我们说我有一个向量列表，如下所示：
[[1]]
[1] -0.36603596 -0.41461025 -0.68573296 -0.55516173 0.05071238 0.47723472 0.10851948
[8] 0.67005116 0.25519780 -0.79428716 0.16506077 0.81905548 0.22808934 -0.39257712
[15] 0.44778539 -0.36149934 -0.90142102 -0.99826169 0.24544167 -0.18989310 -0.67592344
[22] -0.65447808 0.26617179 -0.25020153 0.19562031 0.53520465 -0.47531100 -0.60152887
[29] 0.12012461 -0.68947499 -0.33258301 0.19914520 -0.70396942 0.21574644 -0.67197365
[36] -0.12744723 -0.07113916 0.44497439 0.07592963 -0.29082130 -0.27967624 0.28314801
[43] -0.09840383 -0.55582233 -0.29474315 -0.41717316 0.51017306 -0.31227399 0.39484400
[50] -0.88843530
[[2]]
[1] -0.14763873 -0.69009083 -0.55705599 -0.43779047 0.15626341 -0.00629513 -0.95227841
[8] 0.85645849 -0.40110676 -0.35732008 0.31375323 0.71478975 0.02262899 -0.12802829
[15] 0.58750725 -0.25629463 -0.65609956 -0.83185625 -0.35244759 -0.33287717 -0.99199682
[22] -0.45836093 -0.19431609 -0.41590652 1.06120542 0.20687783 0.13268137 -0.34219985
[29] -0.18096691 -0.24496102 -0.47769117 0.89134577 -0.56128402 0.70825268 0.10426368
[36] -0.13962506 -0.72478276 -0.40178315 0.65943132 -0.82083464 0.22569929 -1.02243310
[43] -0.70983610 -1.36733592 0.68807554 0.09156598 0.76850778 -0.64040433 0.79276407
[50] -0.40297792
[[3]]
[1] 0.34405450 -0.07928067 0.08353835 -0.37919066 -0.47233278 -0.38839824 -0.13269067
[8] 0.17348495 0.42777652 -0.19297300 -0.86438130 0.75787336 -0.34358747 0.47852682
[15] 1.29980892 -0.42527812 -0.25074922 -0.59565850 0.32800193 -0.56109570 -0.72905476
[22] -0.11498356 -0.29827083 -0.21653428 0.78533418 0.64735755 0.31889828 -0.37129803
[29] -0.51252162 0.24192268 -0.29281809 1.03299397 -0.11251429 0.13157698 -0.06404053
[36] 0.01904473 -0.13162565 0.30488937 0.31933970 0.14135025 -0.31501649 0.16738399
[43] -0.19627252 -1.29613018 -0.03572980 -0.72008672 0.13932428 -0.06117093 -0.62665670
[50] -0.12662761
[[4]]
[1] 0.183303468 0.160037845 -0.053473912 0.005199917 -0.126312554 0.116465956 -0.061730281
[8] 0.392903969 -0.008337453 -0.752631038 -0.235599857 0.999534398 0.375208363 0.201100799
[15] 0.444068886 -0.575795949 -0.873388633 -0.863612264 0.076050073 -0.188358603 -0.391865671
[22] -1.726690292 -1.206992567 -0.547175750 0.290255919 1.119834989 0.551360182 -0.510140345
[29] -0.460314706 -0.245835558 -0.315087602 0.947181076 -0.132550448 0.038419545 -0.017929636
[36] 0.041870497 -0.520961791 0.195326850 -0.117783785 -0.427426472 -0.119577158 0.702550914
[43] -0.045789957 -0.794299036 0.181420440 0.407347072 0.571894407 -0.217325835 0.280283391
[50] -0.492866084
[[5]]
[1] -0.40852268 -0.33488615 -0.30609700 -0.67467326 -0.11966383 1.01161858 -0.27108333
[8] 0.92772286 0.39047166 0.29019594 0.24404167 0.07824440 0.32786441 0.21657727
[15] 0.34362648 -0.44996166 -0.27823770 -1.24962127 -0.57241699 -0.30297804 -0.66728157
[22] 0.01783441 0.50773758 -0.31477033 -0.14581338 -0.13827194 -0.25574117 0.40049840
[29] 0.38634920 -0.29027963 -0.03381480 0.48510557 -0.61594522 1.09573928 -0.27992008
[36] -0.41523542 -0.24131548 0.43480320 0.32855110 0.48579320 0.47366867 0.62697303
[43] -0.57792202 -0.81951194 0.21583044 0.15593484 -0.10270703 -0.10206812 -0.25195873
[50] -0.89835763
我想平均相应的矢量项（例如：[[1]] [1]，[[1]] [2]，[[1]] [3]等）得到一个矢量平均值。例如，列表中每个第一个矢量项的平均值为-0.07896788。最好的方法是什么？
答案 0 :(得分：1)
假设列表名为mylist：
mydf=as.data.frame(do.call("rbind",mylist))
colMeans(mydf)
那会是理想的输出吗？