您是否知道一种更有效的方法来生成一个矩阵,该矩阵包含"权重"的所有唯一组合(让权重为w且0 <= w <= 1,并且w的值以0.1的步长分隔),使得权重总和一个,第一个是最高,最后一个是最低的重量.
这是完成工作的代码,但删除行似乎效率低下:
# generate combinations of weights such that w1 >= w2 >= w3 ...
w = seq(0, 1, 0.1) #weights 0, 0.1, ..., 0.9, 1
w = expand.grid(w, w, w, KEEP.OUT.ATTRS = FALSE) #all combinations of 3 weights
w = w[rowSums(w) == 1, ] #make sure the weights sum to one
w = w[!(w[, 1] < w[, 2] | w[, 2] < w[, 3]),] #make sure w1 >= w2 >= w3 ...
w
# Var1 Var2 Var3
# 11 1.0 0.0 0.0
# 21 0.9 0.1 0.0
# 31 0.8 0.2 0.0
# 41 0.7 0.3 0.0
# 51 0.6 0.4 0.0
# 61 0.5 0.5 0.0
# 141 0.8 0.1 0.1
# 151 0.7 0.2 0.1
# 171 0.5 0.4 0.1
# 271 0.6 0.2 0.2
# 281 0.5 0.3 0.2
# 291 0.4 0.4 0.2
# 401 0.4 0.3 0.3
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让我添加一些更一般的信息:在这个问题(上面的3个权重)中,第一个,第二个,第三个值的上限如下:
不可能base:
library(partitions)
step <- 0.1
n_weights <- 3
t(restrictedparts(n = 1/step, m = n_weights) * step)
# [1,] 1.0 0.0 0.0
# [2,] 0.9 0.1 0.0
# [3,] 0.8 0.2 0.0
# [4,] 0.7 0.3 0.0
# [5,] 0.6 0.4 0.0
# [6,] 0.5 0.5 0.0
# [7,] 0.8 0.1 0.1
# [8,] 0.7 0.2 0.1
# [9,] 0.6 0.3 0.1
# [10,] 0.5 0.4 0.1
# [11,] 0.6 0.2 0.2
# [12,] 0.5 0.3 0.2
# [13,] 0.4 0.4 0.2
# [14,] 0.4 0.3 0.3
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