索引0值匹配的总和列表值？

Question

我有一个列表的列表，我想合并它们来求和index[0]值匹配的内部index[1]值。我的名单看起来是这样的：

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，我要像新的结构是：

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

如何在Python实现这一目标？

Answer 1

您可以使用defaultdict(int)从collections模块：

>>> 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从collections。您可以手动为每个键递增值初始化：

from collections import Counter 

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

或者，通过提供的扩展内容的简单列表，Counter会那么做的计数为您提供：

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

在这两种情况下，通过使用列表中理解抓住计数器的内容与c.items()创建结果列表：

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

结果为：

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