比较两个词典列表

(这个问题有点模棱两可。据我所知, a 和 b 这两个列表都包含字典,您要检查这两个列表是否以任何顺序包含相同的字典。)

all 内置的,虽然对于长列表来说这会比较慢,但是有O(n):

>>> lst1 = [{1:2, 3:4}, {5:6, 7:8}]
>>> lst2 = [{7:8, 5:6}, {3:4, 1:2}]
>>> len(lst1) == len(lst2) and all(x in lst2 for x in lst1)
True

更好地转换 list 属于 dict 到 set 属于 frozenset 握着听写项目,有O(n):

>>> set_of_tuples = lambda l: set(frozenset(d.items()) for d in l)
>>> set_of_tuples(lst1)
set([frozenset([(1, 2), (3, 4)]), frozenset([(5, 6), (7, 8)])])
>>> set_of_tuples(lst1) == set_of_tuples(lst2)
True

Counter 如所示 another answer

这是为了更好地比较成员:

a.keys() == b.keys()

keys() 返回字典成员的视图。

要使用两个听写列表,只需将其包装在理解中:

[a.keys() == b.keys() for a, b in zip(list1, list2)]

如果你想知道他们都是平等的,那就用 all()

all([a.keys() == b.keys() for a, b in zip(list1, list2)])

基准

a = {}
b = {}
for i in range(100_0000):
    a[i] = i
    b[i] = i

In [10]: %timeit a==b # wrong, but just for comparison
11.6 ms ± 265 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [11]: %timeit a.keys()==b.keys()
16.9 ms ± 209 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [17]: %timeit _compare(a, b)
30.6 ms ± 633 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [18]: %timeit set(a.keys()) == set(b.keys())
71 ms ± 1.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

一种方法是比较一组值:

da = { 1:'a', 2:'b', 3:'c'}
db = { 3:'a', 1:'b', 2:'c'}

set(db.values()) == set(da.values())

collections.Counter 不按顺序比较两个列表:

from collections import Counter
def compare(a, b):
    return Counter(tuple(d.items()) for d in a) == Counter(tuple(d.items()) for d in b)

因此,以下两个成员级dict列表的比较相等:

compare(
    [{1: 2, 2: 3}, {3: 4, 5: 6}, {3: 4, 5: 6}],
    [{3: 4, 5: 6}, {1: 2, 2: 3}, {3: 4, 5: 6}]
)

回报 True