使用 data.table 有效地处理组内的重复值

Question

从按组内重复（即每行中的相同值）的列 (variable) 获取单个值的首选方法是什么？我应该使用variable[1] 还是应该在by 语句中包含该变量并使用.BY$variable？假设我希望返回值包含 variable 作为列。

从以下测试中可以清楚地看出，在 by 语句中添加额外的变量会减慢速度，甚至可以降低新变量的键控成本（或使用诡计告诉 data.table 不需要额外的键控）。为什么额外的已键入 by 变量会减慢速度？

我想我曾希望包含已经键入的 by 变量将是一种方便的语法技巧，可以将这些变量包含在返回 data.table 中，而无需在 j 语句中显式命名它们，但似乎这是这是不可取的，因为有一些额外的变量相关的开销，即使它们已经被键入。所以我的问题是，是什么导致了这种开销？

一些示例数据：

library(data.table)
n <- 1e8
y <- data.table(sample(1:5,n,replace=TRUE),rnorm(n),rnorm(n))
y[,sumV2:=sum(V2),keyby=V1]

时间表明使用variable[1]（在本例中为sumV2[1]）的方法更快。

x <- copy(y)
system.time(x[, list(out=sum(V3*V2)/sumV2[1],sumV2[1]),keyby=V1])
system.time(x[, list(out=sum(V3*V2)/.BY$sumV2),keyby=list(V1,sumV2)])

我想这并不奇怪，因为data.table 无法知道 setkey(V1) 和 setkey(V1,sumV2) 定义的组实际上是相同的。

令我感到惊讶的是，即使 data.table 键入 setkey(V1,sumV2)（我们完全忽略了设置新密钥所需的时间），使用 sumV2[1] 仍然更快。这是为什么呢？

x <- copy(y)
setkey(x,V1,sumV2)
system.time(x[, list(out=sum(V3*V2)/sumV2[1],sumV2[1]),by=V1])
system.time(x[, list(out=sum(V3*V2)/.BY$sumV2),by=list(V1,sumV2)])

另外，执行setkey(x,V2,sumV2) 所花费的时间是不可忽略的。有什么方法可以欺骗 data.table 通过告诉它密钥实际上并没有发生实质性变化来跳过实际重新键入 x？

x <- copy(y)
system.time(setkey(x,V1,sumV2))

回答我自己的问题，似乎我们可以在设置键时跳过排序，只需分配“排序”属性。这是允许的吗？它会破坏东西吗？

x <- copy(y)
system.time({
  setattr(x, "sorted", c("V1","sumV2"))
  x[, list(out=sum(V3*V2)/.BY$sumV2),by=list(V1,sumV2)]
})

我不知道这是不好的做法还是可能会破坏事情。但是使用setattr 诡计比显式键入要快得多：

x <- copy(y)
system.time({
  setkey(x,V1,sumV2)
  x[, list(out=sum(V3*V2)/.BY$sumV2),by=list(V1,sumV2)]
})

但即使在 by 语句中使用 setattr 诡计并结合使用 sumV2 仍然不如将 sumV2 完全排除在 by 语句之外那么快：

x <- copy(y)
system.time(x[, list(out=sum(V3*V2)/sumV2[1],sumV2[1]),keyby=V1])

在我看来，使用通过属性设置键并使用 sumV2 作为长度为 1 的变量在每个组中应该比仅键入 V1 和使用 sumV2[1] 更快。如果sumV2 未指定为by 变量，则需要为每个组生成sumV2 中重复值的整个向量，然后再将其子集为sumV2[1]。将此与 sumV2 是 by 变量时进行比较，每个组中的 sumV2 只有一个长度为 1 的向量。显然我在这里的推理是不正确的。谁能解释为什么？为什么 sumV2[1] 是最快的选项，甚至与使用 setattr 诡计后将 sumV2 设为变量相比也是最快的选择？

顺便说一句，我很惊讶地发现使用attr<- 并不比setattr 慢（都是瞬时的，意味着根本没有复制）。这与我的理解相反，即基本 R foo<- 函数会复制数据。

x <- copy(y)
system.time(setattr(x, "sorted", c("V1","sumV2")))
x <- copy(y)
system.time(attr(x,"sorted") <- c("V1","sumV2"))

用于此问题的相关SessionInfo()：

data.table version 1.12.2
R version 3.5.3

Answer 1

好的，所以我没有很好的技术答案，但我想我已经在 options(datatable.verbose=TRUE) 的帮助下从概念上解决了这个问题

创建数据

library(data.table)
n <- 1e8

y_unkeyed_5groups <- data.table(sample(1:10000,n,replace=TRUE),rnorm(n),rnorm(n))
y_unkeyed_5groups[,sumV2:=sum(V2),keyby=V1]
y_unkeyed_10000groups <- data.table(sample(1:10000,n,replace=TRUE),rnorm(n),rnorm(n))
y_unkeyed_10000groups[,sumV2:=sum(V2),keyby=V1]

