问题
Say I have a finite iterable X and an equivalence relation ~ on X. We can define a function my_relation(x1, x2) that returns True if x1~x2 and returns False otherwise. I want to write a function that partitions X into equivalence classes. That is, my_function(X, my_relation) should return a list of the equivalence classes of ~.
Is there a standard way to do this in python? Better yet, is there a module designed to deal with equivalence relations?
回答1:
I found this Python recipe by John Reid. It was written in Python 2 and I adapted it to Python 3 to test it. The recipe includes a test to partition the set of integers [-3,5) into equivalence classes based on the relation lambda x, y: (x - y) % 4 == 0.
It seems to do what you want. Here's the adapted version I made in case you want it in Python 3:
def equivalence_partition(iterable, relation):
"""Partitions a set of objects into equivalence classes
Args:
iterable: collection of objects to be partitioned
relation: equivalence relation. I.e. relation(o1,o2) evaluates to True
if and only if o1 and o2 are equivalent
Returns: classes, partitions
classes: A sequence of sets. Each one is an equivalence class
partitions: A dictionary mapping objects to equivalence classes
"""
classes = []
partitions = {}
for o in iterable: # for each object
# find the class it is in
found = False
for c in classes:
if relation(next(iter(c)), o): # is it equivalent to this class?
c.add(o)
partitions[o] = c
found = True
break
if not found: # it is in a new class
classes.append(set([o]))
partitions[o] = classes[-1]
return classes, partitions
def equivalence_enumeration(iterable, relation):
"""Partitions a set of objects into equivalence classes
Same as equivalence_partition() but also numbers the classes.
Args:
iterable: collection of objects to be partitioned
relation: equivalence relation. I.e. relation(o1,o2) evaluates to True
if and only if o1 and o2 are equivalent
Returns: classes, partitions, ids
classes: A sequence of sets. Each one is an equivalence class
partitions: A dictionary mapping objects to equivalence classes
ids: A dictionary mapping objects to the indices of their equivalence classes
"""
classes, partitions = equivalence_partition(iterable, relation)
ids = {}
for i, c in enumerate(classes):
for o in c:
ids[o] = i
return classes, partitions, ids
def check_equivalence_partition(classes, partitions, relation):
"""Checks that a partition is consistent under the relationship"""
for o, c in partitions.items():
for _c in classes:
assert (o in _c) ^ (not _c is c)
for c1 in classes:
for o1 in c1:
for c2 in classes:
for o2 in c2:
assert (c1 is c2) ^ (not relation(o1, o2))
def test_equivalence_partition():
relation = lambda x, y: (x - y) % 4 == 0
classes, partitions = equivalence_partition(
range(-3, 5),
relation
)
check_equivalence_partition(classes, partitions, relation)
for c in classes: print(c)
for o, c in partitions.items(): print(o, ':', c)
if __name__ == '__main__':
test_equivalence_partition()
回答2:
I'm not aware of any python library dealing with equivalence relations.
Perhaps this snippet is useful:
def rel(x1, x2):
return x1 % 5 == x2 % 5
data = range(18)
eqclasses = []
for x in data:
for eqcls in eqclasses:
if rel(x, eqcls[0]):
# x is a member of this class
eqcls.append(x)
break
else:
# x belongs in a new class
eqclasses.append([x])
eqclasses
=> [[0, 5, 10, 15], [1, 6, 11, 16], [2, 7, 12, 17], [3, 8, 13], [4, 9, 14]]
回答3:
The following function takes an iterable a, and an equivalence function equiv, and does what you ask:
def partition(a, equiv):
partitions = [] # Found partitions
for e in a: # Loop over each element
found = False # Note it is not yet part of a know partition
for p in partitions:
if equiv(e, p[0]): # Found a partition for it!
p.append(e)
found = True
break
if not found: # Make a new partition for it.
partitions.append([e])
return partitions
Example:
def equiv_(lhs, rhs):
return lhs % 3 == rhs % 3
a_ = range(10)
>>> partition(a_, equiv_)
[[0, 3, 6, 9], [1, 4, 7], [2, 5, 8]]
回答4:
Would this work ?
def equivalence_partition(iterable, relation):
classes = defaultdict(set)
for element in iterable:
for sample, known in classes.items():
if relation(sample, element):
known.add(element)
break
else:
classes[element].add(element)
return list(classes.values())
I tried it with :
relation = lambda a, b: (a - b) % 2
equivalence_partition(range(4), relation)
Which returned :
[{0, 1, 3}, {2}]
EDIT: If you want this to run as fast as possible, you may :
- wrap it in a Cython module (removing the defaultdict, there aren't much things to change)
- want to try running it with PyPy
- find a dedicated module (wasn't able to find any)
回答5:
In [1]: def my_relation(x):
...: return x % 3
...:
In [2]: from collections import defaultdict
In [3]: def partition(X, relation):
...: d = defaultdict(list)
...: for item in X:
...: d[my_relation(item)].append(item)
...: return d.values()
...:
In [4]: X = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
In [5]: partition(X, my_relation)
Out[5]: [[3, 6, 9, 12], [1, 4, 7, 10], [2, 5, 8, 11]]
For a binary function:
from collections import defaultdict
from itertools import combinations
def partition_binary(Y, relation):
d = defaultdict(list)
for (a, b) in combinations(Y):
l = d[my_relation(a, b)]
l.append(a)
l.append(b)
return d.values()
You can do something like this:
partition_binary(partition(X, rank), my_relation)
Oh, this obviously doesn't work if my_relation returns a boolean. I'd say come up with some abstract way to represent each isomorphism, although I suspect that's the goal of trying to do this in the first place.
来源：https://stackoverflow.com/questions/38924421/is-there-a-standard-way-to-partition-an-interable-into-equivalence-classes-given