两个时间戳之间的R中的左连接

mbjcgjjk 于 12个月前发布在其他

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我的目标是在intervals上执行左连接，其中bike_id匹配，并且records中的created_at时间戳在intervals表中的start和end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

字符串
在本例中，输出如下所示

> 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

型
我试过使用tidyverse的解决方案posted here，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)

型
返回错误：Error: vector memory exhausted (limit reached?)
在data.table中或者甚至在基R中使用merge()的滚动连接是否有替代方法？通过id连接两个嵌套的方法是什么？其中时间戳在连接表中的其他两个之间？
这里是数据

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")

型

来源：https://stackoverflow.com/questions/56688598/left-joining-in-r-between-two-timestamps

3条答案

按热度按时间

xbp102n01#

我们可以使用data.table不等连接

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

字符串

赞(0）回复(0）举报 12个月前

btxsgosb2#

我知道OP要求tidyverse或data.table解决方案，但SQL似乎是完美的工具：

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")

字符串
或者使用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")

型
编辑：正如@G. Grothendieck所指出的，我们可以在阅读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

型

数据：（OP的dput工作，因为从data.table创建的指针）

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))

型

赞(0）回复(0）举报 12个月前

tkclm6bt3#

另一种方法是在bike_id和created_at的日期部分进行连接，然后删除created_at不在start-end区间内的ID。这可能会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)

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它返回：

# 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

型

赞(0）回复(0）举报 12个月前

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两个时间戳之间的R中的左连接

3条答案

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