R 使用data.table包的笛卡尔积

使用R中的data.table包，我尝试使用merge方法创建两个data.tables的笛卡尔乘积，就像在BaseR中一样

在以下工作的基础上：

#assume this order data
orders <- data.frame(date = as.POSIXct(c('2012-08-28','2012-08-29','2012-09-01')),
                     first.name = as.character(c('John','George','Henry')),
                     last.name = as.character(c('Doe','Smith','Smith')),
                     qty = c(10,50,6))

#and these dates
dates <- data.frame(date = seq(from = as.POSIXct('2012-08-28'),
                               to = as.POSIXct('2012-09-07'), by = 'day'))

#get the unique customers
cust<-unique(orders[,c('first.name','last.name')])

#using merge from base R, get the cartesian product
merge(dates, cust, by = integer(0))

我希望结果反映所有日期的所有客户名称，就像在base中一样，但要以数据表为中心。这可能吗？

merge.data.table（x，y）

是一个方便的函数，它包装了对

x[y]

的调用，因此合并需要基于

data.table

中的列。（这就是错误消息试图告诉您的内容）

一种解决方法是向两个data.tables添加一个伪列，其唯一目的是使合并成为可能：

## Add a column "k", and append it to each data.table's vector of keyed columns.
setkeyv(cust.dt[,k:=1], c(key(cust.dt), "k"))
setkeyv(dates.dt[,k:=1], c(key(dates.dt), "k"))

## Merge and then remove the dummy column
res <- merge(dates.dt, cust.dt, by="k")
head(res[,k:=NULL])
#          date first.name last.name
# 1: 2012-08-28     George     Smith
# 2: 2012-08-28      Henry     Smith
# 3: 2012-08-28       John       Doe
# 4: 2012-08-29     George     Smith
# 5: 2012-08-29      Henry     Smith
# 6: 2012-08-29       John       Doe

## Maybe also clean up cust.dt and dates.dt    
# cust.dt[,k:=NULL]
# dates.dt[,k=NULL]

##添加列“k”，并将其附加到每个数据。表的键控列向量。
setkeyv（cust.dt[，k:=1]，c（key（cust.dt），“k”））
setkeyv（dates.dt[，k:=1]，c（key（dates.dt），“k”））
##合并然后删除虚拟列
res如果从数据帧中的第一个和最后一个构建全名，则可以使用
CJ
（交叉连接）。您不能使用所有三个向量，因为将有99个项目

> nrow(CJ(dates$date, cust$first.name, cust$last.name ) )
[1] 99

这将返回一个data.table对象：

> CJ(dates$date,paste(cust$first.name, cust$last.name) )
            V1           V2
 1: 2012-08-28 George Smith
 2: 2012-08-28  Henry Smith
 3: 2012-08-28     John Doe
 4: 2012-08-29 George Smith
 5: 2012-08-29  Henry Smith
 6: 2012-08-29     John Doe
 7: 2012-08-30 George Smith
 8: 2012-08-30  Henry Smith
 9: 2012-08-30     John Doe
10: 2012-08-31     John Doe
11: 2012-08-31 George Smith
12: 2012-08-31  Henry Smith
13: 2012-09-01     John Doe
14: 2012-09-01 George Smith
15: 2012-09-01  Henry Smith
16: 2012-09-02 George Smith
17: 2012-09-02  Henry Smith
18: 2012-09-02     John Doe
19: 2012-09-03  Henry Smith
20: 2012-09-03     John Doe
21: 2012-09-03 George Smith
22: 2012-09-04  Henry Smith
23: 2012-09-04     John Doe
24: 2012-09-04 George Smith
25: 2012-09-05 George Smith
26: 2012-09-05  Henry Smith
27: 2012-09-05     John Doe
28: 2012-09-06 George Smith
29: 2012-09-06  Henry Smith
30: 2012-09-06     John Doe
31: 2012-09-07 George Smith
32: 2012-09-07  Henry Smith
33: 2012-09-07     John Doe
            V1           V2

