r - 聚合嵌套列表对象中不同列表的相似列

Question

我有一个包含许多模拟数据迭代的列表。每次迭代都在它自己的 data.frame 中，并保存在命名列表中。例如，我需要聚合每个列表中每个 data.frame 的第一列，每个列表的第二列，第三列等等，每个列都按年龄分组。下面的代码可以重新创建我拥有的列表的结构。

x <- list(list(
          list(a = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))),
               b = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))),
               c = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))),
               d = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE)))),
               
          list(a = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))),
               b = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))),
               c = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))),
               d = data.frame(p = rnorm(100), age = sort(sample(seq(5,30),100, replace = TRUE))))
          ))

我需要的是 p 的 mean 和 sd 对所有名为 a 的列表的 分组>age，同样适用于所有名为 b 的列表，等等。我想要的结果看起来像这样，也许都是 data.table 或其他方便的东西。我只是不能很好地处理嵌套结构来获得我想要的东西。

$a
      mean_p     sd_p age
1   9.453670 2.034949   5
2  11.881241 1.995676   6
3   9.979276 2.003178   7
4  10.909008 2.104870   8
5   9.338779 1.904653   9
6  11.745993 1.909569  10
7   8.019631 2.050843  11
8   8.875167 2.053025  12
9  10.697181 1.991607  13
10 11.656100 2.005437  14
11 11.960535 2.004246  15
12 10.343899 2.085225  16
13  9.573988 1.975635  17
14  9.038953 2.112180  18
15  9.131533 2.036852  19
16 13.644504 2.160581  20
17 10.284376 1.903301  21
18  9.543758 2.134177  22
19  9.658121 2.202386  23
20 10.633312 1.842427  24
21 11.100520 2.105879  25
22 11.237161 1.871875  26
23 11.530732 1.972589  27
24  9.042670 2.187250  28
25  9.855445 1.970171  29
26 10.649243 2.064264  30

$b
      mean_p     sd_p age
1   9.705460 1.860338   5
2  10.080478 2.109235   6
3   9.712833 2.017124   7
4   9.420388 2.040863   8
5  11.775058 1.955592   9
6   8.124517 2.046651  10
7  10.557953 1.799830  11
8  10.047775 2.001543  12
9   9.229939 1.966953  13
10 11.814084 2.163710  14
11 12.102374 2.105870  15
12  9.870014 1.866519  16
13 10.696258 2.076030  17
14  9.615747 1.987050  18
15  9.781690 1.961923  19
16  9.395733 1.980549  20
17 13.307485 2.115417  21
18  9.589766 2.058452  22
19  7.942926 2.121072  23
20  9.651580 2.178241  24
21 11.736841 1.996304  25
22  8.682040 1.883955  26
23 10.041262 2.143555  27
24 10.834982 2.086041  28
25  9.046422 2.013758  29
26  9.769026 2.023566  30

$c
      mean_p     sd_p age
1  10.022880 2.148975   5
2  12.535348 1.913299   6
3   8.431201 2.252942   7
4   9.930989 1.943403   8
5   9.391383 2.004252   9
6   9.217615 1.897260  10
7  10.974630 2.174417  11
8  10.475837 1.935946  12
9   9.291287 1.917856  13
10  9.191117 1.971489  14
11  9.986106 1.940689  15
12 10.249913 1.984423  16
13 10.802905 2.122448  17
14 10.582817 1.843136  18
15  9.197653 1.864674  19
16 10.648420 2.037330  20
17 10.457500 1.885780  21
18  9.291936 2.050027  22
19 11.137871 1.744456  23
20  9.148791 1.907282  24
21 10.157003 2.183199  25
22 12.019497 1.883032  26
23 10.890207 1.922753  27
24 10.305917 2.070391  28
25  9.355486 2.022310  29
26 10.405735 1.920850  30

$d
      mean_p     sd_p age
1  10.577719 2.157974   5
2  10.557474 2.126788   6
3   9.448008 1.959201   7
4  10.160101 2.021964   8
5   9.664677 2.035892   9
6  10.974770 2.101026  10
7   8.888659 2.026531  11
8  10.185955 2.092113  12
9  10.456310 2.100847  13
10 10.259347 2.091751  14
11  9.150137 2.002525  15
12 11.042025 1.991657  16
13 10.321668 2.102700  17
14  9.537923 1.866761  18
15 10.401667 1.966281  19
16 10.380466 1.934101  20
17  9.947381 1.805547  21
18 10.458567 1.853977  22
19 11.041953 1.970225  23
20  9.826557 1.680464  24
21 10.169353 2.079167  25
22  9.352873 1.907423  26
23  9.084426 2.148295  27
24 10.083584 2.019244  28
25 10.919343 2.099395  29
26 11.621675 2.013150  30

编辑:首先感谢大家的详细反馈。我简化了数据对象的结构，并使用了 Akrun 已在别处提供的一些现有代码。我敢肯定，将来某个地方的某个人可能会发现你们写的一些有用的代码，所以我犹豫是否要结束这个问题，但如果那是我应该做的，那么我会这样做。

Answer 1

我们可以转置(来自purrr)通过循环外部列表(map)将所有'a'元素放在一起， 'b' 在一起，依此类推 ..，然后 flatten list，使用 bind_rows，执行 group_by在 'age' 上并获取 'p' 列的 mean 和 sd

library(dplyr)
library(purrr)
map(x, purrr::transpose) %>% 
   flatten %>%  
   map(~ bind_rows(.x) %>% 
           group_by(age) %>% 
            summarise(mean_p = mean(p), sd_p = sd(p)))

-输出

$a
# A tibble: 26 × 3
     age  mean_p  sd_p
   <int>   <dbl> <dbl>
 1     5  0.182  0.854
 2     6 -0.541  0.575
 3     7 -0.0815 0.962
 4     8  0.372  1.24 
 5     9  0.495  1.17 
 6    10  0.528  1.12 
 7    11 -0.0519 0.696
 8    12  0.439  0.627
 9    13  0.188  0.465
10    14  0.232  1.28 
# … with 16 more rows

$b
# A tibble: 26 × 3
     age  mean_p  sd_p
   <int>   <dbl> <dbl>
 1     5 -0.0386 0.930
 2     6  0.0312 0.961
 3     7 -0.0914 1.12 
 4     8  0.218  0.948
 5     9 -0.155  0.970
 6    10  0.669  1.31 
 7    11 -0.844  0.971
 8    12  0.424  1.21 
 9    13  0.306  1.36 
10    14 -0.0380 0.876
# … with 16 more rows

$c
# A tibble: 26 × 3
     age  mean_p  sd_p
   <int>   <dbl> <dbl>
 1     5 -0.447  1.21 
 2     6  0.0458 0.919
 3     7 -0.0733 1.08 
 4     8 -0.424  1.32 
 5     9 -0.149  0.611
 6    10 -0.0812 0.650
 7    11 -0.182  1.08 
 8    12  0.513  1.00 
 9    13  0.466  0.869
10    14  0.587  1.06 
# … with 16 more rows

$d
# A tibble: 26 × 3
     age  mean_p  sd_p
   <int>   <dbl> <dbl>
 1     5  0.749  1.17 
 2     6  0.597  0.971
 3     7  0.294  1.36 
 4     8  0.536  0.842
 5     9  0.202  1.13 
 6    10 -0.267  0.765
 7    11 -0.338  1.19 
 8    12 -0.0775 0.668
 9    13 -0.416  1.04 
10    14 -0.172  0.943
# … with 16 more rows