【问题标题】：How can I compute statistics by decile groups in data.table如何按 data.table 中的十分位组计算统计信息
【发布时间】：2014-04-29 10:08:03
【问题描述】：

我有一个 data.table，想按组计算统计数据。

R) set.seed(1)
R) DT=data.table(a=rnorm(100),b=rnorm(100))

这些组应该由

定义

R) quantile(DT$a,probs=seq(.1,.9,.1))
           10%            20%            30%            40%            50%            60%            70%            80%            90% 
-1.05265747329 -0.61386923071 -0.37534201964 -0.07670312896  0.11390916079  0.37707993057  0.58121734252  0.77125359976  1.18106507751

我如何计算每个 bin 的平均 b，如果 b=-.5 我在 [-0.61386923071,-0.37534201964] 之内，那么在 bin 3 中

【问题讨论】：

标签： r data.table quantile

【解决方案1】：

怎么样：

> DT[, mean(b), keyby=cut(a,quantile(a,probs=seq(.1,.9,.1)))]
                cut          V1
1:               NA -0.31359818
2:   (-1.05,-0.614] -0.14103182
3:  (-0.614,-0.375] -0.33474492
4: (-0.375,-0.0767]  0.20827735
5:  (-0.0767,0.114]  0.14890251
6:    (0.114,0.377]  0.16685304
7:    (0.377,0.581]  0.07086979
8:    (0.581,0.771]  0.17950572
9:     (0.771,1.18] -0.04951607

为了看看那个 NA（无论如何都要检查结果），我接下来做了：

> DT[, list(mean(b),.N,list(a)), keyby=cut(a,quantile(a,probs=seq(.1,.9,.1)))]
                cut          V1  N                                                                                                                      V3
1:               NA -0.31359818 20                1.59528080213779,1.51178116845085,-2.2146998871775,-1.98935169586337,-1.47075238389927,1.35867955152904,
2:   (-1.05,-0.614] -0.14103182 10        -0.626453810742332,-0.835628612410047,-0.820468384118015,-0.621240580541804,-0.68875569454952,-0.70749515696212,
3:  (-0.614,-0.375] -0.33474492 10        -0.47815005510862,-0.41499456329968,-0.394289953710349,-0.612026393250771,-0.443291873218433,-0.589520946188072,
4: (-0.375,-0.0767]  0.20827735 10      -0.305388387156356,-0.155795506705329,-0.102787727342996,-0.164523596253587,-0.253361680136508,-0.112346212150228,
5:  (-0.0767,0.114]  0.14890251 10 -0.0449336090152309,-0.0161902630989461,0.0745649833651906,-0.0561287395290008,-0.0538050405829051,-0.0593133967111857,
6:    (0.114,0.377]  0.16685304 10             0.183643324222082,0.329507771815361,0.36458196213683,0.341119691424425,0.188792299514343,0.153253338211898,
7:    (0.377,0.581]  0.07086979 10            0.487429052428485,0.575781351653492,0.389843236411431,0.417941560199702,0.387671611559369,0.556663198673657,
8:    (0.581,0.771]  0.17950572 10             0.738324705129217,0.593901321217509,0.61982574789471,0.763175748457544,0.696963375404737,0.768532924515416,
9:     (0.771,1.18] -0.04951607 10              1.12493091814311,0.943836210685299,0.821221195098089,0.918977371608218,0.782136300731067,1.10002537198388,

除此之外：我已经返回了一个list 列（每个单元格本身就是一个向量），以便快速查看进入垃圾箱的值，只是为了检查。 data.table 打印时显示逗号（并且每个单元格仅显示前 6 个项目），但 V3 的每个单元格实际上都有一个数字向量。

因此，第一个和最后一个 break 之外的值被一起编码为 NA。我不清楚如何告诉cut 不要那样做。所以我只是添加了 -Inf 和 +Inf ：

> DT[,list(mean(b),.N),keyby=cut(a,c(-Inf,quantile(a,probs=seq(.1,.9,.1)),+Inf))]
                 cut          V1  N
 1:     (-Inf,-1.05] -0.16938368 10
 2:   (-1.05,-0.614] -0.14103182 10
 3:  (-0.614,-0.375] -0.33474492 10
 4: (-0.375,-0.0767]  0.20827735 10
 5:  (-0.0767,0.114]  0.14890251 10
 6:    (0.114,0.377]  0.16685304 10
 7:    (0.377,0.581]  0.07086979 10
 8:    (0.581,0.771]  0.17950572 10
 9:     (0.771,1.18] -0.04951607 10
10:      (1.18, Inf] -0.45781268 10

这样更好。或者：

> DT[, list(mean(b),.N), keyby=cut(a,quantile(a,probs=seq(0,1,.1)),include=TRUE)]
                 cut          V1  N
 1:    [-2.21,-1.05] -0.16938368 10
 2:   (-1.05,-0.614] -0.14103182 10
 3:  (-0.614,-0.375] -0.33474492 10
 4: (-0.375,-0.0767]  0.20827735 10
 5:  (-0.0767,0.114]  0.14890251 10
 6:    (0.114,0.377]  0.16685304 10
 7:    (0.377,0.581]  0.07086979 10
 8:    (0.581,0.771]  0.17950572 10
 9:     (0.771,1.18] -0.04951607 10
10:       (1.18,2.4] -0.45781268 10

这样您就可以看到最小值和最大值是什么，而不是显示 -Inf 和 +Inf。请注意，您需要将 include=TRUE 传递给 cut 否则将返回 11 个 bin，而第一个 bin 中只有 1 个。

【讨论】：

我错过了那个剪切命令！很酷，这很完美......我也从未使用过 keyby
这里为什么使用keyby而不是by？
@hadley 对垃圾箱进行排序。 by= 按首次出现顺序返回组。
@MattDowle 当我运行DT[, list(mean(b),.N,list(a)), keyby=cut(a,quantile(a,probs=seq(.1,.9,.1)))] 时，我得到Error in setkeyv(ans, names(ans)[seq_along(byval)]) Item 4 of list is not a vector。但是其他 3 个命令工作正常。如果我运行DT[, list(mean(b),.N,paste0(as.character(a),collapse=",")), keyby=cut(a,quantile(a,probs=seq(.1,.9,.1)))]，那就行得通。我在 R3.0.1 Win7 上运行 DT 1.9.2。知道这里发生了什么吗？

【解决方案2】：

我经常做这种事情，所以我在我的 R 包中为它写了一个非常灵活的 bin_data() 方法 - mltools。它完全基于data.table，并利用了新的non-equi joins。

要回答您的具体问题，请将 Bin1 设置为 DT 中的一列，然后按 Bin1 分组

library(data.table)
library(mltools)

DT[, Bin1 := bin_data(vals=a, bins=seq(.1, .9, .1), binType="quantile")]
DT[, list(mean(b)), keyby=Bin1]

                                        Bin1          V1
1:                                        NA -0.31359818
2:   [-1.05265747329296, -0.613869230708978) -0.14103182
3:  [-0.613869230708978, -0.375342019639661) -0.33474492
4: [-0.375342019639661, -0.0767031289639095)  0.20827735
5:  [-0.0767031289639095, 0.113909160788544)  0.14890251
6:    [0.113909160788544, 0.377079930573521)  0.16685304
7:    [0.377079930573521, 0.581217342522697)  0.07086979
8:    [0.581217342522697, 0.771253599758546)  0.17950572
9:      [0.771253599758546, 1.1810650775142] -0.04951607

你也可以做其他很酷的事情

按分位数制作 10 个等间距的 bin

DT[, Bin2 := bin_data(vals=a, bins=10, binType="quantile")]
DT[, list(mean(b)), keyby=Bin2]

                                         Bin2          V1
 1:     [-2.2146998871775, -1.05265747329296) -0.16938368
 2:   [-1.05265747329296, -0.613869230708978) -0.14103182
 3:  [-0.613869230708978, -0.375342019639661) -0.33474492
 4: [-0.375342019639661, -0.0767031289639095)  0.20827735
 5:  [-0.0767031289639095, 0.113909160788544)  0.14890251
 6:    [0.113909160788544, 0.377079930573521)  0.16685304
 7:    [0.377079930573521, 0.581217342522697)  0.07086979
 8:    [0.581217342522697, 0.771253599758546)  0.17950572
 9:      [0.771253599758546, 1.1810650775142) -0.04951607
10:       [1.1810650775142, 2.40161776050478] -0.45781268

使最后一个边界左闭右开

DT[, Bin3 := bin_data(vals=a, bins=10, binType="quantile", boundaryType="lcro)")]  
DT[, list(mean(b)), keyby=Bin2]

 1:                                        NA  0.42510038
 2:     [-2.2146998871775, -1.05265747329296) -0.16938368
 3:   [-1.05265747329296, -0.613869230708978) -0.14103182
 4:  [-0.613869230708978, -0.375342019639661) -0.33474492
 5: [-0.375342019639661, -0.0767031289639095)  0.20827735
 6:  [-0.0767031289639095, 0.113909160788544)  0.14890251
 7:    [0.113909160788544, 0.377079930573521)  0.16685304
 8:    [0.377079930573521, 0.581217342522697)  0.07086979
 9:    [0.581217342522697, 0.771253599758546)  0.17950572
10:      [0.771253599758546, 1.1810650775142) -0.04951607
11:       [1.1810650775142, 2.40161776050478) -0.55591413

指定您自己的显式垃圾箱（注意返回空垃圾箱）

bin_data(dt=DT, binCol="a", bins=seq(-5, 5, 1), returnDT=TRUE)

          Bin         a           b
  1: [-5, -4)        NA          NA
  2: [-4, -3)        NA          NA
  3: [-3, -2) -2.214700 -0.65069635
  4: [-2, -1) -1.989352 -0.17955653
  5: [-2, -1) -1.470752 -0.03763417
 ---                               
100:   [1, 2)  1.586833 -1.20808279
101:   [2, 3)  2.401618  0.42510038
102:   [2, 3)  2.172612  0.20753834
103:   [3, 4)        NA          NA
104:   [4, 5]        NA          NA

使用可变大小的箱子

bin_data(dt=DT, binCol="a", bins=data.table(LB=c(-5, 0, 1), RB=c(0, 1, Inf)), returnDT=TRUE)

          Bin            a           b
  1:  [-5, 0) -0.626453811 -0.62036668
  2:  [-5, 0) -0.835628612 -0.91092165
  3:  [-5, 0) -0.820468384  1.76728727
  4:  [-5, 0) -0.305388387  1.68217608
  5:  [-5, 0) -0.621240581  1.43228224
 ---                               
 95: [1, Inf]  2.172611670  0.20753834
 96: [1, Inf]  1.178086997  0.21992480
 97: [1, Inf]  1.063099837  1.46458731
 98: [1, Inf]  1.207867806  0.40201178
 99: [1, Inf]  1.160402616 -0.73174817
100: [1, Inf]  1.586833455 -1.20808279
          Bin            a           b

【讨论】：

感谢@statquant 我构建它是为了解决不断出现的需求。请注意，它需要一些参数检查和错误处理，但只要您提供适当的输入，它就会起作用。
我会使用它并回来提供建议/改进，，，，通常我会添加一个 by，这样你就可以通过子集化，然后做 bining（我一直这样做，就像 2014 ,2015,...)
@ben。看起来很有趣。我去看看mltools