我可以不使用object.filter python而使用filter吗

我总是在

Question.objects.filter（）中使用filter
但这一次，我想使用filter从类中获取值，例如：

class Expense():
    amount = ''
    currency = ''
    date = '' 

这是我的方法：

def get_expense_list(self, taskId):
    #get expense list
    self.expenseList = [Expense()]
    # get all activity from task 
    self.activiytList = Activity.objects.filter(task_id=taskId)

    for activity in self.activiytList:
        try:    
            self.expenseList.filter(currency=activity.expense_currency, self.expenseList)
            self.expenseList.amount = self.expenseList.expense.amount + activity.expense
        except:
            newExpense = Expense()
            newExpense.currency = activity.currency
            newExpense.amount = activity.amount
            newExpense.date = activity.date
            self.expenseList.append(newExpense)

我正在从活动列表创建费用列表。在
活动.model
中，有以下对象：

amount =  models.DecimalField(max_digits=16, decimal_places=2)
currency =  models.CharField(max_length=5)
date = models.DateField()

我想在费用清单中显示，如果他们的货币和日期相同，金额将增加；如果没有现有的日期和货币，金额将
append（）

请帮助我。
列表中没有
filter（）
方法，因此您正在编写的代码无效

因此，如果您想要的是：

我想在费用列表中显示，如果他们的货币和日期相同，金额将增加；如果没有现有的日期和货币，金额将追加（）

然后你会想做一些类似的事情：

def get_expense_list(self, taskId):
    self.expenseList = []
    self.activityList = Activity.objects.filter(task_id=taskId)

    for activity in self.activityList:
        # all existing expenses for this currency and date
        comparisonFunc = lambda x: x.currency == activity.currency and x.date == activity.date
        expenses = filter(comparisonFunc, self.expenseList)

        # if there are no expenses, create a new
        if len(expenses) == 0:
            expense = Expense()
            expense.amount = activity.amount
            expense.date = activity.date
            expense.currency = activity.currency
            self.expenseList.append(expense)
        # otherwise, increase the amount for expenses that were matching
        else:
            for expense in expenses:
                expense.amount += activity.amount

您也可以通过聚合在数据库中完成所有这些操作。
您好，非常感谢您理解我想展示的内容。实际上，我尝试运行它，并收到错误消息“filter类型的对象没有len（）。非常感谢你的帮助