Question

我尝试根据自己的需要创建完美的字典（dict包含带有值和列表的字典）。但是，似乎我一遍又一遍地指定相同的参考。

brands = ['val1', 'val2', 'val3']
infoBrands = dict.fromkeys(brands,
                dict(dict.fromkeys(['nbOffers', 'nbBestOffers'], 0),
                     **dict.fromkeys(['higherPrice'], [])))

infoBrands['val1']['nbOffers'] += 1
print infoBrands

结果如下：

{'val3':
     {'higherPrice': [],
       'nbOffers': 1,
       'nbBestOffers': 0},
  'val2':
       {'higherPrice': [],
        'nbOffers': 1,
        'nbBestOffers': 0},
  'val1':
       {'higherPrice': [],
        'nbOffers': 1,
        'nbBestOffers': 0}
}

如您所见，val1，val2和val3指的是同一个dict。我不确定我应该怎么处理它？有什么提示吗？

Answer 1

此类事情通常使用dictionary comprehensions而不是dict.fromkeys()：

brands = ['val1', 'val2', 'val3']
infoBrands = {brand: {'nbOffers': 0, 'nbBestOffers': 0, 'higherPrice': []}
                for brand in brands}

infoBrands['val1']['nbOffers'] += 1
print infoBrands

输出：

{'val3': {'higherPrice': [], 'nbOffers': 0, 'nbBestOffers': 0},
 'val2': {'higherPrice': [], 'nbOffers': 0, 'nbBestOffers': 0},
 'val1': {'higherPrice': [], 'nbOffers': 1, 'nbBestOffers': 0}}

Answer 2

以下是使用dict comprehension实现所需结果的另一种方法：

brands = ['val1', 'val2', 'val3']
infoBrands = {g: dict({i: 0 for i in ['nbOffers', 'nbBestOffers']},
                       **{j: [] for j in ['higherPrice']}) for g in brands}

infoBrands['val1']['nbOffers'] += 1
print infoBrands

Answer 3

brands = ['val1', 'val2', 'val3']
init = dict(higher_price=[], nb_offers=0, nb_best_offers=0)
info = {brand: dict(init) for brand in brands}

Answer 4

谢谢大家，

编辑：看看@martineau的回答

我做了以下事情：

brands = ['val1', 'val2', 'val3']
infoBrands = dict.fromkeys(brands)

for brand in brands:
    infoBrands[brand] = dict([('nbOffers', 0), ('nbBestOffers', 0), ('higherPrice', [])])                                                                                                                                                     

infoBrands['val1']['nbOffers'] = 1
print infoBrands

结果：

{'val3'：{'higherPrice'：[]，'nbOffers'：0，'nbBestOffers'：0}， 'val2'：{'higherPrice'：[]，'nbOffers'：0，'nbBestOffers'：0}，'val1'： {'higherPrice'：[]，'nbOffers'：1，'nbBestOffers'：0}}

dict.fromkeys是否一遍又一遍地分配相同的引用？

4 个答案: