嵌套多个“for”循环

Question

我正在使用 Django 1.8.3 和 Python 3.4.3

我的应用程序开始变大，我有几个“for”循环来访问同一模型中的多个对象。虽然我是 Python 新手，但我猜想有一种更简化的方法来准备我的代码。请看下面的小sn-p。该模型是“day_of_week”，但我有几个对象的“for”循环。我还放了一个sn-p的模板代码，来一张全图。感谢您的帮助。

views.py

def get_context_data(self, **kwargs):
        context = super(EmailListView, self).get_context_data(**kwargs)
        days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
        months = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October',
                  'November', 'December']
        subject_type = ['Offer', 'Sell', 'Fear', 'Learn']
        content_type = ['Offer', 'Sell', 'Fear', 'Learn']
        email_list = ['FMGB', 'FMGR', 'AE', 'IBA']

        total_campaigns = {}
        total_campaigns_month = {}
        total_recipients = {}
        total_unsubscribes = {}
        total_bounces = {}
        total_open = {}
        total_clicks = {}

        for day in days:
            total_campaigns[day] = Email.objects.filter(day_of_week=day).count()
        for recipients in days:
            total_recipients[recipients] = Email.objects.filter(day_of_week=recipients).aggregate(
                Sum('recipients')).get('recipients__sum', 0.00)
        for unsubscribes in days:
            total_unsubscribes[unsubscribes] = Email.objects.filter(day_of_week=unsubscribes).aggregate(
                Sum('unsubscribes')).get('unsubscribes__sum', 0.00)
        for bounces in days:
            total_bounces[bounces] = Email.objects.filter(day_of_week=bounces).aggregate(Sum('bounces')).get(
                'bounces__sum', 0.00)
        for open in days:
            total_open[open] = Email.objects.filter(day_of_week=open).aggregate(
                Sum('open')).get('open__sum', 0.00)
        for clicks in days:
            total_clicks[clicks] = Email.objects.filter(day_of_week=clicks).aggregate(
                Sum('clicks')).get('clicks__sum', 0.00)

模板片段... (email.html)

{% if email_list %}
    <tr>
       <td>Monday</td>
       <td>{{ total_campaigns.Monday }}</td>
       <td>{{ total_recipients.Monday }}</td>
       <td>{{ total_unsubscribes.Monday  }}</td>
       <td>{{ total_bounces.Monday  }}</td>
       <td>{{ total_open.Monday  }}</td>
       <td>{{ total_clicks.Monday  }}</td>
       <td>{% average total_open.Monday total_recipients.Monday %}</td>
       <td>{% average total_clicks.Monday total_open.Monday %}</td>
    </tr>
    ...

Answer 1

这应该也是一样的：

for day in days:
    total_campaigns[day] = Email.objects.filter(day_of_week=day).count()
    # recipients
    total_recipients[day] = Email.objects.filter(day_of_week=day).aggregate(
        Sum('recipients')).get('recipients__sum', 0.00)
    # unsubscribes
    total_unsubscribes[day] = Email.objects.filter(day_of_week=day).aggregate(
        Sum('unsubscribes')).get('unsubscribes__sum', 0.00)
    # bounces
    total_bounces[day] = Email.objects.filter(day_of_week=day).aggregate(
        Sum('bounces')).get('bounces__sum', 0.00)
    # open
    total_open[day] = Email.objects.filter(day_of_week=day).aggregate(
        Sum('open')).get('open__sum', 0.00)
    # clicks
    total_clicks[day] = Email.objects.filter(day_of_week=day).aggregate(
        Sum('clicks')).get('clicks__sum', 0.00)

我对 django ORM 并不太熟悉，但您可能可以减少数据库查询的数量

for day in days:
    day_objects = Email.objects.filter(day_of_week=day)
    total_campaigns[day] = day_objects.count()
    # recipients
    ...

...等（并将所有Email.objects.filter(day_of_week=day)替换为新定义的day_objects）

Answer 2

所以我不完全确定这是否都是正确的语法，但也许你可以试试这种方法？

def initialize_dict(dict, days, dict_name):
    for day in days:
        if dict_name='':
            dict[day] = Email.objects.filter(day_of_week=day).count()
        else:
            dict[day] = Email.objects.filter(day_of_week=day)
                       .aggregate(Sum(dict_name))
                       .get(dict_name + '__sum', 0.00)

我不确定这是否是一个更好的解决方案，但它是一个替代方案。您可以决定将天数设为全局变量，而不必将其传递给函数。几个示例调用将是：

initialize_dict(total_campaigns, days, '')
initialize_dict(total_recipients, days, 'recipients')

等等...我希望这会有所帮助。请随时批评此解决方案，说明为什么它是好方法或坏方法！我还在学习和提高自己:)