将字符串列表转换为字典时字典更新序列错误

Question

我有一个字符串列表，列表中的每个元素都有几个用冒号分隔的字符串。我正在尝试将每个元素转换为字典。例如，我的列表中的一个元素如下所示：

attributesList[0]
Out: 'Health Score: A, Happy Hour Specials: Yes, Vegan Options: Yes, Takes Reservations: Yes, Delivery: No, Take-out: Yes, Accepts Credit Cards: Yes, Good For: Brunch, Lunch, Dinner, Parking: Street, Bike Parking: Yes, Wheelchair Accessible: Yes, Good for Kids: No, Good for Groups: Yes, Ambience: Casual, Trendy, Classy, Noise Level: Average, Alcohol: Beer & Wine Only, Good For Happy Hour: Yes, Outdoor Seating: Yes, Wi-Fi: Free, Has TV: No, Waiter Service: Yes, Caters: No, Gender Neutral Restrooms: Yes'

基于link 1和link 2中的解决方案，我尝试了以下方法：

attributesDict = dict(s.split(':') for s in attributesList)
attributesDict = dict(map(str.strip, s.split(':')) for s in attributesList)
attributesDict = dict(map(lambda s : s.split(':') for s in attributesList))

但在每种方法中，我都会收到如下所示的错误消息：

ValueError: dictionary update sequence element #0 has length 24; 2 is required
ValueError: dictionary update sequence element #0 has length 24; 2 is required
TypeError: map() must have at least two arguments.

我研究了一个解决方案here，但我不清楚如何在我的环境中解决这个问题。我还有点担心冒号后面的字符串中出现多个项目，如下例所示：

 Good For: Brunch, Lunch, Dinner,

我可以将冒号后面的三个项目作为值捕获到字典中吗？我怎样才能达到我想要达到的目标

编辑：在下面添加所需的输出

attributesDict[0]
Out: {'Health Score': 'A', 'Happy Hour Specials': 'Yes', 'Vegan Options': 'Yes', 'Takes Reservations': 'Yes', 'Delivery': 'No', 'Take-out': 'Yes', 'Accepts Credit Cards': 'Yes', 'Good For': 'Brunch, Lunch, Dinner', 'Parking': 'Street', 'Bike Parking': 'Yes', 'Wheelchair Accessible': 'Yes', 'Good for Kids': 'No', 'Good for Groups': 'Yes', 'Ambience': 'Casual, Trendy, Classy', 'Noise Level': 'Average', 'Alcohol': 'Beer & Wine Only', 'Good For Happy Hour': 'Yes', 'Outdoor Seating': 'Yes', 'Wi-Fi': 'Free', 'Has TV': 'No', 'Waiter Service': 'Yes', 'Caters': 'No', 'Gender Neutral Restrooms': 'Yes'}

Answer 1

如果您确实想要一个dict列表作为输出，您可以执行以下操作：

def get_dict(l):
    result = {}
    for v in l.split(','):
        if ':' in v:
            key, value = v.split(':')
            result[key.strip()] = value.strip()
        else:
            result[key.strip()] += ', ' + v.strip()
    return result

[get_dict(s) for s in attributesList]

这可能写得更好，但由于可能存在多重价值观，听写理解太复杂了

Answer 2

您可以使用正则表达式：

import re

def gen_key(s):
    yield from (e.group(1).strip() for e in re.finditer(r'([^,]+?):', l[0]))

def gen_values(s):
    yield from (e.group().strip(' ,') for e in re.finditer(r'(?<=[:^])(.+?)(?=[^,]*?:|$)', l[0]))

def gen(s):
    yield from zip(gen_key(s), gen_values(s))

dict(*map(gen, l))

输出：

{'Health Score': 'A',
 'Happy Hour Specials': 'Yes',
 'Vegan Options': 'Yes',
 'Takes Reservations': 'Yes',
 'Delivery': 'No',
 'Take-out': 'Yes',
 'Accepts Credit Cards': 'Yes',
 'Good For': 'Brunch, Lunch, Dinner',
 'Parking': 'Street',
 'Bike Parking': 'Yes',
 'Wheelchair Accessible': 'Yes',
 'Good for Kids': 'No',
 'Good for Groups': 'Yes',
 'Ambience': 'Casual, Trendy, Classy',
 'Noise Level': 'Average',
 'Alcohol': 'Beer & Wine Only',
 'Good For Happy Hour': 'Yes',
 'Outdoor Seating': 'Yes',
 'Wi-Fi': 'Free',
 'Has TV': 'No',
 'Waiter Service': 'Yes',
 'Caters': 'No',
 'Gender Neutral Restrooms': 'Yes'}