整合基於重複字典值

Question

我有一本字典設立這樣

{ 
    "key1" : [1,2,4], 
    "key2" : [2,4], 
    "key3" : [1,2,4], 
    "key4" : [2,4], 
    .... 
} 

我想是這樣的。

[ 
    [ 
    ["key1", "key3"], 
    [1,2,4], 
    ], 
    [ 
    ["key2", "key4"], 
    [2,4], 
    ], 
    ..... 
] 

基於唯一值對的鍵和值列表。我怎樣才能以pythonic的方式做到這一點？

Answer 1

您可以反轉這樣的dictionnary：

orig = { 
    "key1" : [1,2,4], 
    "key2" : [2,4], 
    "key3" : [1,2,4], 
    "key4" : [2,4], 
} 

new_dict = {} 

for k, v in orig.iteritems(): 
    new_dict.setdefault(tuple(v), []).append(k) #need to "freeze" the mutable type into an immutable to allow it to become a dictionnary key (hashable object) 

# Here we have new_dict like this : 
#new_dict = { 
# (2, 4): ['key2', 'key4'], 
# (1, 2, 4): ['key3', 'key1'] 
#} 

# like sverre suggested : 
final_output = [[k,v] for k,v in new_dict.iteritems()] 

Answer 2

這裏是一個列表理解做清潔工作：

[[[key for key in dictionary.keys() if dictionary[key] == value], value] 
    for value in unique(list(dictionary.values()))] 

哪裏unique可以是返回的獨特元素的功能名單。沒有默認設置，但有很多實現（here are some）。

Answer 3

請找我下面的代碼示例，如果它仍然是實際的給你：

orig = { 
    "key1" : [1,2,4], 
    "key2" : [2,4], 
    "key3" : [1,2,4], 
    "key4" : [2,4], 
} 

unique = map(list, set(map(tuple, orig.values()))) 
print map(lambda val: [val, filter(lambda key: orig[key] == val, orig)], unique)