R的滚动日期范围内的唯一值的计数

时间:2016-04-10 17:34:35

标签: sql r data.table time-series correlated-subquery

此问题已有answer for SQL,我可以使用sqldf在R中实现该解决方案。但是,我一直无法找到使用data.table实现它的方法。

问题是计算滚动日期范围内一列的不同值,例如(并直接引用链接问题)如果数据如下所示:

Date   | email 
-------+----------------
1/1/12 | test@test.com
1/1/12 | test1@test.com
1/1/12 | test2@test.com
1/2/12 | test1@test.com
1/2/12 | test2@test.com
1/3/12 | test@test.com
1/4/12 | test@test.com
1/5/12 | test@test.com
1/5/12 | test@test.com
1/6/12 | test@test.com
1/6/12 | test@test.com
1/6/12 | test1@test.com

如果我们使用3天的日期,那么结果集看起来会像这样。

date   | count(distinct email)
-------+------
1/1/12 | 3
1/2/12 | 3
1/3/12 | 3
1/4/12 | 3
1/5/12 | 2
1/6/12 | 2

以下是使用data.table在R中创建相同数据的代码:

date <- as.Date(c('2012-01-01','2012-01-01','2012-01-01',
                  '2012-01-02','2012-01-02','2012-01-03',
                  '2012-01-04','2012-01-05','2012-01-05',
                  '2012-01-06','2012-01-06','2012-01-06'))
email <- c('test@test.com', 'test1@test.com','test2@test.com',
           'test1@test.com', 'test2@test.com','test@test.com',
           'test@test.com','test@test.com','test@test.com',
           'test@test.com','test@test.com','test1@test.com')
dt <- data.table(date, email)

对此的任何帮助将不胜感激。谢谢!

编辑1:

这是一个玩具问题,我想应用于更大的数据集,因此使用笛卡尔积很有问题。相反,我想在SQL中使用等同于相关子查询的东西,例如:我最初链接的问题的解决方案是:

SELECT day
     ,(SELECT count(DISTINCT email)
       FROM   tbl
       WHERE  day BETWEEN t.day - 2 AND t.day -- period of 3 days
      ) AS dist_emails
FROM   tbl t
WHERE  day BETWEEN '2012-01-01' AND '2012-01-06'  
GROUP  BY 1
ORDER  BY 1;

编辑2: 根据@ MichaelChirico的解决方案,这是@jangorecki所要求的时间:

# The data
> dim(temp)
[1] 2627785       4
> head(temp)
         date category1 category2 itemId
1: 2013-11-08         0         2   1713
2: 2013-11-08         0         2  90485
3: 2013-11-08         0         2  74249
4: 2013-11-08         0         2   2592
5: 2013-11-08         0         2   2592
6: 2013-11-08         0         2    765
> uniqueN(temp$itemId)
[1] 13510
> uniqueN(temp$date)
[1] 127

# Timing for data.table
> system.time(dtTime <- temp[,
+   .(count = temp[.(seq.Date(.BY$date - 6L, .BY$date, "day"), 
+   .BY$category1, .BY$category2 ), uniqueN(itemId), nomatch = 0L]), 
+  by = c("date","category1","category2")])
   user  system elapsed 
  6.913   0.130   6.940 
> 
# Time for sqldf
> system.time(sqlDfTime <- 
+ sqldf(c("create index ldx on temp(date, category1, category2)",
+ "SELECT date, category1, category2,
+ (SELECT count(DISTINCT itemId)
+   FROM   temp
+   WHERE category1 = t.category1 AND category2 = t.category2 AND
+   date BETWEEN t.date - 6 AND t.date 
+   ) AS numItems
+ FROM temp t
+ GROUP BY date, category1, category2
+ ORDER BY 1;"))
   user  system elapsed 
 87.225   0.098  87.295 

输出是等效的,但使用data.table而不是sqldf导致12.5倍的加速。相当可观!

2 个答案:

答案 0 :(得分:6)

这里有一些有用的东西,利用了data.table的新非等值连接功能。

dt[dt[ , .(date3=date, date2 = date - 2, email)], 
   on = .(date >= date2, date<=date3), 
   allow.cartesian = TRUE
   ][ , .(count = uniqueN(email)), 
      by = .(date = date + 2)]
#          date V1
# 1: 2011-12-30  3
# 2: 2011-12-31  3
# 3: 2012-01-01  3
# 4: 2012-01-02  3
# 5: 2012-01-03  1
# 6: 2012-01-04  2

说实话,我对这是如何正常工作感到恼火,但我们的想法是在dt上加入date,与之间的任何date相匹配2天前和今天。我不确定为什么我们必须通过之后设置date = date + 2进行清理。

这是一种使用密钥的方法:

setkey(dt, date)

dt[ , .(count = dt[.(seq.Date(.BY$date - 2L, .BY$date, "day")),
                   uniqueN(email), nomatch = 0L]), by = date]

答案 1 :(得分:3)

最近在non-equi的{​​{3}}中实施了data.table, v1.9.7加入功能,可以按照以下方式完成:

dt[.(date3=unique(dt$date2)), .(count=uniqueN(email)), on=.(date>=date3, date2<=date3), by=.EACHI]
#          date      date2 count
# 1: 2011-12-30 2011-12-30     3
# 2: 2011-12-31 2011-12-31     3
# 3: 2012-01-01 2012-01-01     3
# 4: 2012-01-02 2012-01-02     3
# 5: 2012-01-03 2012-01-03     1
# 6: 2012-01-04 2012-01-04     2