试图找出类中现有对象的平均值答案

【问题标题】：Trying to figure out the average of existing objects in a class试图找出类中现有对象的平均值
【发布时间】：2017-11-28 04:43:02
【问题描述】：

我需要帮助来计算每位员工的平均工资。
这是我目前拥有的代码。

class Associate:
    ID = 0;
    avgPay = 0.0

    def __init__(self, ID, name, pay):
        self.ID = ID
        self.name = name
        self.pay = pay
        Associate.ID += 1
        Associate.avgPay = (pay + Associate.avgPay)/Associate.ID

    def speak(self):
        print('Name: ', self.name,'Pay: ', self.pay )

a1 = Associate('A111','Emily',85000)
a1.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))
a2 = Associate('A222','Bob',88000)
a2.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))
a3 = Associate('A333','John',92000)
a3.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))
a4 = Associate('A444','Tom',77000)
a4.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))

我的问题是，每次调用 Associate.avgPay 时，它都会将上次调用的平均值相加，而不是添加 self.pay。
当前输出：

Name: Emily Pay:  85000
ID = 1, avgPay = 85000.0
Name: Bob Pay:  88000
ID = 2, avgPay = 86500.0
Name: John Pay:  92000
ID = 3, avgPay = 59500.0
Name: Tom Pay:  77000
ID = 4, avgPay = 34125.0

正确的输出：

Name: Emily Pay:  85000
ID = 1, avgPay = 85000.0
Name: Bob Pay:  88000
ID = 2, avgPay = 86500.0
Name: John Pay:  92000
ID = 3, avgPay = 88333.33333
Name: Tom Pay:  77000
ID = 4, avgPay = 85500.0

任何帮助将不胜感激。

【问题讨论】：

平均值不是这样的。您需要一份所有付款的清单

标签： python python-3.x class average

【解决方案1】：

如果您为两个以上的对象编写了在纸上尝试的平均算法，您会发现您实际上并没有计算平均值。

您需要一次将所有值相加，然后除，而不是将新值添加到已除的值中，然后再次除...

此外，您正在重复代码这一事实意味着您确实应该考虑使用列表。也不需要任何类属性

这是一个例子

associates = [] 
associates.append(Associate('A111','Emily',85000))
associates.append(Associate('A222','Bob',88000)) 
associates.append(Associate('A333','John',92000)) 
associates.append(Associate('A444','Tom',77000)) 

pays = [a.pay for a in associates] 
avg = sum(pays)/len(pays) 
for id, a in enumerate(associates):
    a.speak()
    print("ID = {0}, avgPay = {1}".format(id, avg))

【讨论】：

【解决方案2】：

至少，平均计算似乎是错误的。应该是，

Associate.avgPay = (pay + Associate.avgPay * (Associate.ID-1))/Associate.ID

【讨论】：

你是乘以取消之前的除法吗？
大概，是的。分子必须是过去支付的总和。
看起来还是不对，因为你还需要在乘以之前减去以前的工资
嗯......我很害怕，但你真的尝试过吗？至少，修改后的方程输出给定值的真实平均值。

【解决方案3】：

正如@cricket_007 所说，平均值不会那样工作。您可以做的是跟踪列表中的所有付款并使用它来计算平均值：

class Associate:
    ID = 0;
    payList = []

    def __init__(self, ID, name, pay):
        self.ID = ID
        self.name = name
        self.pay = pay
        Associate.ID += 1
        Associate.payList.append(pay)
        Associate.avgPay = sum(Associate.payList) / len(Associate.payList)

【讨论】：

【解决方案4】：

如果您不想保留列表，则可以将总工资作为附加类别属性保留：

class Associate:
    ID = 0;
    avgPay = 0.0
    totalPay = 0.0

    def __init__(self, ID, name, pay):
        self.ID = ID
        self.name = name
        self.pay = pay
        Associate.totalPay+=pay
        Associate.ID += 1
        Associate.avgPay = Associate.totalPay / Associate.ID

    def speak(self):
        print('Name: ', self.name,'Pay: ', self.pay )

然后您的一系列对象按您的意愿工作：

a1 = Associate('A111','Emily',85000)
a1.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))
a2 = Associate('A222','Bob',88000)
a2.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))
a3 = Associate('A333','John',92000)
a3.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))
a4 = Associate('A444','Tom',77000)
a4.speak()
print("ID = {0}, avgPay = {1}".format(Associate.ID,Associate.avgPay))

打印：

Name:  Emily Pay:  85000
ID = 1, avgPay = 85000.0
Name:  Bob Pay:  88000
ID = 2, avgPay = 86500.0
Name:  John Pay:  92000
ID = 3, avgPay = 88333.33333333333
Name:  Tom Pay:  77000
ID = 4, avgPay = 85500.0

【讨论】：