蟒蛇嵌套列表字典

问题描述 投票：0回答：4

什么是这个列表转换成字典的最佳方式（my_dict称呼它），因此它可以被索引这样？

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

基本上my_dict[0]会给我约'CalaBar & Grill'一切。

这里的列表：

[['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', '$$'],
 ['Rock’s Chicken & Fries', '4.0 star rating', '$'],
 ['Copeland’s', '3.5 star rating', '$$']]
python dictionary nested-lists
4个回答
1
投票

这应该工作：

# 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'])
1
投票

这里有一个方法与列表理解来做到这一点，在香草蟒蛇。假设你给存储在my_list二维列表：

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

所述zip(k, v)需要两个列表并且将它们映射到一个类似于字典的结构，使得k是键，并且每个v是相应的值。你需要将结果转换为dict，虽然。

1
投票

您可以压缩所需的子类型的字典的键在一个字典的构造函数（假设你的列表存储在变量l）对应的值：

[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': '$$'}, {'name': 'Rock’s Chicken & Fries', 'stars': '4.0 star rating', 'price': '$'}, {'name': 'Copeland’s', 'stars': '3.5 star rating', 'price': '$$'}]
0
投票

使用一个简单的列表理解，你可以创建dict有：

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
$$
South City Kitchen - Midtown
4.5 star rating
$$
...

这会给你一个更强大的venues dict结构，这样就解决您的清单很好。例如，my_dict_entries向你在你的问题寻找名单。

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