慢跑

x <- copy(y)
system.time({
  setattr(x, "sorted", c("V1","sumV2"))
  x[, list(out=sum(V3*V2)/.BY$sumV2),by=list(V1,sumV2)]
})
# Detected that j uses these columns: V3,V2 
# Finding groups using uniqlist on key ... 1.050s elapsed (1.050s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'list(sum(V3 * V2)/.BY$sumV2)'
# GForce is on, left j unchanged
# Old mean optimization is on, left j unchanged.
# Making each group and running j (GForce FALSE) ... 
# memcpy contiguous groups took 0.305s for 6 groups
# eval(j) took 0.254s for 6 calls
# 0.560s elapsed (0.510s cpu) 
# user  system elapsed 
# 1.81    0.09    1.72 

跑得快：

x <- copy(y)
system.time(x[, list(out=sum(V3*V2)/sumV2[1],sumV2[1]),keyby=V1])
# Detected that j uses these columns: V3,V2,sumV2 
# Finding groups using uniqlist on key ... 0.060s elapsed (0.070s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'list(sum(V3 * V2)/sumV2[1], sumV2[1])'
# GForce is on, left j unchanged
# Old mean optimization is on, left j unchanged.
# Making each group and running j (GForce FALSE) ... 
# memcpy contiguous groups took 0.328s for 6 groups
# eval(j) took 0.291s for 6 calls
# 0.610s elapsed (0.580s cpu) 
# user  system elapsed 
# 1.08    0.08    0.82 

finding groups 部分是造成差异的原因。我猜这里发生的事情是设置 key 实际上只是排序（我应该从属性的命名方式中猜到！）并且实际上并没有做任何事情来定义组的开始和结束位置。因此，即使data.table 知道sumV2 已排序，它也不知道它是相同的值，因此必须找到sumV2 中的组的开始和结束位置。

我的猜测是，在技术上可以编写data.table，其中键控不仅排序，而且实际上将每个组的开始/结束行存储在键控变量中，但这可能会占用很多时间具有大量组的 data.tables 的内存。

知道了这一点，似乎建议不要一遍又一遍地重复相同的 by 语句，而是在单个 by 语句中完成您需要做的所有事情。总的来说，这可能是一个很好的建议，但对于少数群体而言，情况并非如此。参见下面的反例：

我以我认为使用 data.table 最快的方式重写了这个（只有一个 by 语句，并使用了 GForce）：

library(data.table)
n <- 1e8
y_unkeyed_5groups <- data.table(sample(1:5,n, replace=TRUE),rnorm(n),rnorm(n))
y_unkeyed_10000groups <- data.table(sample(1:10000,n, replace=TRUE),rnorm(n),rnorm(n))

x <- copy(y_unkeyed_5groups)
system.time({
  x[, product:=V3*V2]
  outDT <- x[,list(sumV2=sum(V2),sumProduct=sum(product)),keyby=V1]
  outDT[,`:=`(out=sumProduct/sumV2,sumProduct=NULL) ]
  setkey(x,V1)
  x[outDT,sumV2:=sumV2,all=TRUE]
  x[,product:=NULL]
  outDT
})

# Detected that j uses these columns: V3,V2 
# Assigning to all 100000000 rows
# Direct plonk of unnamed RHS, no copy.
# Detected that j uses these columns: V2,product 
# Finding groups using forderv ... 0.350s elapsed (0.810s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'list(sum(V2), sum(product))'
# GForce optimized j to 'list(gsum(V2), gsum(product))'
# Making each group and running j (GForce TRUE) ... 1.610s elapsed (4.550s cpu) 
# Detected that j uses these columns: sumProduct,sumV2 
# Assigning to all 5 rows
# RHS for item 1 has been duplicated because NAMED is 3, but then is being plonked. length(values)==2; length(cols)==2)
# forder took 0.98 sec
# reorder took 3.35 sec
# Starting bmerge ...done in 0.000s elapsed (0.000s cpu) 
# Detected that j uses these columns: sumV2 
# Assigning to 100000000 row subset of 100000000 rows
# Detected that j uses these columns: product 
# Assigning to all 100000000 rows
# user  system elapsed 
# 11.00    1.75    5.33 


