假设我在X
上有一个有限的可迭代~
和一个等价关系X
。我们可以定义一个函数my_relation(x1, x2)
,如果True
则返回x1~x2
,否则返回False
。我想编写一个将X
分区为等价类的函数。也就是说,my_function(X, my_relation)
应返回~
的等价类列表。
有没有一种标准的方法在python中执行此操作?更好的是,是否有一个旨在处理等价关系的模块?
答案 0 :(得分:2)
我找到了John Reid的this Python recipe。它是用Python 2编写的,我将它改编为Python 3来测试它。该配方包括一个测试,用于根据关系[-3,5)
将整数集lambda x, y: (x - y) % 4 == 0
划分为等价类。
它似乎做你想要的。这是我在Python 3中想要的改编版本:
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()
答案 1 :(得分:2)
我不知道任何处理等价关系的python库。
也许这段代码很有用:
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]]
答案 2 :(得分:1)
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]]
对于二元函数:
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()
您可以这样做:
partition_binary(partition(X, rank), my_relation)
哦,如果my_relation返回一个布尔值,这显然不起作用。我想说出一些抽象的方式来表示每个同构,尽管我怀疑这是首先尝试这样做的目标。
答案 3 :(得分:1)
以下函数采用可迭代a
和等价函数equiv
,并按您的要求执行:
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
示例:
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]]
答案 4 :(得分:1)
这会有用吗?
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())
我尝试过:
relation = lambda a, b: (a - b) % 2
equivalence_partition(range(4), relation)
返回了:
[{0, 1, 3}, {2}]
编辑:如果您想让它尽快运行,您可以: