Question

我试图找到一种方法来使用替代而不是连接来使用两个键来过滤DT。 dplyr中的解决方案如下所示：

Enter 3 Numbers (Separated By White-space): 1 2 3
The Total Number: 5

我尝试使用filter(DF, A == a | B == b)和data.table上的密钥设置在A中执行相同的操作，但到目前为止还没有运气。

我不想使用B表格，因为矢量搜索效果较差。

让我们以下面的数据为例：

DT[A == a | B == b]

Answer 1

感谢@Frank的回答 - 结果证明这是正确的方法。弗兰克提出了mya = DT[A==a,which=TRUE]; myb = DT[B==b,which=TRUE]; DT[union(mya,myb)]，因为它进行了两次二进制搜索。

我在较大的数据集（97671 x 13）上做了一些基准测试，这就是它的样子（还添加了一些有问题的尝试;为比较添加了连接示例）：

> microbenchmark(filter(ref.transactions, TalentID == talent.id | RecurringProfileID == recurring.profile.id), ref.transactions[TalentID == talent.id | RecurringProfileID == recurring.profile.id], unique(rbindlist(list(ref.transactions[.(talent.id)], ref.transactions[.(unique(c(talent.id, NA)), recurring.profile.id)]))), unique(rbind(ref.transactions[.(talent.id)], ref.transactions[.(unique(c(talent.id, NA)), recurring.profile.id)])), ref.transactions[.(talent.id, recurring.profile.id)], {mya = ref.transactions[TalentID==talent.id,which=TRUE]; myb = ref.transactions[RecurringProfileID==recurring.profile.id,which=TRUE]; ref.transactions[union(mya,myb)]})
Unit: milliseconds
                                                                                                                                                                                                    expr       min        lq      mean    median        uq       max neval
                                                                                                       filter(ref.transactions, TalentID == talent.id | RecurringProfileID ==      recurring.profile.id) 10.039814 11.874223 14.278728 12.560975 13.562596 45.023206   100
                                                                                                               ref.transactions[TalentID == talent.id | RecurringProfileID ==      recurring.profile.id]  6.934124  7.838649  9.323780  8.227186  8.822951 40.115687   100
                                                                       unique(rbindlist(list(ref.transactions[.(talent.id)], ref.transactions[.(unique(c(talent.id,      NA)), recurring.profile.id)])))  9.859269 10.826785 13.546877 11.663016 13.073455 47.173324   100
                                                                                 unique(rbind(ref.transactions[.(talent.id)], ref.transactions[.(unique(c(talent.id,      NA)), recurring.profile.id)]))  9.910144 11.027810 14.633140 11.663457 12.920559 57.256676   100
                                                                                                                                                    ref.transactions[.(talent.id, recurring.profile.id)]  1.196426  1.316740  1.513665  1.470091  1.574857  2.799963   100
 {     mya = ref.transactions[TalentID == talent.id, which = TRUE]     myb = ref.transactions[RecurringProfileID == recurring.profile.id,          which = TRUE]     ref.transactions[union(mya, myb)] }  1.710616  1.978395  3.085824  2.121029  2.370705 30.513052   100
> df.res <- filter(ref.transactions, TalentID == talent.id | RecurringProfileID ==      recurring.profile.id)
> mya = ref.transactions[TalentID==talent.id,which=TRUE]; myb = ref.transactions[RecurringProfileID==recurring.profile.id,which=TRUE]; dt.res <- ref.transactions[union(mya,myb)]
> identical(df.res, dt.res)
[1] TRUE

使用替代（OR）而不是连接使用两列上的键进行二进制搜索DT

1 个答案: