我有一个应用于长度为5的向量的函数返回一个包含4行和5列的矩阵。然后我想使用apply()以便在结果矩阵的每一行上再次调用我的函数,并获得具有16(4 * 4)行和5列的矩阵。不幸的是,apply()将结果合并到4x20矩阵中。如何在不使用列表的情况下更改它?
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
将我的功能应用到我希望拥有的每一行
[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
apply(a,1,function(x) t(matrixFromVector(x)))
[,1] [,2] [,3] [,4]
[1,] 1.0242459 0.9999314 1.0836573 1.0005137
[2,] 2.0484917 1.9998629 2.1673146 2.0010275
[3,] 3.0727376 2.9997943 3.2509719 3.0015412
[4,] 4.0969835 3.9997257 4.3346292 4.0020550
[5,] 5.1212293 4.9996572 5.4182865 5.0025687
[6,] 1.0314108 0.9995248 1.0908017 0.9801833
[7,] 2.0628216 1.9990496 2.1816034 1.9603667
[8,] 3.0942323 2.9985744 3.2724051 2.9405500
[9,] 4.1256431 3.9980992 4.3632069 3.9207333
[10,] 5.1570539 4.9976239 5.4540086 4.9009166
[11,] 0.9697334 0.8484190 0.9120351 0.9596908
[12,] 1.9394669 1.6968380 1.8240703 1.9193816
[13,] 2.9092003 2.5452570 2.7361054 2.8790724
[14,] 3.8789338 3.3936760 3.6481405 3.8387632
[15,] 4.8486672 4.2420950 4.5601756 4.7984540
[16,] 1.0226757 1.0069771 1.0748773 0.9841864
[17,] 2.0453515 2.0139543 2.1497545 1.9683728
[18,] 3.0680272 3.0209314 3.2246318 2.9525592
[19,] 4.0907030 4.0279085 4.2995090 3.9367456
[20,] 5.1133787 5.0348857 5.3743863 4.9209319
答案 0 :(得分:0)
我们可以使用lapply
循环遍历行,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list
将输出放在apply
中,然后执行rbind
do.call(rbind, do.call(c, apply(a, 1, function(x) list(matrixFromVector(x)))))
答案 1 :(得分:0)
为什么不
apply(t(a), 1, matrixFromVector)
或
apply(a, 2, matrixFromVector)