28780 2019-05-08 11:34:32 2019-05-08 12:16:44  3
6   28780 2019-05-08 23:38:39 2019-05-08 23:40:36 NA

数据：（OP 的 

dput
确实有效，因为从 
data.table
创建了指针）

Sys.setenv(TZ = "GMT")

records <- structure(list(bike_id = c(28780L, 28780L, 28780L, 28780L, 28780L, 
28780L), created_at = c("2019-05-03 08:29:18", "2019-05-03 21:05:28", 
"2019-05-04 21:13:39", "2019-05-07 17:24:20", "2019-05-08 11:34:32", 
"2019-05-08 23:38:39"), resolved_at = c("2019-05-03 08:35:37", 
"2019-05-03 21:07:28", "2019-05-04 21:15:40", "2019-05-07 17:26:39", 
"2019-05-08 12:16:44", "2019-05-08 23:40:36")), class = "data.frame", row.names = c(NA, 
-6L))

intervals <- structure(list(bike_id = c(28780L, 28780L, 28780L, 28780L, 28780L, 
28780L), start = c("2019-05-03 04:44:45", "2019-05-04 07:07:39", 
"2019-05-07 23:28:32", "2019-05-10 06:06:21", "2019-05-12 05:21:24", 
"2019-05-13 08:44:54"), end = c("2019-05-03 16:58:56", "2019-05-04 14:48:29", 
"2019-05-08 12:56:24", "2019-05-10 13:12:08", "2019-05-12 11:35:52", 
"2019-05-13 12:28:31"), id = c(1, 2, 3, 4, 5, 6)), class = "data.frame", row.names = c(NA, 
-6L))
1
投票

另一种方法是加入 

bike_id
和 
created_at
的日期部分，然后删除 
created_at
不在区间 
start
-
end
中的 ID。这可以通过将事情分解为单独的步骤来解决内存问题：

library(dplyr)
library(lubridate)
library(purrr)

intervals %>% 
    mutate(date = date(start)) %>% 
    right_join(mutate(records,
                      date = date(created_at)),
                      by = c("bike_id", "date")
              ) %>% 
    mutate(within = created_at %within% interval(start, end),
           within = replace_na(within, F),
           id = map2_dbl(id, within, ~ ifelse(.y, .x, NA))
           ) %>% 
    select(bike_id, id, created_at, resolved_at)

返回：

# A tibble: 6 x 4
  bike_id    id created_at          resolved_at        
    <int> <dbl> <dttm>              <dttm>             
1   28780     1 2019-05-03 08:29:18 2019-05-03 08:35:37
2   28780    NA 2019-05-03 21:05:28 2019-05-03 21:07:28
3   28780    NA 2019-05-04 21:13:39 2019-05-04 21:15:40
4   28780    NA 2019-05-07 17:24:20 2019-05-07 17:26:39
5   28780    NA 2019-05-08 11:34:32 2019-05-08 12:16:44
6   28780    NA 2019-05-08 23:38:39 2019-05-08 23:40:36
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