做dict.fromkeys公司一次又一次地分配同一个参考？

Question

我试图为我的需要创建一个完美的字典（dict包含一个包含值和列表的dict）。不过，我在参考书上似乎是一样的。在

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：

^{pr2}$

如您所见，val1、val2和val3指的是同一个dict。 我不知道该怎么处理？ 有什么提示吗？在

Answer 1

这类事情通常用dictionary comprehensions而不是dict.fromkeys()来完成：

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

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

输出：

^{pr2}$

Answer 2

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

Answer 3

下面是使用^{}获得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