Question

在学习data.table时，我发现无法优雅解决的情况。

最前面的：lm公式的荒谬很明显，我试图确定这种细微差别是否可以通过{{ 1}}生态系统。

data.table

这是因为在块中，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')是长度为1的向量，而不是其余值的列：

cyl

最直接的方法似乎是在内部将其作为临时变量或在需要时从字面上手动加长：

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

问：有没有更优雅的方式来解决这个问题？

各种与松散相关的问题，激发了我的好奇心（将“材料”嵌入DT对象）：

到目前为止，很多人都很好：

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"]

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"]

（使用这个小模型）的性能似乎有一些细微的差异：

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

数据量更大

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

（我猜类似的相对绩效尺度。更好的裁决可能包括更广泛的数据。）

仅凭运行时间的中位数，它看起来就好像是顶部（很小的幅度）：

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

data.table分组列的长度为“ J”中的1

1 个答案: