用于y标签桶中x标签球的Python迭代器 [英] Python iterator for x labeled balls in y labeled buckets

问题描述

我需要一个Python迭代器才能在Y标记的存储桶中产生X标记的球的所有组合，其中X大于或等于Y，并且所有存储桶都包含一个或多个球.

I need a Python iterator to yield all combinations of X labeled balls in Y labeled buckets, where X is greater than or equal to Y, and all buckets contain one or more balls.

对于X = 4和Y = 3的情况，如果球标记为A-B-C-D，而铲斗标记为1-2-3，则一些可能的组合为:

For a case of X = 4 and Y = 3, with balls labeled A-B-C-D, and buckets labeled 1-2-3, some of the possible combinations would be:

时段1:A，B
时段2:C
时段3:D

Bucket 1: A, B
Bucket 2: C
Bucket 3: D

时段1:A
时段2:C，B
时段3:D

Bucket 1: A
Bucket 2: C, B
Bucket 3: D

时段1:A
时段2:C
时段3:D，B

Bucket 1: A
Bucket 2: C
Bucket 3: D, B

时段1:A，C
时段2:B
时段3:D

Bucket 1: A, C
Bucket 2: B
Bucket 3: D

...

推荐答案

AR 帮助找到了答案，除了"X的所有Y分区"之外，我需要的是"X的所有Y分区的所有置换".此处的Knuth算法是"X".我需要创建该版本的Python 3实现(通过将所有 xrange 与 range 交换)，然后将该函数包装在一个应用了 itertools.permutations的新生成器中..下面的代码是结果.

A comment on the original question by AR has helped to find an answer, except instead of "all Y-partitions of X", it is "all permutations of all Y-partitions of X" that I needed. The Knuth algorithm here is a Python 2 implementation of "all Y-partitions of X". I needed to create the Python 3 implementation of that (by swapping all xrange with range), and then wrap the function in a new generator that applies itertools.permutations at each iteration. The code below is the result.

def algorithm_u(ns, m):
    """
    - Python 3 implementation of
        Knuth in the Art of Computer Programming, Volume 4, Fascicle 3B, Algorithm U
    - copied from Python 2 implementation of the same at
        https://codereview.stackexchange.com/questions/1526/finding-all-k-subset-partitions
    - the algorithm returns
        all set partitions with a given number of blocks, as a python generator object

    e.g.
        In [1]: gen = algorithm_u(['A', 'B', 'C'], 2)

        In [2]: list(gen)
        Out[2]: [[['A', 'B'], ['C']],
                  [['A'], ['B', 'C']],
                  [['A', 'C'], ['B']]]
    """

    def visit(n, a):
        ps = [[] for i in range(m)]
        for j in range(n):
            ps[a[j + 1]].append(ns[j])
        return ps

    def f(mu, nu, sigma, n, a):
        if mu == 2:
            yield visit(n, a)
        else:
            for v in f(mu - 1, nu - 1, (mu + sigma) % 2, n, a):
                yield v
        if nu == mu + 1:
            a[mu] = mu - 1
            yield visit(n, a)
            while a[nu] > 0:
                a[nu] = a[nu] - 1
                yield visit(n, a)
        elif nu > mu + 1:
            if (mu + sigma) % 2 == 1:
                a[nu - 1] = mu - 1
            else:
                a[mu] = mu - 1
            if (a[nu] + sigma) % 2 == 1:
                for v in b(mu, nu - 1, 0, n, a):
                    yield v
            else:
                for v in f(mu, nu - 1, 0, n, a):
                    yield v
            while a[nu] > 0:
                a[nu] = a[nu] - 1
                if (a[nu] + sigma) % 2 == 1:
                    for v in b(mu, nu - 1, 0, n, a):
                        yield v
                else:
                    for v in f(mu, nu - 1, 0, n, a):
                        yield v

    def b(mu, nu, sigma, n, a):
        if nu == mu + 1:
            while a[nu] < mu - 1:
                yield visit(n, a)
                a[nu] = a[nu] + 1
            yield visit(n, a)
            a[mu] = 0
        elif nu > mu + 1:
            if (a[nu] + sigma) % 2 == 1:
                for v in f(mu, nu - 1, 0, n, a):
                    yield v
            else:
                for v in b(mu, nu - 1, 0, n, a):
                    yield v
            while a[nu] < mu - 1:
                a[nu] = a[nu] + 1
                if (a[nu] + sigma) % 2 == 1:
                    for v in f(mu, nu - 1, 0, n, a):
                        yield v
                else:
                    for v in b(mu, nu - 1, 0, n, a):
                        yield v
            if (mu + sigma) % 2 == 1:
                a[nu - 1] = 0
            else:
                a[mu] = 0
        if mu == 2:
            yield visit(n, a)
        else:
            for v in b(mu - 1, nu - 1, (mu + sigma) % 2, n, a):
                yield v

    n = len(ns)
    a = [0] * (n + 1)
    for j in range(1, m + 1):
        a[n - m + j] = j - 1
    return f(m, n, 0, n, a)

class algorithm_u_permutations:
    """
    generator for all permutations of all set partitions with a given number of blocks
    e.g.
        In [4]: gen = algorithm_u_permutations(['A', 'B', 'C'], 2)

        In [5]: list(gen)
        Out[5]:
        [(['A', 'B'], ['C']),
         (['C'], ['A', 'B']),
         (['A'], ['B', 'C']),
         (['B', 'C'], ['A']),
         (['A', 'C'], ['B']),
         (['B'], ['A', 'C'])]
    """

    from itertools import permutations

    def __init__(self, ns, m):

        self.au = algorithm_u(ns, m)
        self.perms = self.permutations(next(self.au))

    def __next__(self):

        try:
            return next(self.perms)

        except StopIteration:
            self.perms = self.permutations(next(self.au))
            return next(self.perms)

    def __iter__(self):
        return self

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