Question

我有一个字典，键是整数值，值是字符串列表。我想反过来说，字符串成为键，整数成为值。像：

首先设计：

{1:["a"], 2:["a","b"], 3:["a"], 4:["b", "cd"], 6:["a","cd"]}

第二个设计：

{"a": [1,2,3,6], "b":[2,4], "cd":[4,6]}

任何线索？感谢

Answer 1

对于单行的爱（注意：效率很低！）：

d2 = {val: [key for key in d if val in d[key]] for lst in d.itervalues() for val in lst}

Answer 2

一种简单的方法是创建一个defaultdict，为新键提供一个空列表，然后为当前dict中的每个列表创建一个键，并将原始的dict键附加到新的列表：

from collections import defaultdict

orig = {1: ['a'], 2: ['a', 'b'], 3: ['a'], 4: ['b', 'cd'], 6: ['a', 'cd']}
d = defaultdict(list)

for k, v in orig.items():
    for string in v:
        d[string].append(k)

# d = {'a': [1, 2, 3, 6], 'b': [2, 4], 'cd': [4, 6]}

Answer 3

代码

first_map = {1: ["a"], 2: ["a", "b"], 3: ["a"], 4: ["b", "cd"], 6: ["a", "cd"]}
second_map = {}
for key, value in first_map.items():
    for i in value:
        if i in second_map:
            second_map[i].append(key)
        else:
            second_map[i] = [key]
print(second_map)

Answer 4

z = {1:["a"], 2:["a","b"], 3:["a"], 4:["b", "cd"], 6:["a","cd"]}
d = dict()

for k, v in z.iteritems():
    for i in v: 
        if i in d: 
            d[i].append(k)
        else:   
            d[i] = [k]

使用iteritems()而非items()更好一点，因为前者在Python 2中创建了一个生成器。

Answer 5

dict1= {1:["a"], 2:["a","b"], 3:["a"], 4:["b", "cd"], 6:["a","cd"]}
dict2= {}
for key, value in dict1.items():
   for val in value:
       if val in dict2:
           dict2[val].append(key)
           dict2[val]= list(set(dict2[val]))
       else:
           dict2[val]= [key]

Answer 6

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  </head>
  <body>
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  </body>
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将字典转换为另一个字典

6 个答案: