python - 从 python 中的字典获取键列表/键集的最有效方法是什么？

Question

为了快速比较 2 个字典的键，我使用这种方法创建键集:

dict_1 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_2 = {"file_1":10, "file_2":20, "file_3":30}
set_1 = {file for file in dict_1}
set_2 = {file for file in dict_2}

然后我使用 diff_set = set_1 - set_2 来查看 set_2 中缺少哪些键。

有没有更快的方法？我发现使用 set(dict.keys()) 不太有效，所以我会改用它 - 但它更有效吗？

Answer 1

让我们更正确地衡量(不仅仅是衡量单个执行，也不包括设置)并包括更快的解决方案:

300 ns  300 ns  300 ns  {*dict_1} - {*dict_2}
388 ns  389 ns  389 ns  {file for file in dict_1 if file not in dict_2}
389 ns  390 ns  390 ns  dict_1.keys() - dict_2
458 ns  458 ns  458 ns  set(dict_1) - set(dict_2)
472 ns  472 ns  472 ns  dict_1.keys() - dict_2.keys()
665 ns  665 ns  668 ns  set(dict_1.keys()) - set(dict_2.keys())
716 ns  716 ns  716 ns  {file for file in dict_1} - {file for file in dict_2}

基准代码(Try it online!):

import timeit

setup = '''
dict_1 = {"file_1":10, "file_2":20, "file_3":30, "file_4":40}
dict_2 = {"file_1":10, "file_2":20, "file_3":30}
'''

codes = [
    '{file for file in dict_1} - {file for file in dict_2}',
    'set(dict_1) - set(dict_2)',
    'set(dict_1.keys()) - set(dict_2.keys())',
    'dict_1.keys() - dict_2',
    'dict_1.keys() - dict_2.keys()',
    '{*dict_1} - {*dict_2}',
    '{file for file in dict_1 if file not in dict_2}',
]

exec(setup)
for code in codes:
    print(eval(code))

tss = [[] for _ in codes]
for _ in range(20):
    print()
    for code, ts in zip(codes, tss):
        number = 10000
        t = min(timeit.repeat(code, setup, number=number)) / number
        ts.append(t)
    for code, ts in sorted(zip(codes, tss), key=lambda cs: sorted(cs[1])):
        print(*('%3d ns ' % (t * 1e9) for t in sorted(ts)[:3]), code)