Chris' code now enables unification to produce all possible matches
In [1]: run ../unify/unify_sympy.py
In [2]: expr = Add(1, 2, 3, evaluate=False)
In [3]: a, b = map(Wild, 'ab')
In [4]: pattern = Add(a, b, evaluate=False)
In [5]: for x in unify(expr, pattern, {}): print x
{a_: 1, b_: 2 + 3}
{a_: 2, b_: 1 + 3}
{a_: 3, b_: 1 + 2}
{a_: 1 + 2, b_: 3}
{a_: 2 + 3, b_: 1}
{a_: 1 + 3, b_: 2}
Of course, this quickly leads to computational blowup. Fortunately you can
ask very efficiently for just the first, second, etc...
Note that, as Aaron pointed out, this doesn't handle mathematical matching.
Here is an example which matches a very particular structure.
In [8]: from sympy.abc import x
In [9]: expr = Add(1, 2*x, 3, evaluate=False)
In [10]: pattern = Add(a, 2*b, evaluate=False)
In [11]: for x in unify(expr, pattern, {}): print x
{a_: 1 + 3, b_: x}
Of course, we don't need to match exact structure. Wilds can currently
match any node.
In [6]: pattern = Add(a, b, evaluate=False)
In [7]: for x in unify(expr, pattern, {}): print x
{a_: 1, b_: 2*x + 3}
{a_: 2*x, b_: 1 + 3}
{a_: 3, b_: 2*x + 1}
{a_: 2*x + 1, b_: 3}
{a_: 2*x + 3, b_: 1}
{a_: 1 + 3, b_: 2*x}
On Tue, Oct 30, 2012 at 5:13 PM, Matthew Rocklin <[email protected]> wrote:
> Yes, that works quite well. Thanks Chris!
>
>
> On Tue, Oct 30, 2012 at 4:27 PM, Chris Smith <[email protected]> wrote:
>
>> 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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>>
>
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