Question

我得到了以下代码：

mon = tue = wed = thu = fri = sat = sun = bh = None
for ot in shop_data['opening_time']:
    if ot['day'] == 'Monday':
        mon = ot
    elif ot['day'] == 'Tuesday':
        tue = ot
    elif ot['day'] == 'Wednesday':
        wed = ot
    elif ot['day'] == 'Thursday':
        thu = ot
    elif ot['day'] == 'Friday':
        fri = ot
    elif ot['day'] == 'Saturday':
        sat = ot
    elif ot['day'] == 'Sunday':
        sun = ot
    elif ot['day'] == 'Bank holidays':
        bh = ot

ot变量的结构是：

ot = {'day': 'Monday', 'closed': True}

或

ot = {
      'day': choices.get(operating_time.day),
      'open_time': operating_time.open_time,
      'close_time': operating_time.close_time
     }

用于描述商店的营业时间，以模板

呈现

我想从列表中指定正确的ot以更好的方式更正星期几

这样做的python方式是什么？

修改

这是“星期几”匹配之前的一段代码

shop_operating_time = OperatingTime.objects.filter(
            place=place).order_by('day')
    shop_operating_time_list = []
    choices = dict((x, y) for x, y in DAY_CHOICES)
    days = choices.itervalues()

    operating_time_day = set()
    for operating_time in shop_operating_time:
        data = {
            'day': choices.get(operating_time.day),
            'open_time': operating_time.open_time,
            'close_time': operating_time.close_time,
        }
        shop_operating_time_list.append(data)
        operating_time_day.add(choices.get(operating_time.day))
    remaing_days = set(days) - operating_time_day
    for day in remaing_days:
        data = {
            'day': day,
            'closed': True
        }
        shop_operating_time_list.append(data)
    shop_data = {
        'opening_time': shop_operating_time_list
     }

Answer 1

我创建了一个字典，而不是8个变量：

opening_times = {'mon'：无，'tue'：无，......}

然后您可以根据键轻松分配。您还可以使用课程setattr：

day = 'bh' if ot['day'] == 'Bank Holidays' else ot['day'].lower()[:3]
setattr(opening_times, day, ot)

最后，您甚至可以使用exec（only in Python 2，而且速度很慢！）：

 exec("%s = ot" % var_name)

修改locals为not a good idea（感谢DSM！），即使它似乎有效（有时）。

最佳，pythonic方式匹配星期几与dicts列表

1 个答案: