python嵌套列表到字典

时间：2019-02-04 19:25:18

标签： python dictionary nested-lists

将这个列表转换成字典（称为my_dict）以便以这种方式建立索引的最佳方法是什么？ 

my_dict[i]['name']
my_dict[i]['stars']
my_dict[i]['price']

基本上my_dict[0]会给我有关'CalaBar & Grill'的一切。

以下是列表：

[['CalaBar & Grill', '4.0 star rating', '$$'],
 ['Red Chili Cafe', '4.0 star rating', '$$'],
 ['Gus’s World Famous Fried Chicken', '4.0 star rating', '$$'],
 ['South City Kitchen - Midtown', '4.5 star rating', '$$'],
 ['Mary Mac’s Tea Room', '4.0 star rating', '$$'],
 ['Busy Bee Cafe', '4.0 star rating', '$$'],
 ['Richards’ Southern Fried', '4.0 star rating', '$$'],
 ['Greens & Gravy', '3.5 star rating', '$$'],
 ['Colonnade Restaurant', '4.0 star rating', '$$'],
 ['South City Kitchen Buckhead', '4.5 star rating', '$$'],
 ['Poor Calvin’s', '4.5 star rating', '$$'],
 ['Rock’s Chicken & Fries', '4.0 star rating', '$'],
 ['Copeland’s', '3.5 star rating', '$$']]

4 个答案:

答案 0 :(得分：1)

您可以在dict构造函数中用所需的值压缩所需子字典的键（假设您的列表存储在变量l中）：

[dict(zip(('name', 'stars', 'price'), i)) for i in l]

这将返回：

[{'name': 'CalaBar & Grill', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Red Chili Cafe', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Gus’s World Famous Fried Chicken', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'South City Kitchen - Midtown', 'stars': '4.5 star rating', 'price': '$$'}, {'name': 'Mary Mac’s Tea Room', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Busy Bee Cafe', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Richards’ Southern Fried', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Greens & Gravy', 'stars': '3.5 star rating', 'price': '$$'}, {'name': 'Colonnade Restaurant', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'South City Kitchen Buckhead', 'stars': '4.5 star rating', 'price': '$$'}, {'name': 'Poor Calvin’s', 'stars': '4.5 star rating', 'price': '$$'}, {'name': 'Rock’s Chicken & Fries', 'stars': '4.0 star rating', 'price': '$'}, {'name': 'Copeland’s', 'stars': '3.5 star rating', 'price': '$$'}]

答案 1 :(得分：1)

这是使用香草python进行列表理解的一种方法。假设您给出的2D列表存储在my_list中：

keys = ['name', 'stars', 'price']
my_dict = [dict(zip(keys, values)) for values in my_list]

zip(k, v)接受两个列表，并将它们映射为字典式结构，以k为键，每个v为对应的值。不过，您确实需要将结果转换为dict。

答案 2 :(得分：1)

这应该有效：

# The list
my_list = [['CalaBar & Grill', '4.0 star rating', '$$'], \
 ['Red Chili Cafe', '4.0 star rating', '$$'],\
 ['Gus’s World Famous Fried Chicken', '4.0 star rating', '$$'],\
 ['South City Kitchen - Midtown', '4.5 star rating', '$$'],\
 ['Mary Mac’s Tea Room', '4.0 star rating', '$$'],\
 ['Busy Bee Cafe', '4.0 star rating', '$$'],\
 ['Richards’ Southern Fried', '4.0 star rating', '$$'],\
 ['Greens & Gravy', '3.5 star rating', '$$'],\
 ['Colonnade Restaurant', '4.0 star rating', '$$'],\
 ['South City Kitchen Buckhead', '4.5 star rating', '$$'],\
 ['Poor Calvin’s', '4.5 star rating', '$$'],\
 ['Rock’s Chicken & Fries', '4.0 star rating', '$'],\
 ['Copeland’s', '3.5 star rating', '$$']]

