2

假设我有一个看起来像这样的列表：

a = [(1,2),(3,1),(2,1),(4,5),(9,3),(1,3)]

之后，我想要看起来像这样的东西：

b = [(1,2),(3,1),(4,5),(9,3)]

非常感谢！

4

4 回答 4

4 
b = []
seen = set()
for t in a:
    s = tuple(sorted(t))
    if s not in seen:
        seen.add(s)
        b.append(t)

或者

seen = set()
b = [t for t in a if tuple(sorted(t)) not in seen and not seen.add(tuple(sorted(t)))]
于 2013-04-12T00:50:58.620 回答
3

修改@Pavel 的解决方案，frozenset使其更高效，因为它避免了排序并且通常在原始速度上更快。这应该O(nm)与@Pavel 进行比较，后者应该是O(n * m log(m))列表n的m长度和每个元组的长度。

>>> a = [(1,2),(3,1),(2,1),(4,5),(9,3),(1,3)]
>>> b = []
>>> seen = set()
>>> for t in a:
        s = frozenset(t)
        if s not in seen:
            seen.add(s)
            b.append(t)


>>> b
[(1, 2), (3, 1), (4, 5), (9, 3)]

这是差异的证明：

from timeit import timeit


def dosorted(a):
    b = []
    seen = set()
    for t in a:
        s = tuple(sorted(t))
        if s not in seen:
            seen.add(s)
            b.append(t)
    return b

def dofrozenset(a):
    b = []
    seen = set()
    for t in a:
        s = frozenset(t)
        if s not in seen:
            seen.add(s)
            b.append(t)
    return b

import random
a = [(1,2),(3,1),(2,1),(4,5),(9,3),(1,3)]
b = [tuple(random.randrange(3) for x in range(10)) for x in range(10)]
c = [tuple(random.randrange(3) for x in range(20)) for x in range(20)]

setup = '''
from __main__ import a, b, c, dosorted, dofrozenset'''

print timeit(setup=setup, stmt='dosorted(a)')
print timeit(setup=setup, stmt='dosorted(b)')
print timeit(setup=setup, stmt='dosorted(c)')
print timeit(setup=setup, stmt='dofrozenset(a)')
print timeit(setup=setup, stmt='dofrozenset(b)')
print timeit(setup=setup, stmt='dofrozenset(c)')

9.23814695723 # dosorted(a)
26.8939069072 # dosorted(b)
86.6305864991 # dosorted(c)

5.99305211975 # dofrozenset(a)
10.708619182  # dofrozenset(b)
25.5252673175 # dofrozenset(c)

您可以添加更多调整以使这些更快，例如使用列表理解，但这很快就会变得丑陋。可以与最后一种结合使用的另一种常见技术是：

seen_add, b_append = seen.add, b.append # speeds up name lookup

然后可以从那时起直接调用这些，但请记住，过早优化是邪恶的。

于 2013-04-12T02:53:18.670 回答
1

只需把你的清单，做成一个集合，然后再把它变成一个清单

>>>a = [(1, 2), (3, 1), (2, 1), (4, 5), (9, 3), (1, 3)]
>>>sorted_tuples = [tuple(sorted(tuple_)) for tuple_ in a]
>>>list(set(sorted_tuples))
[(1, 2), (4, 5), (3, 9), (1, 3)]
于 2013-04-12T00:40:52.973 回答
1

在 Python 3.7+（即字典维护插入顺序的版本）中，您可以简单地将列表转换为由元组的 freezesets 键控的字典并获取其值：

a = [(1,2), (3,1), (2,1), (4,5), (9,3), (1,3)]
d = {}

for x in a:
    d.setdefault(frozenset(x), x)

print(list(d.values()))  # [(1, 2), (3, 1), (4, 5), (9, 3)]
于 2019-01-31T10:02:02.327 回答