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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