给字典的笛卡尔积

Question

我有以下列表：

brand=["Audi","Mercedes"]
speed=[130,150]
model=["sport","family"]

我想获得相当于：

ll=[]
ll.append({'brand':'mercedes', 'speed':130, 'model':'family'})
ll.append({'brand':'mercedes', 'speed':130, 'model':'sport'})
ll.append({'brand':'audi', 'speed':130, 'model':'family'})
ll.append({'brand':'audi', 'speed':130, 'model':'sport'})
ll.append({'brand':'mercedes', 'speed':150, 'model':'family'})
ll.append({'brand':'mercedes', 'speed':150, 'model':'sport'})
ll.append({'brand':'audi', 'speed':150, 'model':'family'})
ll.append({'brand':'audi', 'speed':150, 'model':'sport'})

我目前这样做：

from itertools import product
ll=list(product(speed, model, brand))

我有所有需要的组合，但这只是一个列表列表而不是字典列表。不知道有没有直接pythonic的方式来做！

Answer 1

用键压缩你的值：

keys = 'brand', 'speed', 'model'

ll = [dict(zip(keys, combo)) for combo in product(brand, speed, model)]

演示：

>>> from itertools import product
>>> from pprint import pprint
>>> brand = ["Audi", "Mercedes"]
>>> speed = [130, 150]
>>> model = ["sport", "family"]
>>> keys = 'brand', 'speed', 'model'
>>> [dict(zip(keys, combo)) for combo in product(brand, speed, model)]
[{'speed': 130, 'brand': 'Audi', 'model': 'sport'}, {'speed': 130, 'brand': 'Audi', 'model': 'family'}, {'speed': 150, 'brand': 'Audi', 'model': 'sport'}, {'speed': 150, 'brand': 'Audi', 'model': 'family'}, {'speed': 130, 'brand': 'Mercedes', 'model': 'sport'}, {'speed': 130, 'brand': 'Mercedes', 'model': 'family'}, {'speed': 150, 'brand': 'Mercedes', 'model': 'sport'}, {'speed': 150, 'brand': 'Mercedes', 'model': 'family'}]
>>> pprint(_)
[{'brand': 'Audi', 'model': 'sport', 'speed': 130},
 {'brand': 'Audi', 'model': 'family', 'speed': 130},
 {'brand': 'Audi', 'model': 'sport', 'speed': 150},
 {'brand': 'Audi', 'model': 'family', 'speed': 150},
 {'brand': 'Mercedes', 'model': 'sport', 'speed': 130},
 {'brand': 'Mercedes', 'model': 'family', 'speed': 130},
 {'brand': 'Mercedes', 'model': 'sport', 'speed': 150},
 {'brand': 'Mercedes', 'model': 'family', 'speed': 150}]