在键值列表中为每个用户复制字典

Question

我在Python中有一本字典，如下所示：

dicts = [{'uid':'12345', 'markets':'['UK', 'US']'},
         {'uid':'55644', 'markets':'['IN', 'SG', 'IE']'}]

我真正需要的是为“市场”列表中的每一项，为词典列表中的每一本词典制作的任何词典的副本。应该是这样的：

dicts = [{'uid':'12345', 'markets':'['UK']'},
         {'uid':'12345', 'markets':'['US']'},
         {'uid':'55644', 'markets':'['IN']'},
         {'uid':'55644', 'markets':'['IE']'},
         {'uid':'55644', 'markets':'['SG']'}]

有人能帮我理解Python中解决这个问题的最佳方法吗？

Answer 1

下面的代码将创建一个字典列表：

dicts = [{'uid':'12345', 'markets':['UK', 'US']},
         {'uid':'55644', 'markets':['IN', 'SG', 'IE']}]

def flatten(lists):
    res = []
    for d in lists:
        for market in d['markets']:
            res.append({'uid': d['uid'], 'markets': [market]})

    return res

print flatten(dicts)

# [{'markets': ['UK'], 'uid': '12345'}, {'markets': ['US'], 'uid': '12345'},
#  {'markets': ['IN'], 'uid': '55644'}, {'markets': ['SG'], 'uid': '55644'},   
#  {'markets': ['IE'], 'uid': '55644'}]

代码将遍历原始列表中的字典。对于每个字典，代码将在市场上迭代，对于每个市场，它将为结果列表添加一个新字典。除了print之外，该代码将同时在Python2和Python3上工作。

Answer 2

您可以(python 3)尝试：

dicts = [{'uid':'12345', 'markets':['UK', 'US']},
         {'uid':'55644', 'markets':['IN', 'SG', 'IE']}]


def _split_dict(dict_):
    for country in dict_["markets"]:
        yield {'markets': [country], 'uid': dict_['uid']}


def split_dicts(dicts):
    for dict_ in dicts:
        yield from _split_dict(dict_)


print(list(split_dicts(dicts)))
# [{'uid': '12345', 'markets': ['UK']}, {'uid': '12345', 'markets': ['US']}, {'uid': '55644', 'markets': ['IN']}, {'uid': '55644', 'markets': ['SG']}, {'uid': '55644', 'markets': ['IE']}]