相关的this问题，我想知道的算法（和实际code在Java / C / C ++ / Python的/等，如果你有！）生成的所有组合研究与 M 总元素的列表元素。有些 M 元素可以重复。

Related to this question, I am wondering the algorithms (and actual code in java/c/c++/python/etc., if you have!) to generate all combinations of r elements for a list with m elements in total. Some of these m elements may be repeated.

谢谢！

推荐答案

下面是一个递归，我相信有密切的关系让 - 伯纳德·Pellerin的算法，在数学的。

Here is a recursion that I believe is closely related to Jean-Bernard Pellerin's algorithm, in Mathematica.

此获得输入每个类型的元件的数目。输出采用类似的形式。例如：

This takes input as the number of each type of element. The output is in similar form. For example:

{a,a,b,b,c,d,d,d,d} -> {2,2,1,4}

功能：

f[k_, {}, c__] := If[+c == k, {{c}}, {}]

f[k_, {x_, r___}, c___] := Join @@ (f[k, {r}, c, #] & /@ 0~Range~Min[x, k - +c])

使用：

f[4, {2, 2, 1, 4}]

{{0, 0, 0, 4}, {0, 0, 1, 3}, {0, 1, 0, 3}, {0, 1, 1, 2}, {0, 2, 0, 2},
 {0, 2, 1, 1}, {1, 0, 0, 3}, {1, 0, 1, 2}, {1, 1, 0, 2}, {1, 1, 1, 1},
 {1, 2, 0, 1}, {1, 2, 1, 0}, {2, 0, 0, 2}, {2, 0, 1, 1}, {2, 1, 0, 1},
 {2, 1, 1, 0}, {2, 2, 0, 0}}

的

请求该code的说明。这是一个递归函数，它采用可变数量的参数。第一个参数是 K ，子集的长度。第二个是每种类型以供选择的计数的列表。第三个参数及以后在内部使用功能保持子集（组合），因为它是构建。

An explanation of this code was requested. It is a recursive function that takes a variable number of arguments. The first argument is k, length of subset. The second is a list of counts of each type to select from. The third argument and beyond is used internally by the function to hold the subset (combination) as it is constructed.

此定义，因此使用的时候有选择集中没有更多的项目：

This definition therefore is used when there are no more items in the selection set:

f[k_, {}, c__] := If[+c == k, {{c}}, {}]

如果组合（长度）的值的总和等于 K ，然后返回该组合，否则返回一个空集。 （ + C 的缩写加[C] ）

If the total of the values of the combination (its length) is equal to k, then return that combination, otherwise return an empty set. (+c is shorthand for Plus[c])

否则：

f[k_, {x_, r___}, c___] := Join @@ (f[k, {r}, c, #] & /@ 0~Range~Min[x, k - +c])

阅读左至右：

加入用于扁平化嵌套列表的水平，这样的结果是不是越来越深的张量。

Join is used to flatten out a level of nested lists, so that the result is not an increasingly deep tensor.

F [K，{R}，C＃]安培; 调用函数，下探选择集的第一个位置（ X ），并加入了新的元素组合（＃）。

f[k, {r}, c, #] & calls the function, dropping the first position of the selection set (x), and adding a new element to the combination (#).

/ @ 0〜范围〜敏[X，K - + C]）零之间以及每个整数选择的第一个元素的小集，而 K 总组合少，这是可以在不超过组合大小 K 可选择的最大。

/@ 0 ~Range~ Min[x, k - +c]) for each integer between zero and the lesser of the first element of the selection set, and k less total of combination, which is the maximum that can be selected without exceeding combination size k.