在data.table中按组进行回归和汇总统计

我想计算一些汇总统计数据，并在数据表中按组进行不同的回归，并以宽"格式显示结果(即，每组一行，多列).我可以分多个步骤进行操作，但似乎应该可以一次完成所有操作.

I would like to calculate some summary statistics and perform different regressions by group within a data table, and have the results in "wide" format (i.e. one row per group with several columns). I can do it in multiple steps, but it seems like it should be possible to do all at once.

请考虑以下示例数据:

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

假设我要按ID获取中位数，按ID对y〜x进行线性回归，并按ID对y〜x + z进行线性回归.在这里，我得到了中位数:

Let's say I want to get the median by ID, perform linear regression on y ~ x by ID, and perform linear regression on y ~ x + z by ID. Here I get the median:

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

感谢@DWin的此答案，在这里，我得到了两个独立的回归系数集，作为ID的列:

And thanks to this answer by @DWin, here I get the two individual sets of regression coefficients as columns by ID:

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，并重命名各列:

Now I have to join the three result sets, and rename the columns:

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，将它们合并，然后重命名列似乎效率低下.另外，我的实际表比较大，所以我想确保我不使用过多的系统内存. 是否可以在一个" data.table语句中完成所有操作?或者更笼统地说，可以更有效地做到这一点吗?

It seems inefficient to calculate the three results, join them, and then rename the columns. Also, my actual tables are largish so I'd like to make sure I don't use too much system memory. Is it possible to do this all within "one" data.table statement? Or more generally, can this be done more efficiently?

我尝试了不同的事情.这是一个失败的示例，该示例给出了中位数，但忽略了回归系数:

I've tried different things. Here is one failed example that gives the median but ignores the regression coefficients:

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