在 r 中使用 data.table 来消除内部 for 循环

Question

我在 R 中有一个内部 for 循环，我已将其确定为代码中的重大瓶颈。该脚本模拟了时变政策对成年之前个人的影响。外循环遍历我想研究的队列列表（yob = 1910,...,1930 等）。内部循环从 a = 5 到 a = 17 的年龄计数。CSL.details 是一个 data.table，它包含我正在研究的每条法律的详细信息，以我获取的变量的形式出现，这些变量因年份而异= 出生年份 + a.要按出生队列了解该政策的总体影响，我需要跟踪每个 a.ca_years1、ca_years2、ca_years3 和 ca_years4。

ages = seq.int(5,17)
state = "Massachusetts"
yob = seq.int(1910, 1930)
for (birthyear in yob){
    ca_years1 = 0; ca_years2 = 0; ca_years3 = 0; ca_years4 = 0;
    for (a in ages){
      thisyear = birthyear + a
      # Grab each law for given state and year and implement exemption permit
      thislaw <- CSL.details[statename == state & yob == birthyear & thisyear == year]
      if (nrow(thislaw) == 0) next
      exempt_workpermit = (ca_years2 >= thislaw$workyrs & a >= thislaw$workage & thislaw$workage > 0)
      exempt_yearstodropout = (ca_years3 >= thislaw$earlyyrs & a >= thislaw$earlyyrs_condition & thislaw$earlyyrs > 0)
      exempt_cont = ((ca_years2 + ca_years4) >= thislaw$contyrs & thislaw$contyrs > 0)
      # Increment each law when school is required
      if(thislaw$entryage <= a & a < thislaw$exitage){
        ca_years1 = ca_years1 + 1
        if(!exempt_workpermit){ca_years2 = ca_years2 + 1}
        if(!exempt_yearstodropout){ca_years3 = ca_years3 + 1}
      }
      if(thislaw$contage > a & 
         a >= thislaw$workage & 
         !exempt_cont & 
         thislaw$workage > 0 &
         !(thislaw$entryage <= a & a < thislaw$exitage & !exempt_workpermit)
      ){ca_years4 = ca_years4 + 1}
      
    }
    CSL.exposures[statename == state & yob == birthyear]$ca_years1 = ca_years1
    CSL.exposures[statename == state & yob == birthyear]$ca_years2 = ca_years2
    CSL.exposures[statename == state & yob == birthyear]$ca_years3 = ca_years3
    CSL.exposures[statename == state & yob == birthyear]$ca_years4 = ca_years4
  }

是否有data.table替换内循环的解决方案？我是一名中级 R 编码员，想如何入门有点困难。虽然我更喜欢 data.table，但我愿意接受 dplyr 类型的解决方案，如果它们能显着加快代码速度的话。

编辑：这里是 CSL.detail 的示例，作为复制粘贴的 data.table。

statename year  yob statefip entryage exitage earlyyrs earlyyrs_condition workage workyrs contage contyrs statecompschoolyr
    1: Massachusetts 1913 1800       25        7      16        4                 14      14       4      16       0              1852
    2: Massachusetts 1913 1801       25        7      16        4                 14      14       4      16       0              1852
    3: Massachusetts 1913 1802       25        7      16        4                 14      14       4      16       0              1852
    4: Massachusetts 1913 1803       25        7      16        4                 14      14       4      16       0              1852
    5: Massachusetts 1913 1804       25        7      16        4                 14      14       4      16       0              1852

Answer 1

我设法重构代码来解决问题。关键思想是利用 state 和 yob 作为分组变量（因为所有计算都发生在 state 和 yob 对中）。这完全消除了外部循环，只需要一个循环，按年龄迭代。我只是把这个答案保存在这里以供参考，但我不确定 whosebug.com 社区是否有更广泛的教训，所以请随时删除。时间节省大约 95%，主要是因为它减少了调用 data.table.

的开销时间

for(a in ages){
  # grab running total of years of education compelled by state and year of birth
  CSL.details[CSL.exposures, on = .(statename, yob), 
              `:=` (ca_years1 = i.ca_years1,
                    ca_years2 = i.ca_years2,
                    ca_years3 = i.ca_years3,
                    ca_years4 = i.ca_years4)] %>%
    .[year == a + yob, 
      `:=`(
        # create exemptions by age based on number of years of schooling completed
        exempt_workpermit = (ca_years2 >= workyrs & a >= workage & workage > 0),
        exempt_yearstodropout = (ca_years3 >= earlyyrs & a >= earlyyrs_condition & earlyyrs > 0),
        exempt_cont = ((ca_years2 + ca_years4) >= contyrs & contyrs > 0)
      ), by = .(statename, yob)]
  
  CSL.exposures[
    CSL.details[year == a + yob], on = .(yob, statename),
    `:=` (exempt_workpermit = i.exempt_workpermit, exempt_yearstodropout = i.exempt_yearstodropout, 
          exempt_cont = i.exempt_cont, entryage  = i.entryage, 
          exitage = i.exitage, contage = i.contage, workage = i.workage) ] %>%
    .[ , 
       `:=` (
         ca_years1 = 
           fifelse(entryage <= a & a < exitage, 
                   ca_years1 + 1, ca_years1, na = as.numeric(ca_years1)),
         ca_years2 = 
           fifelse(entryage <= a & a < exitage & !exempt_workpermit, 
                   ca_years2 + 1, ca_years2, na = as.numeric(ca_years2)),
         ca_years3 = 
           fifelse(entryage <= a & a < exitage & !exempt_yearstodropout, 
                   ca_years3 + 1, ca_years3, na = as.numeric(ca_years3)),
         ca_years4 = 
           fifelse(contage > a & a >= workage & !exempt_cont & 
                     workage > 0 &
                     !(entryage <= a & a < exitage & !exempt_workpermit),
                   ca_years4 + 1, ca_years4, na = as.numeric(ca_years4))),
       by = .(statename, yob)
    ]
}

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