索引0值匹配的總和列表值？

Question

我有一個列表的列表，我想合併它們來求和index[0]值匹配的內部index[1]值。我的名單看起來是這樣的：

lists = [ 
['Gifts', [4]], 
['Gifts', [4]], 
['Politics', [3]], 
['Supply', [4]], 
['Supply', [4]], 
['Prints', [1]], 
['Prints', [1]], 
['Prints', [1]], 
['Politics', [3]], 
['Politics', [3]], 
['Accounts', [2]], 
['Accounts', [2]], 
['Accounts', [2]], 
['Features', [3]], 
['Features', [2]] 
] 

因此，我要像新的結構是：

new_lists = [ 
['Gifts', 8], ['Politics', 9], ['Supply', 8], ['Prints', 3], ['Accounts', 6], ['Features', 5] 
] 

如何在Python實現這一目標？

Answer 1

您可以使用defaultdict(int)從collections模塊：

>>> from collections import defaultdict 
>>> d = defaultdict(int) 
>>> for key, value in lists: 
...  d[key] += value[0] 
... 
>>> dict(d) 
{'Gifts': 8, 'Prints': 3, 'Accounts': 6, 'Features': 5, 'Supply': 8, 'Politics': 9} 
>>> list(d.items()) 
[('Prints', 3), ('Features', 5), ('Supply', 8), ('Gifts', 8), ('Accounts', 6), ('Politics', 9)] 

Answer 2

使用defaultdict：

In [52]: from collections import defaultdict 

In [53]: d = defaultdict(int) 

In [54]: for name, (val,) in lists: 
    ....:  d[name] += val 
    ....:  

In [55]: d.items() 
Out[55]: dict_items([('Politics', 9), ('Supply', 8), ('Prints', 3), ('Features', 5), ('Gifts', 8), ('Accounts', 6)]) 

Answer 3

或者，使用Counter從collections。您可以手動爲每個鍵遞增值初始化：

from collections import Counter 

c = Counter() 
for i, j in lists: 
    c[i] += l[j] 

或者，通過提供的擴展內容的簡單列表，Counter會那麼做的計數爲您提供：

c = Counter(sum([[i] * j[0] for i,j in lists], [])) 

在這兩種情況下，通過使用列表中理解抓住計數器的內容與c.items()創建結果列表：

r = [[i, j] for i, j in c.items()] 

結果爲：

print(r) 
[['Supply', 8], ['Features', 5], ['Prints', 3], 
['Gifts', 8], ['Accounts', 6], ['Politics', 9]]