我有一组变量（大约21个），我想循环并为每个变量执行以下操作： 1.按年份分为10组，其中组按年度分布的十分位数确定 2.基于这些新组的计算方法（相等和加权）。
测试数据：
set.seed(4)
YR = data.table(yr=1962:2015)
ID = data.table(id=10001:11000)
DT <- YR[,as.list(ID), by = yr] # intentional cartesian join
rm("YR","ID")
# 54,000 obs now add data
DT[,`:=` (ratio = rep(sample(10),each=5400)+rnorm(nrow(DT)),
ratio2 = rep(sample(5),each=10800)+rnorm(nrow(DT)),
weight = abs(rnorm(nrow(DT)))*100,
val = rnorm(nrow(DT))
)]
DT
yr id ratio ratio2 weight val
1: 1962 10001 6.689275 4.895357 129.10487 -0.2022073
2: 1962 10002 4.718753 4.505419 140.70420 -0.0887587
3: 1962 10003 5.786855 4.359488 242.10988 0.9511465
4: 1962 10004 7.896540 4.049974 89.23235 -1.3822148
5: 1962 10005 7.776863 2.233036 177.79650 -1.0671091
---
53996: 2015 10996 10.613272 3.345091 153.81424 0.9269429
53997: 2015 10997 11.260932 1.804315 15.68129 -1.6618414
53998: 2015 10998 8.591909 3.332643 134.80929 -1.1632596
53999: 2015 10999 9.143039 3.012160 178.77301 -0.4761060
54000: 2015 11000 7.470945 4.068919 121.13470 -1.7594423
所以，我想循环比率，然后是ratio2等，计算每个的十分位数，然后用每个新计算的十分位数来总结val。注意，这些不是编号变量，因此我无法使用paste（）和1:21向量重新创建名称。 首先，我写了这个函数来进行分组：
# [function] pctl.grp - order data into groups based on percentil breakpoints
# Number of groups passed
pctl.grp <- function(dat, grp) {
bp <- quantile(dat, probs = c(0,seq(100/grp,100,100/grp))/100)
cut(dat,bp,labels = FALSE, include.lowest = TRUE)
}
然后我可以这样做一次迭代：
# adds in new variable containing 10 groups numbered 1-10
DT[,ratiogrp := lapply(.SD, pctl.grp, 10), by = .(yr), .SDcols = c("ratio")]
DT[,.(ewval = mean(val),
ewratio = mean(ratio),
vwval = weighted.mean(val, weight, na.rm = TRUE),
vwratio = weighted.mean(ratio, weight, na.rm = TRUE)) ,by=ratiogrp][order(ratiogrp)]
这给出了期望的结果：
ratiogrp ewval ewratio vwval vwratio
1: 1 -0.027994385 3.576939 -0.039512050 3.572319
2: 2 -0.001146009 4.329835 0.005093692 4.331433
3: 3 -0.009087386 4.784103 -0.012764902 4.767494
4: 4 -0.014961467 5.094431 -0.015464918 5.110614
5: 5 0.014705294 5.373705 0.015276699 5.364962
6: 6 -0.010195630 5.645182 -0.014102394 5.618484
7: 7 0.001297953 5.949583 -0.012839401 5.925634
8: 8 -0.009300910 6.265297 -0.007141404 6.263371
9: 9 0.012970539 6.651047 0.018474949 6.684825
10: 10 0.003841495 7.363449 -0.004225650 7.351828
但是我如何循环执行每个变量21次呢？我可以轻松地获取我的变量名称：
> grep(c("ratio"), names(DT))
[1] 3 4
> names(DT)[grep(c("ratio"), names(DT))]
[1] "ratio" "ratio2"
所以认为for (z in 1:length(namelist)) {}或其他东西可行。但我不知道如何在data.table结构中引用这些名称（或数字）来重新创建我上面所做的。
答案 0 :(得分：2)
采用长格式......
mDT = melt(DT, meas=patterns("ratio"), value.name = "ratio")
setorder(mDT, variable, yr, ratio)
mDT[, dec := cut(.I, 10, labels = FALSE), by=.(yr, variable)]
mDT[, .(
mval = mean(val),
mrat = mean(ratio),
wmval = weighted.mean(val, weight),
wmrat = weighted.mean(ratio, weight)
), keyby=.(variable, dec)]
variable dec mval mrat wmval wmrat
1: ratio 1 -0.0279943849 3.576939 -0.039512050 3.572319
2: ratio 2 -0.0011460087 4.329835 0.005093692 4.331433
3: ratio 3 -0.0090873863 4.784103 -0.012764902 4.767494
4: ratio 4 -0.0149614666 5.094431 -0.015464918 5.110614
5: ratio 5 0.0147052939 5.373705 0.015276699 5.364962
6: ratio 6 -0.0101956297 5.645182 -0.014102394 5.618484
7: ratio 7 0.0012979528 5.949583 -0.012839401 5.925634
8: ratio 8 -0.0093009096 6.265297 -0.007141404 6.263371
9: ratio 9 0.0129705386 6.651047 0.018474949 6.684825
10: ratio 10 0.0038414948 7.363449 -0.004225650 7.351828
11: ratio2 1 -0.0120823787 1.195964 -0.016154551 1.199026
12: ratio2 2 -0.0072534833 1.904354 -0.030409684 1.908494
13: ratio2 3 -0.0283728080 2.282277 -0.028168936 2.301685
14: ratio2 4 -0.0068901529 2.590815 0.002836866 2.585152
15: ratio2 5 -0.0035769658 2.880104 0.002391468 2.872702
16: ratio2 6 0.0087575593 3.147469 0.004565452 3.134459
17: ratio2 7 -0.0052354409 3.412187 -0.005866282 3.426711
18: ratio2 8 0.0123337036 3.704371 0.009488475 3.701694
19: ratio2 9 0.0027419978 4.071582 -0.008958386 4.076264
20: ratio2 10 -0.0002925368 4.786477 0.003691116 4.772209