在Python中，是否有一种方式可以将集合或无序元组作为字典(或等效的数据 struct )的键提供

您可以使用帮助器函数以一种更便宜的方式创建排序的元组:

def key(s, t):
    return (s, t) if s < t else (t, s)

d = {key('w1', 'w2'): 4}

print(d[key('w1','w2')])
print(d[key('w2','w1')])

函数调用开销使其比内联表达式慢，但仍快于到目前为止提到的替代方法.

下面是计算它们并使用它们来查找dict值的各种表达式及其时间:

 132.7 ± 1.8 ns  (s, t) if s < t else (t, s)
 171.7 ± 1.5 ns  key(s, t)
 236.5 ± 0.9 ns  frozenset({s, t})
 254.9 ± 2.5 ns  frozenset((s, t))
 273.3 ± 2.0 ns  frozenset([s, t])
 343.2 ± 3.6 ns  tuple(sorted([s, t]))
 767.0 ± 9.7 ns  T(s, t)

基准代码(Attempt This Online!):

from timeit import timeit
from statistics import mean, stdev

setup = '''
s, t = 'w1', 'w2'
sts = [(s, t), (t, s)] * 5000

def key(s, t):
    return (s, t) if s < t else (t, s)

# from Anentropic's answer
class T(tuple):
    def __new__(cls, *args):
        return super().__new__(cls, tuple(sorted(args)))

d = {KEY: 4}
'''

funcs = [
    "tuple(sorted([s, t]))",
    "frozenset([s, t])",
    "frozenset((s, t))",
    "frozenset({s, t})",
    "key(s, t)",
    "(s, t) if s < t else (t, s)",
    "T(s, t)",
]

times = {f: [] for f in funcs}
def stats(f):
    ts = [t * 1e9 for t in sorted(times[f])[:10]]
    return f'{mean(ts):6.1f} ± {stdev(ts):3.1f} ns '

for _ in range(100):
    for f in funcs:
        code = f'for s, t in sts: d[{f}]'
        setup_ = setup.replace('KEY', f)
        t = timeit(code, setup_, number=10**0) / 1e4
        times[f].append(t)

for f in sorted(funcs, key=stats):
    print(stats(f), f)