如何计算R中矩阵列表的平均值 [英] how to calculate the mean of list of list of matrices in r
问题描述
我有一个类似于以下的列表，它是包含矩阵的列表的列表(因此"ftable"是十个列表的列表，每个内部列表都包含七个矩阵).我需要计算可能也具有NA值的关联矩阵的均值.我尝试了几种方法，但都出错了.
I have a list like below, which is a list of lists containing matrices(so "ftable" is a list of ten lists and each of the internal lists contains seven matrices). I need to calculate the mean of associated matrices which may also have NA values as well.I have tried several ways but I got errors.
for(i in 1:10){
for(j in 1:7){
ftable[[i]][[j]] <- matrix (x,nrow=8,ncol=8, byrow=TRUE)
}
}
> str(ftable)
List of 10
$ :List of 7
.......
.......
作为结果，我需要有一个包含七个矩阵的列表，这些矩阵中的每一个都是对ftable [[1]] [[i]]，ftable [[2]] [[i]]应用均值的结果，...，ftable [[10]] [[i]]和i在1:7中.
as the result I need to have a list containing seven matrices that each of these matrices are the result of applying mean to ftable[[1]][[i]], ftable[[2]][[i]], ... , ftable[[10]][[i]] and i in 1:7.
我已经尝试过，但是出现错误:
I have tried this but I got error:
meanTable <- list()
for (i in 1:7)
meanTable[[i]] <- matrix (0, nrow=8,ncol=8)
> meanTable <- lapply(1:7, function(i) Reduce(mean, list(ftable[[1]][i],ftable[[2]][i],ftable[[3]][i],ftable[[4]][i],ftable[[5]][i],ftable[[6]][i],ftable[[7]][i],ftable[[8]][i],ftable[[9]][i],ftable[[10]][i])))
Error in mean.default(init, x[[i]]) :
'trim' must be numeric of length one
In addition: Warning message:
In mean.default(init, x[[i]]) :
argument is not numeric or logical: returning NA
矩阵的一个示例:
> ftable[[1]][[1]]
1 2 3 4 5 6 7 8
1 NA 0.924 0.835 -0.336 0.335 -0.948 0.285 0.749
2 NA NA 0.772 -0.333 0.333 -0.892 0.127 0.715
3 NA NA NA -0.476 0.475 -0.756 0.258 0.749
4 NA NA NA NA -0.999 0.399 -0.150 -0.399
5 NA NA NA NA NA -0.399 0.151 0.399
6 NA NA NA NA NA NA -0.134 -0.715
7 NA NA NA NA NA NA NA 0.144
8 NA NA NA NA NA NA NA NA
推荐答案
我认为这是您所需要的.最简单的方法是取消列出外部列表，然后按如下所示应用Reduce:
I think this is what you require. The easiest way would be to unlist your outer list and then apply Reduce as follows:
我将创建来自user1317221_G
set.seed(45)
mat1 <- matrix(c(sample(10),NA,NA),nrow=2)
matlist1 <- list(mat1,mat1,mat1)
mat2 <- matrix(c(sample(11:20),NA,NA),nrow=2)
matlist2 <- list(mat2,mat2,mat2)
bigmatlist <- list(matlist1,matlist2)
> bigmatlist
# [[1]]
# [[1]][[1]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 7 2 10 1 4 NA
# [2,] 3 9 8 5 6 NA
#
# [[1]][[2]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 7 2 10 1 4 NA
# [2,] 3 9 8 5 6 NA
#
# [[1]][[3]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 7 2 10 1 4 NA
# [2,] 3 9 8 5 6 NA
#
#
# [[2]]
# [[2]][[1]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 14 13 19 17 15 NA
# [2,] 18 20 16 11 12 NA
#
# [[2]][[2]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 14 13 19 17 15 NA
# [2,] 18 20 16 11 12 NA
#
# [[2]][[3]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 14 13 19 17 15 NA
# [2,] 18 20 16 11 12 NA
现在解决.
# in your case, outer.len = 10 and inner.len = 7
outer.len <- 2
inner.len <- 3
prod.len <- outer.len * inner.len
list.un <- unlist(bigmatlist, recursive = FALSE)
o <- lapply(1:inner.len, function(idx) {
Reduce('+', list.un[seq(idx, prod.len, by = inner.len)])/outer.len
})
> o
# [[1]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 10.5 7.5 14.5 9 9.5 NA
# [2,] 10.5 14.5 12.0 8 9.0 NA
#
# [[2]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 10.5 7.5 14.5 9 9.5 NA
# [2,] 10.5 14.5 12.0 8 9.0 NA
#
# [[3]]
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 10.5 7.5 14.5 9 9.5 NA
# [2,] 10.5 14.5 12.0 8 9.0 NA
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