What is said works, I believe...see the docstring
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; if None, only unique permutations for a given
partition will be given.
Examples
========
>>> from sympy.utilities.iterables import kbin
>>> 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,)]
>>> for p in kbin(range(3), 2, ordered=None):
... print p
...
[[0], [1, 2]]
[[1], [2, 0]]
[[2], [0, 1]]
[[0, 1], [2]]
[[1, 2], [0]]
[[2, 0], [1]]
"""
from sympy.utilities.iterables import partitions, permutations
def rotations(seq):
for i in range(len(seq)):
yield seq
seq.append(seq.pop(0))
if ordered is None:
func = rotations
else:
func = 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:
template = [len(i) for i in rv]
for pp in func(l):
rvp = []
ii = 0
for t in template:
jj = ii + t
rvp.append(pp[ii: jj])
ii = jj
yield rvp
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