Question

我们说我有大量的词典（可能是10到39种词典）。我想计算所有词典中每个键的数量。即如果我有3个词典：

{1: 'url1', 3: 'url2', 7: 'url3', 5: 'url4'}
{1: 'url1', 7: 'url3'}
{5: 'url4', 10: 'url5'}

然后在结果中我应该得到{1: [2, 'url1'], 10: [1, 'url5'], 3: [1, 'url2'], 5: [2, 'url4'], 7: [2, 'url3']}。

我来到以下代码：

lists = [{1: 'url1', 3: 'url2', 7: 'url3', 5: 'url4'}, {1: 'url1', 7: 'url3'}, {5: 'url4', 10: 'url5'}]
result = {}
for l in lists:
    for i in l:
        if i in result:
            result[i][0] += 1
        else:
            result[i] = [1, l[i]]

有更好（更快）的方法吗？

Answer 1

如果您可以接受稍微不同的输出，这可能对您有用：

from collections import Counter

dicts = [
    {1: 'url1', 3: 'url2', 7: 'url3', 5: 'url4'},
    {1: 'url1', 7: 'url3'},
    {5: 'url4', 10: 'url5'},
]

result = Counter()
for d in dicts:
    result.update(d.keys())

print dict(result)

请注意，它有键和计数，但没有值。

可替换地：

from collections import Counter
from itertools import chain

dicts = [
    {1: 'url1', 3: 'url2', 7: 'url3', 5: 'url4'},
    {1: 'url1', 7: 'url3'},
    {5: 'url4', 10: 'url5'},
]

result = Counter(chain.from_iterable(dicts))

print dict(result)

最终版本：此版本完全生成您要求的输出：

from collections import Counter
from itertools import chain

dicts = [
    {1: 'url1', 3: 'url2', 7: 'url3', 5: 'url4'},
    {1: 'url1', 7: 'url3'},
    {5: 'url4', 10: 'url5'},
]

result = Counter(chain.from_iterable(d.items() for d in dicts))
result = {k:[n,v] for ((k,v),n) in result.items()}

print dict(result)

如何计算几个词典中重复键的数量？

1 个答案: