R – 给定矩阵和幂,生成包含矩阵列的所有唯一组合的多个矩阵

根据我在下面链接的相关问题(参见@Aleh解决方案):我希望在给定功率的矩阵中的列之间仅计算唯一的产品.

例如,对于N = 5,M = 3,p = 2,我们得到列(1,1),(1,2),(1,3),(2,1),(2,2)的乘积,(2,3),(3,1),(3,2),(3,3).我想修改(@Aleh’s)代码只计算列(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)之间的乘积.但是我希望为每个p阶执行此操作.

有人可以帮我在R中完成这个吗?

提前谢谢了!

相关问题提问:R – Given a matrix and a power, produce multiple matrices containing all combinations of matrix columns

最佳答案
如果我理解正确,那么这就是你要找的:

# all combinations of p elements out of M with repetiton 
# c.f. http://www.mathsisfun.com/combinatorics/combinations-permutations.html
comb_rep <- function(p, M) {
  combn(M + p - 1, p) - 0:(p - 1)
}

# use cols from mat to form a new matrix
# take row products
col_prod <- function(cols, mat) {
  apply(mat[ ,cols], 1, prod)
}

N <- 5
M <- 3
p <- 3
mat <- matrix(1:(N*M),N,M)

col_comb <- lapply(2:p, comb_rep, M)
col_comb
#> [[1]]
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]    1    1    1    2    2    3
#> [2,]    1    2    3    2    3    3
#> 
#> [[2]]
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,]    1    1    1    1    1    1    2    2    2     3
#> [2,]    1    1    1    2    2    3    2    2    3     3
#> [3,]    1    2    3    2    3    3    2    3    3     3

# prepend original matrix
res_mat <- list()
res_mat[[1]] <- mat
c(res_mat, 
  lapply(col_comb, function(cols) apply(cols, 2, col_prod, mat)))
#> [[1]]
#>      [,1] [,2] [,3]
#> [1,]    1    6   11
#> [2,]    2    7   12
#> [3,]    3    8   13
#> [4,]    4    9   14
#> [5,]    5   10   15
#> 
#> [[2]]
#>      [,1] [,2] [,3] [,4] [,5] [,6]
#> [1,]    1    6   11   36   66  121
#> [2,]    4   14   24   49   84  144
#> [3,]    9   24   39   64  104  169
#> [4,]   16   36   56   81  126  196
#> [5,]   25   50   75  100  150  225
#> 
#> [[3]]
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#> [1,]    1    6   11   36   66  121  216  396  726  1331
#> [2,]    8   28   48   98  168  288  343  588 1008  1728
#> [3,]   27   72  117  192  312  507  512  832 1352  2197
#> [4,]   64  144  224  324  504  784  729 1134 1764  2744
#> [5,]  125  250  375  500  750 1125 1000 1500 2250  3375

但是,它并不是真正有效的,例如,从原始矩阵的三列而不是原始矩阵的一列和第二功率的一列计算第三功率.

编辑:评论中提到的实际尺寸测试表明,@ Moody_Mudskipper的乘法方法要快得多,而我的组合方法要快一些.因此将两者结合起来是有意义的:

# original function from @Moody_Mudskipper's answer
fun <- function(mat,p) {
  mat <- as.data.frame(mat)
  combs <- do.call(expand.grid,rep(list(seq(ncol(mat))),p)) # all combinations including permutations of same values
  combs <- combs[!apply(combs,1,is.unsorted),]              # "unique" permutations only
  rownames(combs) <- apply(combs,1,paste,collapse="-")      # Just for display of output, we keep info of combinations in rownames
  combs <- combs[order(rownames(combs)),]                   # sort to have desired column order on output
  apply(combs,1,function(x) Reduce(`*`,mat[,x]))            # multiply the relevant columns
}
combined <- function(mat, p) {
  mat <- as.data.frame(mat)
  combs <- combn(ncol(mat) + p - 1, p) - 0:(p - 1)          # all combinations with repetition
  colnames(combs) <- apply(combs, 2, paste, collapse = "-") # Just for display of output, we keep info of combinations in colnames
  apply(combs, 2, function(x) Reduce(`*`, mat[ ,x]))        # multiply the relevant columns
}
N <- 10000
M <- 25
p <- 4
mat <- matrix(runif(N*M),N,M)
microbenchmark::microbenchmark(
  fun(mat, p),
  combined(mat, p),
  times = 10
)
#> Unit: seconds
#>              expr      min       lq     mean   median       uq      max neval
#>       fun(mat, p) 3.456853 3.698680 4.067995 4.032647 4.341944 4.869527    10
#>  combined(mat, p) 2.543994 2.738313 2.870446 2.793768 3.090498 3.254232    10

注意,对于M> 1,这两个函数不会产生相同的结果.因为列排序不同,因为词汇排序是1-10 <1. 1-2在乐趣中工作.如果将组合中的相同词法排序插入,则结果将是相同的.

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