R 应用-返回按行而不是按列绑定的结果

我有一个函数，它应用于长度为5的向量，返回一个4行5列的矩阵。然后我想使用apply（）对结果矩阵的每一行再次调用我的函数，并获得16（4*4）行和5列的矩阵。unfortunely apply（）将结果合并到4x20矩阵中。如何在不使用列表的情况下进行更改

matrixFromVector = function(x){
 return(rbind(x*rnorm(1,1,.01),x*rnorm(1,1,.01),x*rnorm(1,1,.1),x*rnorm(1,1,.01))) }

a = matrixFromVector(1:5)
t(a)
         [,1]     [,2]     [,3]      [,4]
[1,] 1.008391 1.005974 1.077223 0.9865611
[2,] 2.016782 2.011947 2.154445 1.9731222
[3,] 3.025173 3.017921 3.231668 2.9596833
[4,] 4.033565 4.023894 4.308890 3.9462444
[5,] 5.041956 5.029868 5.386113 4.9328055

在将我的函数应用于a的每一行之后，我希望

            [1,]        [2,]        [3,]        [4,]        [5,]
[1,]    1.0242459   2.0484917   3.0727376   4.0969835   5.1212293
[2,]    0.9999314   1.9998629   2.9997943   3.9997257   4.9996572
[3,]    1.0836573   2.1673146   3.2509719   4.3346292   5.4182865
[4,]    1.0005137   2.0010275   3.0015412   4.0020550   5.0025687
[5,]    1.0314108   2.0628216   3.0942323   4.1256431   5.1570539
[6,]    0.9995248   1.9990496   2.9985744   3.9980992   4.9976239
[7,]    1.0908017   2.1816034   3.2724051   4.3632069   5.4540086
[8,]    0.9801833   1.9603667   2.9405500   3.9207333   4.9009166
[9,]    0.9697334   1.9394669   2.9092003   3.8789338   4.8486672
[10,]   0.8484190   1.6968380   2.5452570   3.3936760   4.2420950
[11,]   0.9120351   1.8240703   2.7361054   3.6481405   4.5601756
[12,]   0.9596908   1.9193816   2.8790724   3.8387632   4.7984540
[13,]   1.0226757   2.0453515   3.0680272   4.0907030   5.1133787
[14,]   1.0069771   2.0139543   3.0209314   4.0279085   5.0348857
[15,]   1.0748773   2.1497545   3.2246318   4.2995090   5.3743863
[16,]   0.9841864   1.9683728   2.9525592   3.9367456   4.9209319

相反，我得到了

apply(a,1,matrixFromVector)
       [,1]      [,2]     [,3]      [,4]
 [1,] 1.0262524 1.0237143 1.074673 0.9885002
 [2,] 0.9990472 1.0189053 1.062644 0.9965570
 [3,] 0.9464976 0.8973152 1.138847 0.8639614
 [4,] 1.0063561 1.0080947 1.080825 1.0033793

 [5,] 2.0525048 2.0474286 2.149346 1.9770004
 [6,] 1.9980944 2.0378107 2.125288 1.9931140
 [7,] 1.8929952 1.7946303 2.277693 1.7279229
 [8,] 2.0127121 2.0161895 2.161650 2.0067587

 [9,] 3.0787573 3.0711429 3.224019 2.9655005
[10,] 2.9971416 3.0567160 3.187933 2.9896710
[11,] 2.8394929 2.6919455 3.416540 2.5918843
[12,] 3.0190682 3.0242842 3.242475 3.0101380

[13,] 4.1050097 4.0948572 4.298693 3.9540007
[14,] 3.9961888 4.0756214 4.250577 3.9862280
[15,] 3.7859905 3.5892607 4.555386 3.4558457
[16,] 4.0254242 4.0323789 4.323300 4.0135174

[17,] 5.1312621 5.1185715 5.373366 4.9425009
[18,] 4.9952359 5.0945267 5.313221 4.9827850
[19,] 4.7324881 4.4865759 5.694233 4.3198072
[20,] 5.0317803 5.0404736 5.404125 5.0168967

或

---

我们可以使用

lappy

循环行，然后执行此操作

do.call(rbind, lapply(seq_len(nrow(a)), function(i)  matrixFromVector(a[i,])))

或者我们使用

apply

将输出放入

列表中，然后执行
rbind

do.call(rbind, do.call(c, apply(a, 1, function(x) list(matrixFromVector(x)))))

我们可以使用
lappy
循环行，然后执行此操作

do.call(rbind, lapply(seq_len(nrow(a)), function(i)  matrixFromVector(a[i,])))

或者我们使用
apply
将输出放入
列表中，然后执行
rbind

do.call(rbind, do.call(c, apply(a, 1, function(x) list(matrixFromVector(x)))))

为什么不呢

apply(t(a), 1, matrixFromVector)

或

为什么不呢

apply(t(a), 1, matrixFromVector)

或