来自@JoshO'Brien的解决方案使用了
merge
，但下面是一个类似的替代方案，没有（AFAIK）

如果我正确理解了
？data.table:：merge
中的文档，
X[Y]
应该比
data.table:：merge（X，Y）
稍快一些（从1.8.7版开始）。为了解决这个问题，参考FAQ 2.12，但是FAQ有点混乱。首先，正确的参考值应该是1.12，而不是2.12。并且它们不表示它们是引用merge的基本版本还是data.table one，或者两者都引用。因此，这可能最终成为一个看起来更混乱的解决方案，或者它可能更快。

[编辑自Matthew]谢谢：现在在v1.8.7中有所改进（
？merge.data.table
，FAQ 1.12和新增的FAQ 2.24）

DT_ordersOk，CJ是我一直在寻找的，但我认为唯一的id字段更适合我的实际情况，而不是串联名称。谢谢+非常好@ZachWaite——借用了DWin的想法，您还可以尝试合并（CJ（date=dates$date，first.name=cust$first.name），cust，“first.name”）
，然后您可以轻松地对列重新排序，如果您愿意的话。谢谢。现在提出改进？合并和常见问题解答，链接返回此处。
## Add a column "k", and append it to each data.table's vector of keyed columns.
setkeyv(cust.dt[,k:=1], c(key(cust.dt), "k"))
setkeyv(dates.dt[,k:=1], c(key(dates.dt), "k"))

## Merge and then remove the dummy column
res <- merge(dates.dt, cust.dt, by="k")
head(res[,k:=NULL])
#          date first.name last.name
# 1: 2012-08-28     George     Smith
# 2: 2012-08-28      Henry     Smith
# 3: 2012-08-28       John       Doe
# 4: 2012-08-29     George     Smith
# 5: 2012-08-29      Henry     Smith
# 6: 2012-08-29       John       Doe

## Maybe also clean up cust.dt and dates.dt    
# cust.dt[,k:=NULL]
# dates.dt[,k=NULL]

> nrow(CJ(dates$date, cust$first.name, cust$last.name ) )
[1] 99

> CJ(dates$date,paste(cust$first.name, cust$last.name) )
            V1           V2
 1: 2012-08-28 George Smith
 2: 2012-08-28  Henry Smith
 3: 2012-08-28     John Doe
 4: 2012-08-29 George Smith
 5: 2012-08-29  Henry Smith
 6: 2012-08-29     John Doe
 7: 2012-08-30 George Smith
 8: 2012-08-30  Henry Smith
 9: 2012-08-30     John Doe
10: 2012-08-31     John Doe
11: 2012-08-31 George Smith
12: 2012-08-31  Henry Smith
13: 2012-09-01     John Doe
14: 2012-09-01 George Smith
15: 2012-09-01  Henry Smith
16: 2012-09-02 George Smith
17: 2012-09-02  Henry Smith
18: 2012-09-02     John Doe
19: 2012-09-03  Henry Smith
20: 2012-09-03     John Doe
21: 2012-09-03 George Smith
22: 2012-09-04  Henry Smith
23: 2012-09-04     John Doe
24: 2012-09-04 George Smith
25: 2012-09-05 George Smith
26: 2012-09-05  Henry Smith
27: 2012-09-05     John Doe
28: 2012-09-06 George Smith
29: 2012-09-06  Henry Smith
30: 2012-09-06     John Doe
31: 2012-09-07 George Smith
32: 2012-09-07  Henry Smith
33: 2012-09-07     John Doe
            V1           V2

DT_orders<-data.table(date=as.POSIXct(c('2012-08-28','2012-08-29','2012-08-29','2012-09-01')),
                      first.name=as.character(c('John','John','George','Henry')),
                      last.name=as.character(c('Doe','Doe','Smith','Smith')),
                      qty=c(10,2,50,6),
                      key="first.name,last.name")

# Note that I added a second record to the orders table for John Doe, to make sure it could handle duplicate first/last name combinations.

DT_dates<-data.table(date=seq(from=as.POSIXct('2012-08-28'),
                              to=as.POSIXct('2012-09-07'),by='day'),
                     key="date")

DT_custdates<-data.table(k=1,unique(DT_dates),key="k")[unique(DT_orders)[,list(k=1,first.name,last.name)]][,k:=NULL]