如何对对象的所有实例的属性求和

Question

我想对一个对象的所有实例的 costsum 属性求和。

class ActivityCenter:

    def __init__(self, costpool, costsum, costdriver, cdunits):
        self.costpool = costpool
        self.costsum = costsum
        self.costdriver = costdriver
        self.cdunits = cdunits

cp1 = ActivityCenter("Material Handling", 480000, "Pounds", 160000)
cp2 = ActivityCenter("Production orders", 90000, "Num of POs", 200)

# I am looking to add up the costsum values for all instances, similar to:
costsumtotal = (cp1.__dict__.get("costsum")) + (cp2.__dict__.get("costsum"))

到目前为止，我已经尝试使用 sum() 理解如下，参考this solution：

B = []
for i in range(10):
    B.append(ActivityCenter())

s = sum(i.costsum for i in B)

但我无法克服缺少 4 个必需的位置参数的 TypeError。

Answer 1

要使用 Python 中的 sum 内置函数作为对象的成员变量，您需要对对象的成员变量进行序列（例如，元组或列表）。下面的 sn-p 显示了如何制作对象成员变量的列表。您发布的代码省略了comprehension expression。我希望它会有所帮助:)

class ActivityCenter:

    def __init__(self, costpool, costsum, costdriver, cdunits):
        self.costpool = costpool
        self.costsum = costsum
        self.costdriver = costdriver
        self.cdunits = cdunits

"""
Create some objects

objs = []
for i in range(num_obj):
    objs.append(ActivityCenter(<some value>,<...>,<...>,<...>))

Or use objects to make a list
"""
cp1 = ActivityCenter("Material Handling", 480000, 160000, "Pounds")
cp2 = ActivityCenter("Production orders", 90000, 200, "Num of POs")
cp3 = ActivityCenter("Marketing", 120000, 1000, "Num of events")

objs = [cp1, cp2, cp3]

total_cost = sum([obj.costsum for obj in objs])  # List comprehension
print("Total cost: ", total_cost)