# initialize an empty list
my_dict = []

# create list of dictionary
for elem in my_list:
    temp_dict = {}
    temp_dict['name'] = elem[0]
    temp_dict['stars'] = elem[1]
    temp_dict['price'] = elem[2]
    my_dict.append(temp_dict)


# testing
print(my_dict[1]['stars'])
print(my_dict[5]['price'])
print(my_dict[0]['name'])
print(my_dict[7]['stars'])

答案 3 :(得分：0)

使用更简单的列表理解，您可以创建带有以下内容的dict：

list = [['CalaBar & Grill', '4.0 star rating', '$$'],
 ['Red Chili Cafe', '4.0 star rating', '$$'],
 ['Gus’s World Famous Fried Chicken', '4.0 star rating', '$$'],
 ['South City Kitchen - Midtown', '4.5 star rating', '$$'],
 ['Mary Mac’s Tea Room', '4.0 star rating', '$$'],
 ['Busy Bee Cafe', '4.0 star rating', '$$'],
 ['Richards’ Southern Fried', '4.0 star rating', '$$'],
 ['Greens & Gravy', '3.5 star rating', '$$'],
 ['Colonnade Restaurant', '4.0 star rating', '$$'],
 ['South City Kitchen Buckhead', '4.5 star rating', '$$'],
 ['Poor Calvin’s', '4.5 star rating', '$$'],
 ['Rock’s Chicken & Fries', '4.0 star rating', '$'],
 ['Copeland’s', '3.5 star rating', '$$']]

my_dict = {'venues': [{'name': item[0], 'stars': item[1], 'price': item[2]} for item in list]}

my_dict_entries = my_dict['venues']

for i in range(len((my_dict_entries))):
    print(my_dict_entries[i]['name'])
    print(my_dict_entries[i]['stars'])
    print(my_dict_entries[i]['price'])

dict：

{"venues": [{"name": "CalaBar & Grill", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "Red Chili Cafe", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "Gus\u2019s World Famous Fried Chicken", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "South City Kitchen - Midtown", "rating": "4.5 star rating", "pricing": "$$"}, {"name": "Mary Mac\u2019s Tea Room", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "Busy Bee Cafe", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "Richards\u2019 Southern Fried", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "Greens & Gravy", "rating": "3.5 star rating", "pricing": "$$"}, {"name": "Colonnade Restaurant", "rating": "4.0 star rating", "pricing": "$$"}, {"name": "South City Kitchen Buckhead", "rating": "4.5 star rating", "pricing": "$$"}, {"name": "Poor Calvin\u2019s", "rating": "4.5 star rating", "pricing": "$$"}, {"name": "Rock\u2019s Chicken & Fries", "rating": "4.0 star rating", "pricing": "$"}, {"name": "Copeland\u2019s", "rating": "3.5 star rating", "pricing": "$$"}]}

dict_entries：

[{'name': 'CalaBar & Grill', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Red Chili Cafe', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Gus’s World Famous Fried Chicken', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'South City Kitchen - Midtown', 'stars': '4.5 star rating', 'price': '$$'}, {'name': 'Mary Mac’s Tea Room', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Busy Bee Cafe', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Richards’ Southern Fried', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'Greens & Gravy', 'stars': '3.5 star rating', 'price': '$$'}, {'name': 'Colonnade Restaurant', 'stars': '4.0 star rating', 'price': '$$'}, {'name': 'South City Kitchen Buckhead', 'stars': '4.5 star rating', 'price': '$$'}, {'name': 'Poor Calvin’s', 'stars': '4.5 star rating', 'price': '$$'}, {'name': 'Rock’s Chicken & Fries', 'stars': '4.0 star rating', 'price': '$'}, {'name': 'Copeland’s', 'stars': '3.5 star rating', 'price': '$$'}]

输出[截断]：

CalaBar & Grill
4.0 star rating
$$
Red Chili Cafe
4.0 star rating
$$
Gus’s World Famous Fried Chicken
4.0 star rating
$$
South City Kitchen - Midtown
4.5 star rating
$$
...

这将为您提供更健壮的venues dict结构，使您可以很好地处理列表。例如，my_dict_entries为您提供了您要查找的问题清单。

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