本文整理汇总了Python中sympy.combinatorics.permutations.Permutation.rmul_with_af方法的典型用法代码示例。如果您正苦于以下问题：Python Permutation.rmul_with_af方法的具体用法？Python Permutation.rmul_with_af怎么用？Python Permutation.rmul_with_af使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sympy.combinatorics.permutations.Permutation的用法示例。
在下文中一共展示了Permutation.rmul_with_af方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: test_ranking
# 需要导入模块: from sympy.combinatorics.permutations import Permutation [as 别名]
# 或者: from sympy.combinatorics.permutations.Permutation import rmul_with_af [as 别名]
def test_ranking():
assert Permutation.unrank_lex(5, 10).rank() == 10
p = Permutation.unrank_lex(15, 225)
assert p.rank() == 225
p1 = p.next_lex()
assert p1.rank() == 226
assert Permutation.unrank_lex(15, 225).rank() == 225
assert Permutation.unrank_lex(10, 0).is_Identity
p = Permutation.unrank_lex(4, 23)
assert p.rank() == 23
assert p.array_form == [3, 2, 1, 0]
assert p.next_lex() is None
p = Permutation([1, 5, 2, 0, 3, 6, 4])
q = Permutation([[1, 2, 3, 5, 6], [0, 4]])
a = [Permutation.unrank_trotterjohnson(4, i).array_form for i in range(5)]
assert a == [[0, 1, 2, 3], [0, 1, 3, 2], [0, 3, 1, 2], [3, 0, 1,
2], [3, 0, 2, 1] ]
assert [Permutation(pa).rank_trotterjohnson() for pa in a] == list(range(5))
assert Permutation([0, 1, 2, 3]).next_trotterjohnson() == \
Permutation([0, 1, 3, 2])
assert q.rank_trotterjohnson() == 2283
assert p.rank_trotterjohnson() == 3389
assert Permutation([1, 0]).rank_trotterjohnson() == 1
a = Permutation(list(range(3)))
b = a
l = []
tj = []
for i in range(6):
l.append(a)
tj.append(b)
a = a.next_lex()
b = b.next_trotterjohnson()
assert a == b is None
assert {tuple(a) for a in l} == {tuple(a) for a in tj}
p = Permutation([2, 5, 1, 6, 3, 0, 4])
q = Permutation([[6], [5], [0, 1, 2, 3, 4]])
assert p.rank() == 1964
assert q.rank() == 870
assert Permutation([]).rank_nonlex() == 0
prank = p.rank_nonlex()
assert prank == 1600
assert Permutation.unrank_nonlex(7, 1600) == p
qrank = q.rank_nonlex()
assert qrank == 41
assert Permutation.unrank_nonlex(7, 41) == Permutation(q.array_form)
a = [Permutation.unrank_nonlex(4, i).array_form for i in range(24)]
assert a == [
[1, 2, 3, 0], [3, 2, 0, 1], [1, 3, 0, 2], [1, 2, 0, 3], [2, 3, 1, 0],
[2, 0, 3, 1], [3, 0, 1, 2], [2, 0, 1, 3], [1, 3, 2, 0], [3, 0, 2, 1],
[1, 0, 3, 2], [1, 0, 2, 3], [2, 1, 3, 0], [2, 3, 0, 1], [3, 1, 0, 2],
[2, 1, 0, 3], [3, 2, 1, 0], [0, 2, 3, 1], [0, 3, 1, 2], [0, 2, 1, 3],
[3, 1, 2, 0], [0, 3, 2, 1], [0, 1, 3, 2], [0, 1, 2, 3]]
N = 10
p1 = Permutation(a[0])
for i in range(1, N+1):
p1 = p1*Permutation(a[i])
p2 = Permutation.rmul_with_af(*[Permutation(h) for h in a[N::-1]])
assert p1 == p2
ok = []
p = Permutation([1, 0])
for i in range(3):
ok.append(p.array_form)
p = p.next_nonlex()
if p is None:
ok.append(None)
break
assert ok == [[1, 0], [0, 1], None]
assert Permutation([3, 2, 0, 1]).next_nonlex() == Permutation([1, 3, 0, 2])
assert [Permutation(pa).rank_nonlex() for pa in a] == list(range(24))