Question

背景

（问题不是必需的，但可能对阅读有用）

Rolling join on data.table with duplicate keys

Odd behaviour when joining with multiple conditions

数据

library(data.table)   ## using version 1.9.6
## arrival timetable
dt_arrive <- structure(list(txn_id = c(1L, 1L, 1L, 1L, 1L), place = c("place_a", 
"place_a", "place_a", "place_a", "place_a"), arrival_minutes = c(515, 
534, 547, 561, 581), journey_id = 1:5), .Names = c("txn_id", 
"place", "arrival_minutes", "journey_id"), class = c("data.table", 
"data.frame"), row.names = c(NA, -5L), sorted = c("txn_id", 
"place"))

## departure timetable
dt_depart <- structure(list(txn_id = c(1L, 1L, 1L, 1L), place = c("place_a", 
"place_a", "place_a", "place_a"), arrival_minutes = c(489, 507, 
519, 543), journey_id = 10:13), .Names = c("txn_id", "place", 
"arrival_minutes", "journey_id"), sorted = c("txn_id", "place"
), class = c("data.table", "data.frame"), row.names = c(NA, -4L
))

> dt_arrive
   txn_id   place arrival_minutes journey_id
1:      1 place_a             515          1
2:      1 place_a             534          2
3:      1 place_a             547          3
4:      1 place_a             561          4
5:      1 place_a             581          5

> dt_depart
   txn_id   place arrival_minutes journey_id
1:      1 place_a             489         10
2:      1 place_a             507         11
3:      1 place_a             519         12
4:      1 place_a             543         13

问题

我希望加入那些在<{strong> dt_depart$journey_id之后出现 dt_arrive$journey_id arrival_minutes的人员txn_id （即place＆amp; txn_id place journey_in_id journey_out_id journey_place_arrive journey_place_depart 1 place_a 1 12 515 519 1 place_a 1 13 515 543 1 place_a 2 13 534 543上的内部联接）

例如，我想要的输出是：

setkey(dt_arrive, txn_id, place) setkey(dt_depart, txn_id, place) dt_join <- dt_arrive[dt_depart, { idx = (i.arrival_minutes > arrival_minutes) .(journey_in_id = journey_id[idx], journey_out_id = i.journey_id, journey_place_arrive = arrival_minutes[idx], journey_place_depart = i.arrival_minutes ) }, by=.EACHI]

尝试

使用我构建的两个相关问题的方法

dt_depart

但这会提供NA的所有内容，因此请在结果中包含txn_id place journey_in_id journey_out_id journey_place_arrive journey_place_depart 1: 1 place_a NA 10 NA 489 2: 1 place_a NA 11 NA 507 3: 1 place_a 1 12 515 519 4: 1 place_a 1 13 515 543 5: 1 place_a 2 13 534 543 - 这表示'正确加入'：

nomatch=0

我尝试使用complete.cases强制它进行'内连接'，但这没有用。

我可以使用NA删除=CHAR(RANDBETWEEN(80,90))&CHAR(RANDBETWEEN(80,90))&CHAR(RANDBETWEEN(80,90))&CHAR(RANDBETWEEN(80,90))&CHAR(RANDBETWEEN(80,90))&IF(LEN(A1)=4,CHAR(RANDBETWEEN(80,90)),"")&A1行，但我想知道是否有办法在查询中执行此操作？

Answer 1

以下是unclever方法：采用交叉/笛卡尔连接，然后过滤。

merge(dt_arrive, dt_depart, allow.cartesian=TRUE)[arrival_minutes.y > arrival_minutes.x]

#    txn_id   place arrival_minutes.x journey_id.x arrival_minutes.y journey_id.y
# 1:      1 place_a               515            1               519           12
# 2:      1 place_a               515            1               543           13
# 3:      1 place_a               534            2               543           13

通过采用笛卡尔式加入，我们可能会占用大量内存。

Answer 2

一个可能的解决方案是通过制作一些任意间隔列来使用foverlaps

setDT(dt_arrive)
setDT(dt_depart)

dt_arrive[, `:=`(arrival_minutes_copy = arrival_minutes)]
## reorder columns
dt_arrive <- dt_arrive[, .(txn_id, place, journey_id, arrival_minutes, arrival_minutes_copy)]

dt_depart[, `:=`(arrival_minutes_copy = min(arrival_minutes))]
## reorder columns
dt_depart <- dt_depart[, .(txn_id, place, journey_id, arrival_minutes_copy, arrival_minutes)]

setkey(dt_arrive, arrival_minutes, arrival_minutes_copy)
setkey(dt_depart, arrival_minutes_copy, arrival_minutes)

foverlaps(dt_arrive, 
          dt_depart,
          type = "within",
          nomatch=0L)


#      place txn_id journey_id arrival_minutes_copy arrival_minutes i.txn_id i.journey_id i.arrival_minutes i.arrival_minutes_copy
# 1: place_a      1         12                  489             519        1            1               515                    515
# 2: place_a      1         13                  489             543        1            1               515                    515
# 3: place_a      1         13                  489             543        1            2               534                    534

<强>基准

library(microbenchmark)

fun_foverlap <- function(dt_a, dt_d){
    dt <- foverlaps(dt_a, 
                                    dt_d,
                        type = "within",
                        nomatch=0L)
    return(dt)
}

fun_merge <- function(dt_a, dt_d){
    dt <- merge(dt_a, dt_d, allow.cartesian=TRUE)[arrival_minutes.y > arrival_minutes.x]
    return(dt)
}

fun_nomatch <- function(dt_a, dt_d){
    dt <- dt_a[dt_d, nomatch=0, allow.cartesian=TRUE][i.arrival_minutes > arrival_minutes]
    return(dt)
}

microbenchmark(fun_foverlap(dt_arrive_foverlap, dt_depart_foverlap),
                             fun_merge(dt_arrive_merge, dt_depart_merge),
                             fun_nomatch(dt_arrive_nomatch, dt_depart_nomatch))

# Unit: microseconds
                                                  expr      min       lq      mean   median        uq      max neval cld
 # fun_foverlap(dt_arrive_foverlap, dt_depart_foverlap) 3538.189 3717.077 3967.6648 3872.586 4006.7205 5812.355   100   c
 #          fun_merge(dt_arrive_merge, dt_depart_merge)  883.697  925.655  980.4159  942.877  967.9745 2223.147   100  b 
 #    fun_nomatch(dt_arrive_nomatch, dt_depart_nomatch)  593.082  625.471  682.8975  643.034  665.4125 2077.748   100 a

使用不等式表达式的内连接

2 个答案: