Question

如何从下面的dict列表中删除重复项，同时总结相应的值谢谢

{'VIC': [['City West ', 10.0], ['City West ', 30.0], ['Thornbury ',
         0.0]], 'NSW': [['Queanbeyan ', 5.0],['Queanbeyan ', 20.0]], 
         'ACT': [['Civic ', 8.0], ['Civic ', 60.0], ['Civic ', 1.0]]}

期望的输出

{'VIC': [['City West ', 40.0], ['Thornbury ', 0.0]], 'NSW':
        [['Queanbeyan ', 25.0]], 'ACT': [['Civic ', 69.0]]}

Answer 1

您可以使用dict-comprehension重建dict。在重建期间，您可以使用itertools.groupby将嵌套列表分组到每个子列表的第一个值上，并使用列表推导对第二个值求和：

>>> from itertools import groupby

>>> d = {'VIC': [['City West ', 10.0], ['City West ', 30.0], ['Thornbury ', 0.0]], 'NSW': [['Queanbeyan ', 5.0],['Queanbeyan ', 20.0]], 'ACT': [['Civic ', 8.0], ['Civic ', 60.0], ['Civic ', 1.0]]}
>>> {k:[[ke,sum([i[1] for i in list(g)])] for ke,g in groupby(sorted(v), key= lambda x: x[0])] for k,v in d.items()}
>>> {'NSW': [['Queanbeyan ', 25.0]], 'VIC': [['City West ', 40.0], ['Thornbury ', 0.0]], 'ACT': [['Civic ', 69.0]]}

也可以使用operator.itemgetter代替lambda重写上述表达式，以获得一些性能

>>> from operator import itemgetter
>>> {k:[[ke,sum([i[1] for i in list(g)])] for ke,g in groupby(sorted(v), key=itemgetter(0))] for k,v in d.items()}

Answer 2

您可以使用以下

获得所需的输出

dictionary = {'VIC': [['City West ', 10.0], ['City West ', 30.0], ['Thornbury ', 0.0]], 'NSW': [['Queanbeyan ', 5.0],['Queanbeyan ', 20.0]], 'ACT': [['Civic ', 8.0], ['Civic ', 60.0], ['Civic ', 1.0]]}

dic = {}
for i in dictionary:
    temp = {}
    for j in dictionary[i]:
        if not j[0] in temp: temp.update({j[0]: j[1]})
        else: temp[j[0]] += j[1]
    dic.update({i: [[j, temp[j]] for j in temp]})
dic

{＆＃39; ACT＆＃39;：[[＆＃39; Civic＆＃39;，69.0]]，＆＃39; NSW＆＃39;：[[＆＃39; Queanbeyan＆＃39;，25.0]]，＆＃39; VIC＆＃39;： [[＆＃39; City West＆＃39;，40.0]，[＆＃39; Thornbury＆＃39;，0.0]]}

但是我会建议你唱一本字典词典以简化你可以做以下事情

dictionary = {'VIC': [['City West ', 10.0], ['City West ', 30.0], ['Thornbury ', 0.0]], 'NSW': [['Queanbeyan ', 5.0],['Queanbeyan ', 20.0]], 'ACT': [['Civic ', 8.0], ['Civic ', 60.0], ['Civic ', 1.0]]}

dic = {}
for i in dictionary:
    temp = {}
    for j in dictionary[i]:
        if not j[0] in temp: temp.update({j[0]: j[1]})
        else: temp[j[0]] += j[1]
    dic.update({i: temp})

print(dic)

{＆＃39; ACT＆＃39;：{＆＃39; Civic＆＃39;：69.0}，＆＃39; NSW＆＃39;：{＆＃39; Queanbeyan＆＃39; ：25.0}，＆＃39; VIC＆＃39;： {＆＃39; City West＆＃39;：40.0，＆＃39; Thornbury＆＃39;：0.0}}

Answer 3

使用defaultdict

将我的帽子扔进戒指

In [6]: from collections import defaultdict

In [7]: tree = lambda: defaultdict(tree)

In [8]: a
Out[8]:  
{'ACT': [['Civic ', 8.0], ['Civic ', 60.0], ['Civic ', 1.0]],
 'NSW': [['Queanbeyan ', 5.0], ['Queanbeyan ', 20.0]],       
 'VIC': [['City West ', 10.0], ['City West ', 30.0], ['Thornbury ', 0.0]]}

In [9]: b = tree()

In [10]: for key in a:
    ...:     for entry in a[key]:
    ...:         if b[key][entry[0]]:
    ...:             b[key][entry[0]] += entry[1]
    ...:         else:                           
    ...:             b[key][entry[0]] = entry[1] 
    ...:                                         

In [11]: b
Out[11]:  
defaultdict(<function __main__.<lambda>>,
            {'ACT': defaultdict(<function __main__.<lambda>>,
                         {'Civic ': 69.0}),                  
             'NSW': defaultdict(<function __main__.<lambda>>,
                         {'Queanbeyan ': 25.0}),             
             'VIC': defaultdict(<function __main__.<lambda>>,
                         {'City West ': 40.0, 'Thornbury ': 0.0})})

In [12]: c = {}

In [14]: for key in b:
    ...:     for entry in b[key]:
    ...:         if key in c:
    ...:             c[key].append([entry, b[key][entry]])
    ...:         else:
    ...:             c[key] = [[entry, b[key][entry]]]
    ...:

In [15]: c
Out[15]:
{'ACT': [['Civic ', 69.0]],
 'NSW': [['Queanbeyan ', 25.0]],
 'VIC': [['City West ', 40.0], ['Thornbury ', 0.0]]}

我使用了一个活泼的哈希映射来执行此操作，然后使用defaultdict。我喜欢它创建的自动分组功能。

删除dict列表的副本

3 个答案: