如何用Python基于多个关联列表生成符合预期结构的字典列表

你的原代码没有处理list_leagues中的嵌套列表，直接把国家和对应的联赛集合配对，没有拆分多个联赛生成独立条目，修改方案如下：

修改后代码（易读版本） #

list_countries=['italy','france']
list_leagues=[['serie-a','serie-b'], 'ligue-1']

keys=['Country','League']
spam = []

for country, leagues in zip(list_countries, list_leagues):
    # 统一转为列表格式，避免单个字符串被遍历为单个字符
    league_items = leagues if isinstance(leagues, list) else [leagues]
    for league in league_items:
        spam.append(dict(zip(keys, [country, league])))

print(spam)

运行输出 #

[{'Country': 'italy', 'League': 'serie-a'},
 {'Country': 'italy', 'League': 'serie-b'},
 {'Country': 'france', 'League': 'ligue-1'}]

如果你偏好更简洁的写法，也可以用嵌套列表推导式实现：

spam = [dict(zip(keys, [country, league])) 
        for country, leagues in zip(list_countries, list_leagues) 
        for league in (leagues if isinstance(leagues, list) else [leagues])]

内容的提问来源于stack exchange，提问作者luka