Python 如何从两个单独的列表中为每个键创建多个值的字典？

我正在尝试创建一个每个键都有多个值的字典。例如：

top_10 = ['Volkswagen_Golf_1.4', 'BMW_316i', 'Ford_Fiesta', 'BMW_318i', 'Volkswagen_Polo', 'BMW_320i', 'Opel_Corsa', 'Renault_Twingo', 'Volkswagen_Golf', 'Opel_Corsa_1.2_16V']

common_brands = ['volkswagen', 'bmw', 'opel', 'mercedes_benz', 'audi', 'ford']

我想创建一个如下所示的词典：

{'volkswagen': ['Volkswagen_Golf_1.4', 'Volkswagen_Polo', 'Volkswagen_Golf'], 'bmw': ['BMW_316i', 'BMW_318i', 'BMW_320i'], 'opel': ['Opel_Corsa', 'Opel_Corsa_1.2_16V'],'ford': ['Ford_Fiesta'], 'Renault': ['Reanault_Twingo']}

使用我尝试过的代码，每个品牌只能获得一个型号，而且我找不到添加不在common_brands列表中的品牌的方法

models_by_brand = {}

for brand in common_brands:
    for model in top_10:
        if brand in model.lower():
            models_by_brand[brand] = [model]

models_by_brand

输出：

{'bmw': ['BMW_320i'],
 'ford': ['Ford_Fiesta'],
 'opel': ['Opel_Corsa_1.2_16V'],
 'volkswagen': ['Volkswagen_Golf']}

{'volkswagen': ['Volkswagen_Golf_1.4', 'Volkswagen_Polo', 'Volkswagen_Golf'],
 'bmw': ['BMW_316i', 'BMW_318i', 'BMW_320i'],
 'opel': ['Opel_Corsa', 'Opel_Corsa_1.2_16V'],
 'ford': ['Ford_Fiesta']}

您可以使用并拆分车辆名称以获得品牌（如果这些名称是标准化的）：

---

通过使用

defaultdict

，您可以按品牌[brand]编写

模型。附加（模型）

，如果词典中当前没有品牌的模型，将创建并使用空列表。

如果要保留代码的结构，请使用列表：

models_by_brand = {}

for brand in common_brands:
    model_list=[]
    for model in top_10:
        if brand in model.lower():
            model_list.append(model)
    models_by_brand[brand] = model_list
models_by_brand = {k:v for k,v in models_by_brand.items() if v!=[]}

输出：

{'bmw': ['BMW_320i'],
 'ford': ['Ford_Fiesta'],
 'opel': ['Opel_Corsa_1.2_16V'],
 'volkswagen': ['Volkswagen_Golf']}

{'volkswagen': ['Volkswagen_Golf_1.4', 'Volkswagen_Polo', 'Volkswagen_Golf'],
 'bmw': ['BMW_316i', 'BMW_318i', 'BMW_320i'],
 'opel': ['Opel_Corsa', 'Opel_Corsa_1.2_16V'],
 'ford': ['Ford_Fiesta']}

---

@霍尔特的答案将是最有效的答案。

解决问题的传统方法是检查字典中是否有关键字。如果没有，则将其初始化为空列表，否则附加到该列表

# The following code should work just fine.
top_10 = ['Volkswagen_Golf_1.4', 'BMW_316i', 'Ford_Fiesta', 'BMW_318i', 'Volkswagen_Polo', 'BMW_320i', 'Opel_Corsa',
          'Renault_Twingo', 'Volkswagen_Golf', 'Opel_Corsa_1.2_16V']

common_brands = ['volkswagen', 'bmw', 'opel', 'mercedes_benz', 'audi', 'ford']

result = {}
cars = []
# For each car brand
for k in common_brands:
    # For each car model
    for c in top_10:
        # if car brand present in car model name append it to list
        if k.lower() in c.lower():
            cars.append(c)
    # if cars list is not empty copy it to the dictionary with key k
    if len(cars) > 0:
        result[k] = cars.copy()
    # Reset cars list for next iteration
    cars.clear()

print(result)

models_by_brand = {}

for brand in common_brands:
    for model in top_10:
        if brand in model.lower():
            if brand not in models_by_brand:
                models_by_brand[brand] = list()
            models_by_brand[brand].append(model)

print(models_by_brand)

另一种解决方案是使用defaultdict。Defaultdict用于预先初始化任何键的默认值

from collections import defaultdict
models_by_brand = defaultdict(list)

for brand in common_brands:
    for model in top_10:
        if brand in model.lower():
            models_by_brand[brand].append(model)

print(models_by_brand)

---

您想如何添加不在

普通品牌中的品牌？您确定车辆名称始终为
品牌名称
？您可以使用
设置默认值
将列表设置为不知道的钥匙的值。无论哪种方式，我都认为最好使用
defaultdict
，但对某些人来说，它们有点让人困惑