如何統計多個字典中重複鍵的數量？

Question

假設我有大量的字典（可能是10'000字典）。我想統計所有字典中每個鍵的數量。即如果我有3點字典：

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

然後在結果我應該得到{1: [2, 'url1'], 10: [1, 'url5'], 3: [1, 'url2'], 5: [2, 'url4'], 7: [2, 'url3']}。

我來到了下面的代碼：

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]] 

是在沒有更好的（快）的方式來做到這一點？

Answer 1

如果你能接受輸出略有不同，這可能會爲你工作：

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) 

注已經鍵和數量，但沒有價值。

或者：

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) 

最後版本：這一個正好產生您所要求的輸出：

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)