dplyr left_join小于,大于条件

时间:2016-05-18 02:48:05

标签: sql r postgresql left-join dplyr

此问题与问题Efficiently merging two data frames on a non-trivial criteriaChecking if date is between two dates in r有些相关。我在这里发布的请求该功能是否存在: GitHub issue

我希望使用dplyr::left_join()加入两个数据帧。我用来加入的条件是小于,大于<=>dplyr::left_join()是否支持此功能?或者只在它们之间使用=运算符。这很容易从SQL运行(假设我在数据库中有数据帧)

这是一个MWE:我有两个数据集,一个公司年(fdata),第二个是每五年发生一次的调查数据。因此,对于两个调查年之间fdata的所有年份,我都会​​加入相应的调查年度数据。

id <- c(1,1,1,1,
        2,2,2,2,2,2,
        3,3,3,3,3,3,
        5,5,5,5,
        8,8,8,8,
        13,13,13)

fyear <- c(1998,1999,2000,2001,1998,1999,2000,2001,2002,2003,
       1998,1999,2000,2001,2002,2003,1998,1999,2000,2001,
       1998,1999,2000,2001,1998,1999,2000)

byear <- c(1990,1995,2000,2005)
eyear <- c(1995,2000,2005,2010)
val <- c(3,1,5,6)

sdata <- tbl_df(data.frame(byear, eyear, val))

fdata <- tbl_df(data.frame(id, fyear))

test1 <- left_join(fdata, sdata, by = c("fyear" >= "byear","fyear" < "eyear"))

我得到了

Error: cannot join on columns 'TRUE' x 'TRUE': index out of bounds 

除非left_join可以处理条件,但我的语法遗漏了什么?

4 个答案:

答案 0 :(得分:16)

看起来这是打包 fuzzyjoin 地址的任务。包的各种功能看起来和工作类似于 dplyr 连接功能。

在这种情况下,fuzzy_*_join函数之一将适合您。 dplyr::left_joinfuzzyjoin::fuzzy_left_join之间的主要区别在于,您提供了在match.fun参数的匹配过程中使用的函数列表。请注意,by参数的编写方式与left_join中的参数相同。

以下是一个例子。我用来匹配的函数分别是>=<fyearbyear比较的fyeareyear。在

library(fuzzyjoin)

fuzzy_left_join(fdata, sdata, 
             by = c("fyear" = "byear", "fyear" = "eyear"), 
             match_fun = list(`>=`, `<`))

Source: local data frame [27 x 5]

      id fyear byear eyear   val
   (dbl) (dbl) (dbl) (dbl) (dbl)
1      1  1998  1995  2000     1
2      1  1999  1995  2000     1
3      1  2000  2000  2005     5
4      1  2001  2000  2005     5
5      2  1998  1995  2000     1
6      2  1999  1995  2000     1
7      2  2000  2000  2005     5
8      2  2001  2000  2005     5
9      2  2002  2000  2005     5
10     2  2003  2000  2005     5
..   ...   ...   ...   ...   ...

答案 1 :(得分:15)

使用filter。 (但请注意,此答案会生成正确的LEFT JOIN;但MWE会使用INNER JOIN来提供正确的结果。)

如果要求合并两个表没有要合并的东西,dplyr包不满意,所以在下面,我为此目的在两个表中创建一个虚拟变量,然后过滤,然后删除{{1 }}:

dummy

请注意,如果您在PostgreSQL中执行此操作(例如),查询优化器会查看fdata %>% mutate(dummy=TRUE) %>% left_join(sdata %>% mutate(dummy=TRUE)) %>% filter(fyear >= byear, fyear < eyear) %>% select(-dummy) 变量,如以下两个查询说明所示:

dummy

使用SQL更干净地完成完全相同的结果:

> fdata %>% 
+     mutate(dummy=TRUE) %>%
+     left_join(sdata %>% mutate(dummy=TRUE)) %>%
+     filter(fyear >= byear, fyear < eyear) %>%
+     select(-dummy) %>%
+     explain()
Joining by: "dummy"
<SQL>
SELECT "id" AS "id", "fyear" AS "fyear", "byear" AS "byear", "eyear" AS "eyear", "val" AS "val"
FROM (SELECT * FROM (SELECT "id", "fyear", TRUE AS "dummy"
FROM "fdata") AS "zzz136"

LEFT JOIN 

(SELECT "byear", "eyear", "val", TRUE AS "dummy"
FROM "sdata") AS "zzz137"

USING ("dummy")) AS "zzz138"
WHERE "fyear" >= "byear" AND "fyear" < "eyear"


<PLAN>
Nested Loop  (cost=0.00..50886.88 rows=322722 width=40)
  Join Filter: ((fdata.fyear >= sdata.byear) AND (fdata.fyear < sdata.eyear))
  ->  Seq Scan on fdata  (cost=0.00..28.50 rows=1850 width=16)
  ->  Materialize  (cost=0.00..33.55 rows=1570 width=24)
        ->  Seq Scan on sdata  (cost=0.00..25.70 rows=1570 width=24)

答案 2 :(得分:13)

data.table从v 1.9.8

开始添加非等联接
library(data.table) #v>=1.9.8
setDT(sdata); setDT(fdata) # converting to data.table in place

fdata[sdata, on = .(fyear >= byear, fyear < eyear), nomatch = 0,
      .(id, x.fyear, byear, eyear, val)]
#    id x.fyear byear eyear val
# 1:  1    1998  1995  2000   1
# 2:  2    1998  1995  2000   1
# 3:  3    1998  1995  2000   1
# 4:  5    1998  1995  2000   1
# 5:  8    1998  1995  2000   1
# 6: 13    1998  1995  2000   1
# 7:  1    1999  1995  2000   1
# 8:  2    1999  1995  2000   1
# 9:  3    1999  1995  2000   1
#10:  5    1999  1995  2000   1
#11:  8    1999  1995  2000   1
#12: 13    1999  1995  2000   1
#13:  1    2000  2000  2005   5
#14:  2    2000  2000  2005   5
#15:  3    2000  2000  2005   5
#16:  5    2000  2000  2005   5
#17:  8    2000  2000  2005   5
#18: 13    2000  2000  2005   5
#19:  1    2001  2000  2005   5
#20:  2    2001  2000  2005   5
#21:  3    2001  2000  2005   5
#22:  5    2001  2000  2005   5
#23:  8    2001  2000  2005   5
#24:  2    2002  2000  2005   5
#25:  3    2002  2000  2005   5
#26:  2    2003  2000  2005   5
#27:  3    2003  2000  2005   5
#    id x.fyear byear eyear val

您还可以通过更多的努力在1.9.6中使用foverlaps

答案 3 :(得分:2)

一个选项是按列顺序连接列表列,然后取消列:

# evaluate each row individually
fdata %>% rowwise() %>% 
    # insert list column of single row of sdata based on conditions
    mutate(s = list(sdata %>% filter(fyear >= byear, fyear < eyear))) %>% 
    # unnest list column
    tidyr::unnest()

# Source: local data frame [27 x 5]
# 
#       id fyear byear eyear   val
#    (dbl) (dbl) (dbl) (dbl) (dbl)
# 1      1  1998  1995  2000     1
# 2      1  1999  1995  2000     1
# 3      1  2000  2000  2005     5
# 4      1  2001  2000  2005     5
# 5      2  1998  1995  2000     1
# 6      2  1999  1995  2000     1
# 7      2  2000  2000  2005     5
# 8      2  2001  2000  2005     5
# 9      2  2002  2000  2005     5
# 10     2  2003  2000  2005     5
# ..   ...   ...   ...   ...   ...