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[英]I want to group my list of lists by date and sum the values where the dates match
[英]Sum list values where index 0 values match?
我有一個列表列表,我想將它們合並以將index[0]
值匹配的內部index[1]
值求和。 我的清單如下所示:
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,我希望新結構為:
new_lists = [
['Gifts', 8], ['Politics', 9], ['Supply', 8], ['Prints', 3], ['Accounts', 6], ['Features', 5]
]
如何在Python中實現?
您可以從collections
模塊使用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)]
或者,使用來自collections
的Counter
。 您可以通過增加每個鍵的值來手動初始化它:
from collections import Counter
c = Counter()
for i, j in lists:
c[i] += l[j]
或者,通過提供擴展內容的平面列表, Counter
然后將為您進行計數:
c = Counter(sum([[i] * j[0] for i,j in lists], []))
在這兩種情況下,都使用列表理解通過c.items()
獲取計數器的內容來創建結果列表:
r = [[i, j] for i, j in c.items()]
結果是:
print(r)
[['Supply', 8], ['Features', 5], ['Prints', 3],
['Gifts', 8], ['Accounts', 6], ['Politics', 9]]
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)])
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