假设我们有一个集合列表
set_list
,在哪里
k[i]
项目从中选择
set_list[[i]]
,那么从数学上讲,我们可以这样来解决这个问题:
-
生成每个集合的所有组合;
-
-
为每个组合创建所有排列。
函数
MixedCombnPerm
RcppAlgos
对于第1步和第3步。目前第2步没有使用最优算法。它是一种“野蛮的力量”,依赖于更快地执行法律
expand.grid
rbind
mgcv
)这可以在Rcpp中进行编码,但由于时间原因,我现在不会这么做。
library(RcppAlgos)
MixedCombnPerm <- function (set_list, k, perm = FALSE) {
###################
## mode checking ##
###################
if (!all(vapply(set_list, is.vector, TRUE)))
stop("All sets must be 'vectors'!")
if (length(unique(vapply(set_list, mode, ""))) > 1L)
stop("Please ensure that all sets have the same mode!")
################
## basic math ##
################
## size of each sets
n <- lengths(set_list, FALSE)
## input validation
if (length(n) != length(k)) stop("length of 'k' different from number of sets!")
if (any(k > n)) stop("can't choose more items than set size!")
## number of sets
n_sets <- length(n)
## total number of items
n_items <- sum(k)
## number of combinations
n_combinations_by_set <- choose(n, k)
n_combinations <- prod(n_combinations_by_set)
#################################
## step 1: combinations by set ##
#################################
## generate `n_combinations[i]` combinations on set i
combinations_by_set <- vector("list", n_sets)
for (i in seq_len(n_sets)) {
## each column of combinations_by_set[[i]] is a record
combinations_by_set[[i]] <- t.default(comboGeneral(set_list[[i]], k[i]))
}
################################
## step 2: merge combinations ##
################################
## merge combinations from all sets
## slow_expand_grid <- function (m) expand.grid(lapply(m, seq_len))
fast_expand_grid <- function (m) {
n_sets <- length(m) ## number of sets
mm <- c(1L, cumprod(m)) ## cumulative leading dimension
grid_size <- mm[n_sets + 1L] ## size of the grid
grid_ind <- vector("list", n_sets)
for (i in seq_len(n_sets)) {
## grid_ind[[i]] <- rep_len(rep(seq_len(m[i]), each = mm[i]), M)
grid_ind[[i]] <- rep_len(rep.int(seq_len(m[i]), rep.int(mm[i], m[i])), grid_size)
}
grid_ind
}
grid_ind <- fast_expand_grid(n_combinations_by_set)
## each column is a record
combinations_grid <- mapply(function (x, j) x[, j, drop = FALSE],
combinations_by_set, grid_ind,
SIMPLIFY = FALSE, USE.NAMES = FALSE)
all_combinations <- do.call("rbind", combinations_grid)
########################################################
## step 3: generate permutations for each combination ##
########################################################
if (!perm) return(all_combinations)
else {
## generate `factorial(n_items)` permutations for each combination
all_permutations <- vector("list", n_combinations)
for (i in seq_len(n_combinations)) {
all_permutations[[i]] <- permuteGeneral(all_combinations[, i], n_items)
}
return(all_permutations)
}
}
## note the "as.character(1:3)"
set_list <- list(LETTERS[1:2], letters[1:6], as.character(1:3))
k <- c(1, 3, 2)
函数返回矩阵中的组合(每列都是一条记录)if参数
perm = FALSE
请尝试以下示例:
combinations <- MixedCombnPerm(set_list, k)
permutations <- MixedCombnPerm(set_list, k, TRUE)
combinations[, 1:6]
# [,1] [,2] [,3] [,4] [,5] [,6]
#[1,] "A" "B" "A" "B" "A" "B"
#[2,] "a" "a" "a" "a" "a" "a"
#[3,] "b" "b" "b" "b" "b" "b"
#[4,] "c" "c" "d" "d" "e" "e"
#[5,] "1" "1" "1" "1" "1" "1"
#[6,] "2" "2" "2" "2" "2" "2"
permutations[[1]][1:6, ]
# [,1] [,2] [,3] [,4] [,5] [,6]
#[1,] "A" "a" "b" "c" "1" "2"
#[2,] "A" "a" "b" "c" "2" "1"
#[3,] "A" "a" "b" "1" "c" "2"
#[4,] "A" "a" "b" "1" "2" "c"
#[5,] "A" "a" "b" "2" "c" "1"
#[6,] "A" "a" "b" "2" "1" "c"
