r数据中的循环.表:通过可变分布创建组,然后按新组计算均值



我有一组变量(约21个(,我想遍历每个变量,并为每个变量进行以下操作:1.每年分为10组,分组由年度分配确定确定2.根据这些新组的计算平均值(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,我想循环循环,然后是比率2等,计算每个分数,然后按每个新计算的十分位数总结val。请注意,这些不是编号变量,因此我无法使用iste((和1:21向量重新创建名称。首先,我写了此功能来进行分组:

# [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个迭代:

# 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)]

给出所需结果:

    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,如何通过每个变量进行21次循环?我可以很容易地获得这样的变量的名称:

> grep(c("ratio"), names(DT))
[1] 3 4
> names(DT)[grep(c("ratio"), names(DT))]
[1] "ratio"  "ratio2"

因此,请认为for (z in 1:length(namelist)) {}或某些东西会起作用。但是我不确定如何在数据中引用这些名称(或数字(。

要长时间格式...

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

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