在data.table中按组进行回归和汇总统计

Question

我想计算一些摘要统计数据,并在数据表中按组执行不同的回归,并将结果以"宽"格式(即每组包含多列的一行).我可以通过多个步骤完成它,但似乎应该可以一次完成所有操作.

考虑这个示例数据:

set.seed=46984
dt <- data.table(ID=c(rep('Frank',5),rep('Tony',5),rep('Ed',5)), y=rnorm(15), x=rnorm(15), z=rnorm(15),key="ID")
dt
#       ID          y          x            z
# 1:    Ed  0.2129400 -0.3024061  0.845335632
# 2:    Ed  0.4850342 -0.5159197 -0.087965415
# 3:    Ed  1.8917489  1.7803220  0.760465271
# 4:    Ed -0.4330460 -2.1720944  0.973812545
# 5:    Ed  0.7685060  0.7947470  1.279761200
# 6: Frank  0.4978475 -0.2906851  0.568101004
# 7: Frank  0.6323386 -0.5596599  1.537133025
# 8: Frank -0.8243218 -0.4354885  0.057818033
# 9: Frank  1.2402488  0.3229422  0.005995249
#10: Frank  0.2436210 -0.2651422  0.349532173
#11:  Tony  0.4179568  0.1418463  0.142380549
#12:  Tony  0.7036613  0.4402572  0.141237901
#13:  Tony -0.1978720 -0.9553784  0.480425820
#14:  Tony -1.7269375 -0.1881292  0.370583351
#15:  Tony  1.1064903  0.4375014 -0.798221750

假设我想通过ID获得中位数,通过ID对y~x执行线性回归,并通过ID对y~x + z执行线性回归.在这里,我得到中位数:

dt.med <- dt[,list(y.med=median(y)),by=ID]
dt.med
#      ID     y.med
#1:    Ed 0.4850342
#2: Frank 0.4978475
#3:  Tony 0.4179568

并且由于@DWin的这个答案,我在这里得到两组单独的回归系数作为ID列:

dt.reg.1 <- dt[,as.list(coef(lm(y ~ x))), by=ID]
dt.reg.1
#      ID (Intercept)         x
#1:    Ed  0.63057884 0.5482373
#2: Frank  0.69720351 1.3813007
#3:  Tony  0.08588421 1.0179131

dt.reg.2 <- dt[,as.list(coef(lm(y ~ x + z))), by=ID]
dt.reg.2
#      ID (Intercept)         x          z
#1:    Ed   0.8262577 0.5587170 -0.2582699
#2: Frank   0.4317538 2.7221024  1.1807442
#3:  Tony   0.1494439 0.3166547 -1.2029693

现在我必须加入三个结果集,并重命名列:

dt.ans <- dt.med[dt.reg.1][dt.reg.2]
setnames(dt.ans,c("ID","y.med","reg.1.c0","reg.1.c1","reg.2.c0","reg.2.c1","reg.2.c2"))

最后,这是示例的所需输出:

dt.ans
#      ID     y.med   reg.1.c0  reg.1.c1  reg.2.c0  reg.2.c1   reg.2.c2
#1:    Ed 0.4850342 0.63057884 0.5482373 0.8262577 0.5587170 -0.2582699
#2: Frank 0.4978475 0.69720351 1.3813007 0.4317538 2.7221024  1.1807442
#3:  Tony 0.4179568 0.08588421 1.0179131 0.1494439 0.3166547 -1.2029693

计算三个结果,连接它们,然后重命名列似乎效率低下.另外,我的实际表格很大,所以我想确保不要使用太多的系统内存. 是否可以在"one"data.table语句中完成所有操作？ 或者更一般地说,这可以更有效地完成吗？

我尝试了不同的东西.这是一个失败的例子,它给出了中位数,但忽略了回归系数:

dt[,as.list(median(y),coef(lm(y ~ x))), by=ID]
#      ID        V1
#1:    Ed 0.4850342
#2: Frank 0.4978475
#3:  Tony 0.4179568

Answer 1

dt[,c(y.med = median(y),
      reg.1 = as.list(coef(lm(y ~ x))),
      reg.2 = as.list(coef(lm(y ~ x + z)))), by=ID]
#      ID     y.med reg.1.(Intercept)   reg.1.x reg.2.(Intercept)      reg.2.x   reg.2.z
#1:    Ed 0.7280448        0.75977555 0.1132509        0.83322290 -0.484348116 0.7655563
#2: Frank 0.6100339       -0.07830664 0.2700846        0.04720686  0.004027939 0.7168521
#3:  Tony 0.2710623       -0.78319379 0.9166601       -0.35836990  0.622822617 0.4161102