.SD in data.table join 以引用 i 中的任意列列表

Question

问题：根据连接键使用另一个 table 中的权重计算一个 table 列的加权平均值。

以下是 reprex 中的步骤：

library(data.table)
#DT1 table of values - here just 2 columns, but may be an arbitrary number
DT1 <- data.table(k1 = c('A1','A2','A3'), 
                  k2 = c('X','X','Y'), 
                  v1 = c(10,11,12), 
                  v2 = c(.5, .6, 1.7))
#DT2 table of weights - columns correspond to value columns in table 1
DT2 <- data.table(k2 = c('X','Y'), 
                  w1 = c(5,2), 
                  w2 = c(1,7))
#Vectors of corresponding column names (could be any number of columns)
vals <- c('v1','v2')
weights <- c('w1','w2')
i.weights <- paste0('i.', weights)

#1. This returns all columns
DT1[DT2, on=.(k2)]
#>    k1 k2 v1  v2 w1 w2
#> 1: A1  X 10 0.5  5  1
#> 2: A2  X 11 0.6  5  1
#> 3: A3  Y 12 1.7  2  7
#2. This use of SD is standard
DT1[DT2, on=.(k2), .SD, .SDcols = vals, by=.(k1)]
#>    k1 v1  v2
#> 1: A1 10 0.5
#> 2: A2 11 0.6
#> 3: A3 12 1.7
#3. But refer to the columns of i (DT2) and it fails, both without and with the i. prefix
DT1[DT2, on=.(k2), .SD, .SDcols = weights, by=.(k1)]
#> Error in `[.data.table`(DT1, DT2, on = .(k2), .SD, .SDcols = weights, : Some items of .SDcols are not column names: [w1, w2]
DT1[DT2, on=.(k2), .SD, .SDcols = i.weights, by=.(k1)]
#> Error in `[.data.table`(DT1, DT2, on = .(k2), .SD, .SDcols = i.weights, : Some items of .SDcols are not column names: [i.w1, i.w2]
#4. So following suggestion in 
# turn to mget() - in one command it fails
DT1[DT2, on=.(k2), c(mget(vals), mget(weights)), by=.(k1,k2)]
#> Error: value for 'w1' not found
#5. But by exploiting 1. above and splitting into chained queries we get success!
DT1[DT2, on=.(k2),][, c(mget(vals), mget(weights)), by=.(k1,k2)]
#>    k1 k2 v1  v2 w1 w2
#> 1: A1  X 10 0.5  5  1
#> 2: A2  X 11 0.6  5  1
#> 3: A3  Y 12 1.7  2  7
#6. Now we can turn to the original intention, but no luck
DT1[DT2, on=.(k2)][, .(wmean = weighted.mean(mget(vals), mget(weights))), by=.(k1,k2)]
#> Error in x * w: non-numeric argument to binary operator
#7. One more step - turn the lists returned by mget to data.tables - hurrahh!
DT1[DT2, on=.(k2)][, .(wmean = weighted.mean(setDT(mget(vals)), setDT(mget(weights)))), by=.(k1,k2)]
#>    k1 k2    wmean
#> 1: A1  X 8.416667
#> 2: A2  X 9.266667
#> 3: A3  Y 3.988889

^{由 reprex package (v2.0.0)}

于 2021-11-26 创建

真的有这么难吗？有没有更直接（最好是性能更高）的方法来做到这一点？

推论 - 我实际上想用这个计算在 DT1 中创建一个新列，但由于这以两个链接查询结束，我无法在此命令中进行分配。我必须创建一个新的 table 并将其连接回原始以添加该列。有没有避免这个额外步骤的解决方案？

Answer 1

另一种方法是将数据从宽融合到长，然后相互连接。

molten_dt1 = melt(DT1, measure.vars = vals)[, variable := as.integer(substring(variable, 2))]
molten_dt2 = melt(DT2, measure.vars = weights)[, variable := as.integer(substring(variable, 2))]

molten_dt1[molten_dt2, 
           on = .(k2, variable)
           ][,
             weighted.mean(value, i.value),
             by = .(k1, k2)]

之所以不简单，是因为我们需要进行并行列查找（即 v1 * w1 和 v2 * w2），复杂性总是会增加，因为我们需要考虑这种关系列之间。融化数据使我们能够简化我们的方法，因为数据结构允许我们加入，而且我们在 weighted.mean 中使用向量而不是 data.frames.

另一个注意事项是，如果您为列表创建一个新的 weighted.mean() 方法，这允许我们跳过 setDT 要求，那么您可以简化原始方法。

## slight changes made to stats:::weighted.mean.default
weighted.mean.list = function (x, w, ..., na.rm = FALSE) 
{
  x = unlist(x)
  if (missing(w)) {
    if (na.rm) 
      x <- x[!is.na(x)]
    return(sum(x)/length(x))
  }
  w = unlist(w)
  if (length(w) != length(x)) 
    stop("'x' and 'w' must have the same length")
  if (na.rm) {
    i <- !is.na(x)
    w <- w[i]
    x <- x[i]
  }
  sum((x * w)[w != 0])/sum(w)
}

DT1[DT2, on=.(k2)][, .(wmean = weighted.mean(mget(vals), mget(weights))), by=.(k1,k2)]

.SD in data.table join 以引用 i 中的任意列列表

.SD in data.table join to refer to arbitrary list of columns in i

join

data.table