高效计算 data.table 列的线性组合 [英] Efficiently computing a linear combination of data.table columns

问题描述

我在 data.table 中有 nc 列，在向量中有 nc 标量.我想对列进行 线性组合，但我不知道我将使用哪些列.最有效的方法是什么?

I have nc columns in a data.table, and nc scalars in a vector. I want to take a linear combination of the columns, but I don't know ahead of time which columns I will be using. What is the most efficient way to do this?

require(data.table)
set.seed(1)

n  <- 1e5
nc <- 5
cf <- setNames(rnorm(nc),LETTERS[1:nc])
DT <- setnames(data.table(replicate(nc,rnorm(n))),LETTERS[1:nc])

方法

假设我想使用前四列.我可以手动写:

ways to do it

Suppose I want to use the first four columns. I can manually write:

DT[,list(cf['A']*A+cf['B']*B+cf['C']*C+cf['D']*D)]

我可以想到两种自动方式(在不知道应该全部使用 A-E 的情况下工作):

I can think of two automatic ways (that work without knowing that A-E should all be used):

mycols <- LETTERS[1:4] # the first four columns
DT[,list(as.matrix(.SD)%*%cf[mycols]),.SDcols=mycols]
DT[,list(Reduce(`+`,Map(`*`,cf[mycols],.SD))),.SDcols=mycols]

基准测试

我希望 as.matrix 会使第二个选项变慢，并且对 Map-Reduce 组合的速度真的没有直觉.

benchmarking

I expect the as.matrix to make the second option slow, and really have no intuition for the speed of Map-Reduce combinations.

require(rbenchmark)
options(datatable.verbose=FALSE) # in case you have it turned on

benchmark(
    manual=DT[,list(cf['A']*A+cf['B']*B+cf['C']*C+cf['D']*D)],
    coerce=DT[,list(as.matrix(.SD)%*%cf[mycols]),.SDcols=mycols],
    maprdc=DT[,list(Reduce(`+`,Map(`*`,cf[mycols],.SD))),.SDcols=mycols]
)[,1:6]

    test replications elapsed relative user.self sys.self
2 coerce          100    2.47    1.342      1.95     0.51
1 manual          100    1.84    1.000      1.53     0.31
3 maprdc          100    2.40    1.304      1.62     0.75

当我重复 benchmark 调用时，相对于手动方法，我的速度会降低 5% 到 40%.

I get anywhere from a 5% to 40% percent slowdown relative to the manual approach when I repeat the benchmark call.

这里的尺寸 - n 和 length(mycols) - 与我正在使用的尺寸接近，但我将多次运行这些计算，改变系数向量，cf.

The dimensions here -- n and length(mycols) -- are close to what I am working with, but I will be running these computations many times, altering the coefficient vector, cf.

推荐答案

这对我来说比你的手动版本快 2 倍:

This is almost 2x faster for me than your manual version:

Reduce("+", lapply(names(DT), function(x) DT[[x]] * cf[x]))

benchmark(manual = DT[, list(cf['A']*A+cf['B']*B+cf['C']*C+cf['D']*D)],
          reduce = Reduce('+', lapply(names(DT), function(x) DT[[x]] * cf[x])))
#    test replications elapsed relative user.self sys.self user.child sys.child
#1 manual          100    1.43    1.744      1.08     0.36         NA        NA
#2 reduce          100    0.82    1.000      0.58     0.24         NA        NA

若要仅迭代 mycols，请将 lapply 中的 names(DT) 替换为 mycols.

And to iterate over just mycols, replace names(DT) with mycols in lapply.

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