将元组列表转换为di列表

Question

我有一个元组列表：

[('Lebron James', datetime.date(2017, 12, 23), 500),
 ('Julie Peralta', datetime.date(2017, 12, 13), 1500),
 ('Reynaldo Pahay', datetime.date(2017, 12, 11), 2500)]

我想把它转换成字典列表：

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最好和有效的方法是什么？在

Answer 1

import datetime

data = [('Lebron James', datetime.date(2017, 12, 23), 500), 
        ('Julie Peralta', datetime.date(2017, 12, 13), 1500), 
        ('Reynaldo Pahay', datetime.date(2017, 12, 11), 2500)]

print([{'Name': name, 'date': date, 'penalty': penalty} for name, date, penalty in data])
# [{'Name': 'Lebron James', 'date': datetime.date(2017, 12, 23), 'penalty': 500}, {'Name': 'Julie Peralta', 'date': datetime.date(2017, 12, 13), 'penalty': 1500}, {'Name': 'Reynaldo Pahay', 'date': datetime.date(2017, 12, 11), 'penalty': 2500}]

Answer 2

{a1}你可以利用。在

import datetime

def get_list_of_dict(keys, list_of_tuples):
     """
     This function will accept keys and list_of_tuples as args and return list of dicts
     """
     list_of_dict = [dict(zip(keys, values)) for values in list_of_tuples]
     return list_of_dict

演示

^{pr2}$

Answer 3

import datetime
lista = [('Lebron James', datetime.date(2017, 12, 23), 500), ('Julie Peralta', datetime.date(2017, 12, 13), 1500), ('Reynaldo Pahay', datetime.date(2017, 12, 11), 2500)]
output = [{"name":a[0], "date":a[1],'penalty':a[2]} for a in lista]
print(output)