R Data.table：合并列（聚合的单个函数）

我有一些类似的数据

foo另一种方法是：

foo1 <- CJ(var1 = c(T,F), var2 = c(T,F))[, var3 := c('z', 'y', 'x', 'w')]
setkey(foo, var1, var2)

foo[foo1, var3 :=i.var3][order(uid)][, c(1L, 4L)]
#   uid var3
#1:   a    x
#2:   b    y
#3:   c    z

foo1这可能也没有经过优化，但我发现它比@David Arenburg的解决方案更容易阅读

foo[, `:=` (var3 = ifelse(var1 & var2, "w", ifelse(var1, "x", ifelse(var2, "y", "z"))), 
            var1 = NULL, var2 = NULL)]
foo
#    uid var3
# 1:   a    x
# 2:   b    y
# 3:   c    z

我个人不喜欢
ifelse
，尤其是嵌套的
ifelse
s:）。我想，在这种情况下，我们可以不用它，用这样的东西，也许

foo[, `:=`(var3 = factor(2*var1+var2, levels=3:0, labels=c("w","x","y","z")), 
           var2 = NULL, var1 = NULL)]
#    uid var3
# 1:   a    x
# 2:   b    y
# 3:   c    z

我做了一些基准测试，@akrun的解决方案似乎是更大数据集的最快解决方案。为了使结果具有可比性，必须进行一些更改（在代码中对所有更改进行注释），但这不会对性能产生太大影响

# setup of data
require(data.table)
require(microbenchmark)
set.seed(1)
Nsims <- 1e4 # size of dataset
foo <- data.table(uid = 1:Nsims,
                  var1 = sample(c(TRUE, FALSE), Nsims, TRUE), 
                  var2 = sample(c(TRUE, FALSE), Nsims, TRUE))
# benchmarktest
microbenchmark(
{ #@shadow
  foo1 <- copy(foo)
  foo1[, `:=` (var3=ifelse(var1&var2, "w", ifelse(var1, "x", ifelse(var2, "y", "z"))), 
               var1=NULL, var2=NULL)]
}
, 
{ #@Arun
  foo2 <- copy(foo)
  foo2[, `:=`(var3 = as.character(factor(2*var1+var2, levels=3:0, labels=c("w","x","y","z"))), 
              # used as.character to give same result as other solutions
              var2 = NULL, var1 = NULL)]
},
{ #@akrun
  foo3 <- copy(foo)
  foo.index <- CJ(var1 = c(T,F), var2 = c(T,F))[, var3 := c('z', 'y', 'x', 'w')]
  setkey(foo3, var1, var2)
  foo3 <- foo3[foo.index, var3 := i.var3][, `:=` (var1=NULL, var2=NULL)][order(uid)]
  # assigned to foo3 to get same result as other solutions and used var1:=NULL, etc to achieve OP's
  # requirement "Moreover, any additional column that foo has besides (var1, var2) should be taken over into the new data.table"
}
)
#        min        lq   median        uq      max neval
#  19.635460 19.801922 19.93224 20.814533 22.57868   100
#  12.611448 12.762514 12.79219 12.864043 48.10415   100
#   4.691303  4.945683  4.98808  5.084922  7.21636   100
#
# making sure they give the same solutions
all.equal(foo1, foo2)
# [1] TRUE
all.equal(foo1, foo3)
# [1] TRUE

#数据设置
要求（数据表）
要求（微基准）
种子（1）
Nsims@eddi。谢谢你的编辑。我尝试了
CJ
方法，但不知何故，var3列没有正确排序。
CJ
和
expand.grid
的排序不同-可能是您遇到问题的地方。我第一次看到在
r
中使用了中间“值矩阵”，这是一项很好的技能。@davidernburg不是真的，但我确实发现这种方法非常有趣：）@shadow的解决方案是您应该将
集
与之进行比较的解决方案。
# setup of data
require(data.table)
require(microbenchmark)
set.seed(1)
Nsims <- 1e4 # size of dataset
foo <- data.table(uid = 1:Nsims,
                  var1 = sample(c(TRUE, FALSE), Nsims, TRUE), 
                  var2 = sample(c(TRUE, FALSE), Nsims, TRUE))
# benchmarktest
microbenchmark(
{ #@shadow
  foo1 <- copy(foo)
  foo1[, `:=` (var3=ifelse(var1&var2, "w", ifelse(var1, "x", ifelse(var2, "y", "z"))), 
               var1=NULL, var2=NULL)]
}
, 
{ #@Arun
  foo2 <- copy(foo)
  foo2[, `:=`(var3 = as.character(factor(2*var1+var2, levels=3:0, labels=c("w","x","y","z"))), 
              # used as.character to give same result as other solutions
              var2 = NULL, var1 = NULL)]
},
{ #@akrun
  foo3 <- copy(foo)
  foo.index <- CJ(var1 = c(T,F), var2 = c(T,F))[, var3 := c('z', 'y', 'x', 'w')]
  setkey(foo3, var1, var2)
  foo3 <- foo3[foo.index, var3 := i.var3][, `:=` (var1=NULL, var2=NULL)][order(uid)]
  # assigned to foo3 to get same result as other solutions and used var1:=NULL, etc to achieve OP's
  # requirement "Moreover, any additional column that foo has besides (var1, var2) should be taken over into the new data.table"
}
)
#        min        lq   median        uq      max neval
#  19.635460 19.801922 19.93224 20.814533 22.57868   100
#  12.611448 12.762514 12.79219 12.864043 48.10415   100
#   4.691303  4.945683  4.98808  5.084922  7.21636   100
#
# making sure they give the same solutions
all.equal(foo1, foo2)
# [1] TRUE
all.equal(foo1, foo3)
# [1] TRUE