r - data.table 分组列的长度为 "J"中的 1

Question

学习中 data.table ，我发现了一个我无法优雅地解决的情况。

前面:lm 的荒谬公式很明显，我试图确定是否可以使用 data.table 中的关键字或特殊运算符轻松解决此细微差别。生态系统。

library(data.table)
mt <- as.data.table(mtcars)
mt[, list(model = list(lm(mpg ~ disp))), by = "cyl"]
#    cyl model
# 1:   6  <lm>
# 2:   4  <lm>
# 3:   8  <lm>
mt[, list(model = list(lm(mpg ~ disp + cyl))), by = "cyl"]
# Error in model.frame.default(formula = mpg ~ disp + cyl, drop.unused.levels = TRUE) : 
#   variable lengths differ (found for 'cyl')

这是因为在区块内部， cyl是长度为 1 的向量，而不是像其余值一样的列:

mt[, list(model = { browser(); list(lm(mpg ~ cyl+disp)); }), by = "cyl"]
# Called from: `[.data.table`(mt, , list(model = {
#     browser()
#     list(lm(mpg ~ cyl + disp))
#   ...
# Browse[1]> 
# debug at #1: list(lm(mpg ~ cyl + disp))
# Browse[2]> 
disp
# [1] 160.0 160.0 258.0 225.0 167.6 167.6 145.0
# Browse[2]> 
cyl
# [1] 6

最直接的似乎是在内部手动延长它作为临时变量或字面上需要的地方:

mt[, list(model = { cyl2 <- rep(cyl, nrow(.SD)); list(lm(mpg ~ cyl2+disp)); }), by = "cyl"]
mt[, list(model = list(lm(mpg ~ rep(cyl, nrow(.SD))+disp))), by = "cyl"]

问:有没有更优雅的方法来处理这个问题？

各种松散相关的问题，激发了我的好奇心(在 DT 对象中嵌入“东西”):

Setting column name in "group by" operation with data.table

Run a function inside data.table in R

Using data.table to create a column of regression coefficients

到目前为止，候选人很多:

mt[, .(model = .(lm(mpg ~ cyl + disp, data = mt[.I]))), by = .(cyl)]
mt[, .(model = .(lm(mpg ~ cyl + disp))), by =.(cylgroup=cyl)]
mt[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=names(mt)]
mt[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=TRUE]
mt[, .(model = .(lm(mpg ~ cyl + disp, data = cbind(.SD, as.data.table(.BY))))), by = "cyl"]

Answer 1

感谢所有候选人。

mt[, .(model = .(lm(mpg ~ cyl + disp, data = mt[.I]))), by = .(cyl)]
mt[, .(model = .(lm(mpg ~ cyl + disp))), by =.(cylgroup=cyl)]
mt[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=names(mt)]
mt[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=TRUE]
mt[, .(model = .(lm(mpg ~ cyl + disp, data = cbind(.SD, as.data.table(.BY))))), by = "cyl"]

性能(使用这个小模型)似乎有一些小的差异:

library(microbenchmark)
microbenchmark(
  c1 = mt[, .(model = .(lm(mpg ~ cyl + disp, data = mt[.I]))), by = .(cyl)],
  c2 = mt[, .(model = .(lm(mpg ~ cyl + disp))), by =.(cylgroup=cyl)],
  c3 = mt[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=names(mt)],
  c4 = mt[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=TRUE],
  c5 = mt[, .(model = .(lm(mpg ~ cyl + disp, data = cbind(.SD, as.data.table(.BY))))), by = "cyl"]
)
# Unit: milliseconds
#  expr    min      lq     mean  median      uq     max neval
#    c1 3.7328 4.21745 4.584591 4.43485 4.57465  9.8924   100
#    c2 2.6740 3.11295 3.244856 3.21655 3.28975  5.6725   100
#    c3 2.8219 3.30150 3.618646 3.46560 3.81250  6.8010   100
#    c4 2.9084 3.27070 3.620761 3.44120 3.86935  6.3447   100
#    c5 5.6156 6.37405 6.832622 6.54625 7.03130 13.8931   100

有更大的数据

mtbigger <- rbindlist(replicate(1000, mtcars, simplify=FALSE))
microbenchmark(
  c1 = mtbigger[, .(model = .(lm(mpg ~ cyl + disp, data = mtbigger[.I]))), by = .(cyl)],
  c2 = mtbigger[, .(model = .(lm(mpg ~ cyl + disp))), by =.(cylgroup=cyl)],
  c3 = mtbigger[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=names(mtbigger)],
  c4 = mtbigger[, .(model = .(lm(mpg ~ cyl + disp, .SD))), by=cyl, .SDcols=TRUE],
  c5 = mtbigger[, .(model = .(lm(mpg ~ cyl + disp, data = cbind(.SD, as.data.table(.BY))))), by = "cyl"]
)
# Unit: milliseconds
#  expr     min       lq     mean  median       uq      max neval
#    c1 27.1635 30.54040 33.98210 32.2859 34.71505  76.5064   100
#    c2 23.9612 25.83105 28.97927 27.5059 30.02720  67.9793   100
#    c3 25.7880 28.27205 31.38212 30.2445 32.79030 105.4742   100
#    c4 25.6469 27.84185 30.52403 29.8286 32.60805  37.8675   100
#    c5 29.2477 32.32465 35.67090 35.0291 37.90410  68.5017   100

(我猜相对性能比例类似。更好的裁决可能包括更广泛的数据。)

仅通过中值运行时间，看起来顶部(以很小的幅度)是:

mtbigger[, .(model = .(lm(mpg ~ cyl + disp))), by =.(cylgroup=cyl)]