列表中具有相同索引的 `rbind` 元素

Question

我有一个 list，它有三个 list，每个有 2 个 data.frames，如下所示：-

d <- list(list(data.frame(height = rnorm(10), weight = runif(10)), data.frame(nr = rnorm(10), qr = rchisq(10, 10))),
          list(data.frame(height = rnorm(6), weight = runif(6)), data.frame(nr = rnorm(6), qr = rchisq(6, 10))),
          list(data.frame(height = rnorm(8), weight = runif(8)), data.frame(nr = rnorm(8), qr = rchisq(8, 10))))


[[1]]
[[1]][[1]]
        height      weight
1  -0.49424331 0.023996582
2  -0.80320654 0.029460558
3  -0.89797434 0.932508002
4  -0.25267069 0.790625104
5   0.27474082 0.859495769
6   0.14285128 0.009731295
7  -0.86224008 0.343969165
8   0.07358127 0.106006154
9  -1.61474408 0.302890840
10  2.23920173 0.133115944

[[1]][[2]]
            nr        qr
1  -1.24342871 10.278033
2   1.37520549 13.246929
3  -0.06046197  6.267480
4   0.73643661 14.084240
5  -0.01897590  1.323470
6   1.10877385 11.739945
7  -1.09511298 10.616714
8  -1.03525533  9.992008
9  -0.04301281 12.943073
10 -0.79446848  8.670066


[[2]]
[[2]][[1]]
      height    weight
1 -2.7323741 0.8825884
2  0.4745896 0.1813869
3  0.9158570 0.9660507
4  0.8927806 0.1156805
5 -0.8443665 0.3079322
6 -0.4703602 0.2345349

[[2]][[2]]
          nr        qr
1 -1.6915651  9.532319
2 -1.9810859 15.145930
3 -0.4890531 10.013549
4  0.2163449 13.407265
5  1.0770555  6.676846
6 -0.5431102  7.688177


[[3]]
[[3]][[1]]
      height     weight
1 -1.2671410 0.48468152
2 -0.7792946 0.04499799
3 -0.6976782 0.10917336
4  0.8274744 0.69698260
5 -0.9456592 0.64183451
6 -1.2882436 0.29868696
7  0.6424889 0.86165232
8 -0.8255187 0.16430852

[[3]][[2]]
          nr        qr
1 -0.4160331  5.341376
2 -1.0321303  8.947948
3 -0.3380597  7.937599
4 -2.1520878 11.740298
5 -0.8979710  2.393419
6 -1.1172138 11.780884
7 -1.0309391  2.673642
8  0.8822399 12.351724

我想对其进行转换，以便所有具有 (height, weight) 列的 data.frame 都 rbinded 在一起，并且所有 (nr, qr) data.frames 都是rbind在一起。所以基本上列表中每个列表的第一个元素应该绑定在一起，列表中每个列表的第二个元素应该绑定在一起。

预期输出将是另一个列表，其中包含两个 data.frame，如下所示：-

[[1]]
        height      weight
1  -0.49424331 0.023996582
2  -0.80320654 0.029460558
3  -0.89797434 0.932508002
4  -0.25267069 0.790625104
5   0.27474082 0.859495769
6   0.14285128 0.009731295
7  -0.86224008 0.343969165
8   0.07358127 0.106006154
9  -1.61474408 0.302890840
10  2.23920173 0.133115944
11 -2.7323741 0.8825884
12  0.4745896 0.1813869
13  0.9158570 0.9660507
14  0.8927806 0.1156805
15 -0.8443665 0.3079322
16 -0.4703602 0.2345349
17 -1.2671410 0.48468152
18 -0.7792946 0.04499799
19 -0.6976782 0.10917336
20  0.8274744 0.69698260
21 -0.9456592 0.64183451
22 -1.2882436 0.29868696
23  0.6424889 0.86165232
24 -0.8255187 0.16430852

[[2]]
            nr        qr
1  -1.24342871 10.278033
2   1.37520549 13.246929
3  -0.06046197  6.267480
4   0.73643661 14.084240
5  -0.01897590  1.323470
6   1.10877385 11.739945
7  -1.09511298 10.616714
8  -1.03525533  9.992008
9  -0.04301281 12.943073
10 -0.79446848  8.670066
11 -1.6915651  9.532319
12 -1.9810859 15.145930
13 -0.4890531 10.013549
14  0.2163449 13.407265
15  1.0770555  6.676846
16 -0.5431102  7.688177
17 -0.4160331  5.341376
18 -1.0321303  8.947948
19 -0.3380597  7.937599
20 -2.1520878 11.740298
21 -0.8979710  2.393419
22 -1.1172138 11.780884
23 -1.0309391  2.673642
24  0.8822399 12.351724

Answer 1

这应该可以做到。

dd <- list(do.call(rbind, lapply(d, "[[", 1)), do.call(rbind, lapply(d, "[[", 2)))

列表中具有相同索引的 `rbind` 元素

`rbind` elements with same index in a list

r

rbind