> The commutativity problem can be solved if this stackoverflow problem can be
> solved
> http://stackoverflow.com/questions/13131491/partition-n-items-into-k-bins-in-python-lazily
I think you've got all you need in the treasure trove that is
iterables. Tim Peter's routine for partitions is very fast. You just
harness that to your sequence (commutative or non-commutative) that
you want to partition, something like:
def kbin(l, k, ordered=True):
"""
Return sequence ``l`` partitioned into ``k`` bins.
If ordered is True then the order of the items in the
flattened partition will be the same as the order of the
items in ``l``; if False, all permutations of the items will
be given.
Examples
========
>>> for p in kbin(range(3), 2):
... print p
...
[[0], [1, 2]]
[[0, 1], [2]]
>>> for p in kbin(range(3), 2, ordered=False):
... print p
...
[(0,), (1, 2)]
[(0,), (2, 1)]
[(1,), (0, 2)]
[(1,), (2, 0)]
[(2,), (0, 1)]
[(2,), (1, 0)]
[(0, 1), (2,)]
[(0, 2), (1,)]
[(1, 0), (2,)]
[(1, 2), (0,)]
[(2, 0), (1,)]
[(2, 1), (0,)]
"""
from sympy.utilities.iterables import partitions
from sympy.core.compatibility import permutations
for p in partitions(len(l), k):
if sum(p.values()) != k:
continue
for pe in permutations(p.keys()):
rv = []
i = 0
for part in pe:
for do in range(p[part]):
j = i + part
rv.append(l[i: j])
i = j
if ordered:
yield rv
else:
take = [len(i) for i in rv]
for pp in permutations(l):
rvp = []
ii = 0
for t in take:
jj = ii + t
rvp.append(pp[ii: jj])
ii = jj
yield rvp
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