Question

我有一个列表列表，我希望将它们合并以对index[1]值匹配的内部index[0]值求和。我的列表如下所示：

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，我希望新结构为：

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

我如何在Python中实现这一目标？

Answer 1

您可以使用collections模块中的defaultdict(int)：

>>> 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

或者，使用Counter中的collections。您可以通过递增每个键的值来手动初始化它：

from collections import Counter

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

或者，通过提供展开内容的平面列表，Counter将为您进行计数：

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

在这两种情况下，使用列表理解来创建结果列表，使用c.items()抓取计数器的内容：

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

结果是：

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

Answer 3

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)])

索引0值匹配的总和列表值？

3 个答案: