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Martin Hope
Daniil Demchenko
Asked: 2022-09-04 16:43:52 +0000 UTC
  • 0

请帮我修复代码。您需要使用 itertools 模块中的类和工具显示比当前日期少一年的名称/日期元组列表（即，相对于当前日期的日期已经过去一年）。

from itertools import chain
import datetime
from datetime import timedelta


class User:
    username = str
    register_date = datetime

    def __init__(self, username, register_date):
        self.username = username
        self.register_date = register_date

    def sort(self):
        if datetime.datetime.now() - timedelta(days=365) > self.register_date:
            return User(f'username={self.username}, register_date={self.register_date}')


user_reg = [User(username='Chess', register_date=datetime.datetime(2021, 9, 5, 4, 10, 43, 712296)),
            User(username='Robert', register_date=datetime.datetime(2021, 9, 4, 4, 10, 43, 712304)),
            User(username='Monica', register_date=datetime.datetime(2021, 9, 3, 4, 10, 43, 712286)),
            User(username='Sara', register_date=datetime.datetime(2021, 9, 3, 4, 10, 43, 712233)),
            User(username='Alice', register_date=datetime.datetime(2021, 8, 30, 4, 10, 43, 712313)),
            User(username='Mila', register_date=datetime.datetime(2021, 8, 28, 4, 10, 43, 712324)),
            User(username='John', register_date=datetime.datetime(2021, 8, 27, 4, 10, 43, 712273))]

for i in user_reg:
    print(i.sort())
>>>TypeError: __init__() missing 1 required positional argument: 'register_date'

通常，生成的代码应该与下面的代码相同，只使用 itertools 模块中的类和工具。还想知道：是否可以仅使用str方法来实现上面的代码？我尝试在其中添加一个 if，但它给出了错误“ str返回了非字符串（类型 NoneType）”。我还请您指出我在使用该课程时的错误，并告诉我如何正确使用。谢谢！

user_reg2 = [('Chess', datetime.datetime(2021, 9, 5, 4, 10, 43, 712296)),
            ('Robert', datetime.datetime(2021, 9, 4, 4, 10, 43, 712304)),
            ('Monica', datetime.datetime(2021, 9, 3, 4, 10, 43, 712286)),
            ('Sara', datetime.datetime(2021, 9, 3, 4, 10, 43, 712233)),
            ('Alice', datetime.datetime(2021, 8, 30, 4, 10, 43, 712313)),
            ('Mila', datetime.datetime(2021, 8, 28, 4, 10, 43, 712324)),
            ('John', datetime.datetime(2021, 8, 27, 4, 10, 43, 712273))]

result = []
for i in user_reg2:
    if datetime.datetime.now() - datetime.timedelta(days=365) > i[1]:
        result.append(i)
print(result)
>>>[('Robert', datetime.datetime(2021, 9, 4, 4, 10, 43, 712304)), 
    ('Monica', datetime.datetime(2021, 9, 3, 4, 10, 43, 712286)), 
    ('Sara', datetime.datetime(2021, 9, 3, 4, 10, 43, 712233)), 
    ('Alice', datetime.datetime(2021, 8, 30, 4, 10, 43, 712313)), 
    ('Mila', datetime.datetime(2021, 8, 28, 4, 10, 43, 712324)), 
    ('John', datetime.datetime(2021, 8, 27, 4, 10, 43, 712273))]
python
Martin Hope
Daniil Demchenko
Asked: 2022-09-01 17:06:13 +0000 UTC
  • 0

请帮我写代码。有一个字典列表：

slovar = [{code: 228, log: '08', time: datetime.datetime(2021, 9, 1, 7, 6, 3, 787903)}, 
          {code: 227, log: '18', time: datetime.datetime(2016, 1, 2, 7, 8, 1, 787903)}, 
          {code: 228, log: '56', time: datetime.datetime(2019, 5, 1, 2, 3, 3, 787903)}, 
          {code: 226, log: '82', time: datetime.datetime(2015, 4, 3, 7, 1, 3, 787903)}, 
          {code: 226, log: '76', time: datetime.datetime(2018, 9, 1, 5, 6, 1, 787903)}]

有必要根据代码键将字典分成虚构的组（即，使它们看起来像是在分组。一组有字典，代码：228，另一组代码：227，等等）

之后，需要将这些“假想组”按时间键排序，从每个组中获取最新的字典，然后从每个字典中取出代码和日志键的值，形成一个字典形式{代码：日志，代码：日志等}

为了便于理解，所有操作后的输出应该是这样的：

>>>{228: '08', 
    227: '18', 
    226: '76'}
python
Martin Hope
Daniil Demchenko
Asked: 2022-08-20 03:55:56 +0000 UTC
  • 0

写了这段代码：

from collections import defaultdict
import datetime


slovar = [{'date': datetime.date(2022, 8, 20), 'type': 'WRITE'},
          {'date': datetime.date(2022, 8, 18), 'type': 'READ'},
          {'date': datetime.date(2022, 8, 18), 'type': 'WRITE'}]

historic = defaultdict(list)

for i in slovar:
    date, type = i['date'], i['type']
    historic[date].append(type)

print(dict(historic))
>>>{datetime.date(2022, 8, 20): ['WRITE'], datetime.date(2022, 8, 18): ['READ', 'WRITE']}

我不知道如何将它带到这种形式（即，使字典在字典中 + 按日期排序 + 计数 READ 和 WRITE 并将它们绑定到日期）：

>>>{{datetime.date(2022, 8, 18): {'READ': 1, 'WRITE': 1}, datetime.date(2022, 8, 20): {'WRITE': 1}}

请帮帮我

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