x2 <- copy(y_unkeyed_5groups)
system.time({
  x2[,sumV2:=sum(V2),keyby=V1]
  outDT2 <- x2[, list(sumV2=sumV2[1],out=sum(V3*V2)/sumV2[1]),keyby=V1]
})
# Detected that j uses these columns: V2 
# Finding groups using forderv ... 0.310s elapsed (0.700s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'sum(V2)'
# Old mean optimization is on, left j unchanged.
# Making each group and running j (GForce FALSE) ... 
# collecting discontiguous groups took 0.714s for 5 groups
# eval(j) took 0.079s for 5 calls
# 1.210s elapsed (1.160s cpu) 
# setkey() after the := with keyby= ... forder took 1.03 sec
# reorder took 3.21 sec
# 1.600s elapsed (3.700s cpu) 
# Detected that j uses these columns: sumV2,V3,V2 
# Finding groups using uniqlist on key ... 0.070s elapsed (0.070s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'list(sumV2[1], sum(V3 * V2)/sumV2[1])'
# GForce is on, left j unchanged
# Old mean optimization is on, left j unchanged.
# Making each group and running j (GForce FALSE) ... 
# memcpy contiguous groups took 0.347s for 5 groups
# eval(j) took 0.265s for 5 calls
# 0.630s elapsed (0.620s cpu) 
# user  system elapsed 
# 6.57    0.98    3.99 

all.equal(x,x2)
# TRUE
all.equal(outDT,outDT2)
# TRUE

好吧，事实证明，当只有 5 个组时，不重复语句和使用 GForce 所获得的效率并不重要。但是对于更多的组来说，这确实会有所不同，（尽管我没有写出这样一种方式来区分仅使用一个 by 语句而不是 GForce 与使用 GForce 和多个 by 语句的好处）：

x <- copy(y_unkeyed_10000groups)
system.time({
  x[, product:=V3*V2]
  outDT <- x[,list(sumV2=sum(V2),sumProduct=sum(product)),keyby=V1]
  outDT[,`:=`(out=sumProduct/sumV2,sumProduct=NULL) ]
  setkey(x,V1)
  x[outDT,sumV2:=sumV2,all=TRUE]
  x[,product:=NULL]
  outDT
})
# 
# Detected that j uses these columns: V3,V2 
# Assigning to all 100000000 rows
# Direct plonk of unnamed RHS, no copy.
# Detected that j uses these columns: V2,product 
# Finding groups using forderv ... 0.740s elapsed (1.220s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'list(sum(V2), sum(product))'
# GForce optimized j to 'list(gsum(V2), gsum(product))'
# Making each group and running j (GForce TRUE) ... 0.810s elapsed (2.390s cpu) 
# Detected that j uses these columns: sumProduct,sumV2 
# Assigning to all 10000 rows
# RHS for item 1 has been duplicated because NAMED is 3, but then is being plonked. length(values)==2; length(cols)==2)
# forder took 1.97 sec
# reorder took 11.95 sec
# Starting bmerge ...done in 0.000s elapsed (0.000s cpu) 
# Detected that j uses these columns: sumV2 
# Assigning to 100000000 row subset of 100000000 rows
# Detected that j uses these columns: product 
# Assigning to all 100000000 rows
# user  system elapsed 
# 18.37    2.30    7.31 

x2 <- copy(y_unkeyed_10000groups)
system.time({
  x2[,sumV2:=sum(V2),keyby=V1]
  outDT2 <- x[, list(sumV2=sumV2[1],out=sum(V3*V2)/sumV2[1]),keyby=V1]
})

# Detected that j uses these columns: V2 
# Finding groups using forderv ... 0.770s elapsed (1.490s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'sum(V2)'
# Old mean optimization is on, left j unchanged.
# Making each group and running j (GForce FALSE) ... 
# collecting discontiguous groups took 1.792s for 10000 groups
# eval(j) took 0.111s for 10000 calls
# 3.960s elapsed (3.890s cpu) 
# setkey() after the := with keyby= ... forder took 1.62 sec
# reorder took 13.69 sec
# 4.660s elapsed (14.4s cpu) 
# Detected that j uses these columns: sumV2,V3,V2 
# Finding groups using uniqlist on key ... 0.070s elapsed (0.070s cpu) 
# Finding group sizes from the positions (can be avoided to save RAM) ... 0.000s elapsed (0.000s cpu) 
# lapply optimization is on, j unchanged as 'list(sumV2[1], sum(V3 * V2)/sumV2[1])'
# GForce is on, left j unchanged
# Old mean optimization is on, left j unchanged.
# Making each group and running j (GForce FALSE) ... 
# memcpy contiguous groups took 0.395s for 10000 groups
# eval(j) took 0.284s for 10000 calls
# 0.690s elapsed (0.650s cpu) 
# user  system elapsed 
# 20.49    1.67   10.19 

all.equal(x,x2)
# TRUE
all.equal(outDT,outDT2)
# TRUE

更一般地说，data.table 的速度非常快，但为了提取最快和最有效的计算以充分利用底层 C 代码，您需要特别注意 data.table 的内部工作原理。我最近了解了 data.table 中的 GForce 优化，似乎特定形式的 j 语句（涉及简单函数，如 mean 和 sum）在有 by 语句时直接在 C 中解析和执行。