Question

给定100x100矩阵，我们想要计算一个20x20矩阵，它的每个单元代表原始矩阵的5x5平方的平均值。

如何表演？（如果您对此操作有任何更好的名称，请注释重命名问题。）

Answer 1

以下是几个选项。

直接的方法是使用栅格包中的aggregate()：

m <- matrix(1:10000, ncol=100)

library(raster)
r <- raster(m)
as.matrix(aggregate(r, 5))

## aggregate() also supports non-square aggregation windows
as.matrix(aggregate(r, c(20, 50)))
#        [,1]   [,2]   [,3]   [,4]   [,5]
# [1,]  975.5 2975.5 4975.5 6975.5 8975.5
# [2,] 1025.5 3025.5 5025.5 7025.5 9025.5

对于更优雅或模糊的方法（取决于您的观点），使用几个矩阵乘法：

m <- matrix(1:10000, ncol=100)

mm <- suppressWarnings(matrix(rep(c(1, rep(0, 20)), each=5), ncol=20, nrow=100))
(t(mm) %*% m %*% mm)/25

Answer 2

让我做一个小玩具示例：

R > mat = matrix(1:36, 6,6)
R > mat
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    7   13   19   25   31
[2,]    2    8   14   20   26   32
[3,]    3    9   15   21   27   33
[4,]    4   10   16   22   28   34
[5,]    5   11   17   23   29   35
[6,]    6   12   18   24   30   36

R > A = matrix(paste(ceiling(col(mat)/2), ceiling(row(mat)/2), sep = "-"), nc = ncol(mat))
R > A
     [,1]  [,2]  [,3]  [,4]  [,5]  [,6] 
[1,] "1-1" "1-1" "2-1" "2-1" "3-1" "3-1"
[2,] "1-1" "1-1" "2-1" "2-1" "3-1" "3-1"
[3,] "1-2" "1-2" "2-2" "2-2" "3-2" "3-2"
[4,] "1-2" "1-2" "2-2" "2-2" "3-2" "3-2"
[5,] "1-3" "1-3" "2-3" "2-3" "3-3" "3-3"
[6,] "1-3" "1-3" "2-3" "2-3" "3-3" "3-3"

R > matrix(tapply(mat, A, mean), 3, 3)
     [,1] [,2] [,3]
[1,]  4.5 16.5 28.5
[2,]  6.5 18.5 30.5
[3,]  8.5 20.5 32.5

这样，对于6 * 6矩阵，计算每个2 * 2块矩阵，我们得到一个3 * 3汇总矩阵。

Reference

Answer 3

晚会，但是你走了：

# width/height of the sub-squares
side <- 5
# test matrix
test <- outer(1:100,1:100)
# make a selection matrix categorising each cell
select <- matrix(
                  rep(1:(length(test)/(side^2)),each=side),
                  nrow=nrow(test)
                )[,rep(1:(ncol(test)/side),each=side)]
# get the result
matrix(
        sapply(1:(length(test)/(side^2)),function(x) mean(test[select==x])),
        nrow=(ncol(test)/side)
      )

矩阵横向扩展

3 个答案: