python - python对象的访问名称

Question

我制作了一个使用电热丝的类(class):

class Heating_wire:

    def __init__(self, ro, L,d,alpha):

        self.ro = ro
        self.L = L
        self.d = d
        self.alpha = alpha
        self.RT = [1]
        self.vector_T = [1]
    def get_R20(self):
        self.R_20 =  self.ro*self.L/(np.pi*(self.d/2)**2)

    def  calcular_RT(self,vector_temp):
        self.vector_T = vector_temp
        self.RT =  [self.R_20*(1 + temp*self.alpha) for temp in vector_temp ]
        return self.RT

实例化一些对象:

kantal = Heating_wire(1.45,0.25,0.3,4e-5)
nicromo = Heating_wire(1.18,0.25,0.3,0.0004)
ferroniquel = Heating_wire(0.86,0.25,0.3,9.3e-4)

wires = [kantal,nicromo,ferroniquel]

并画了个图:

leg = []
vector_temp = np.linspace(20,1000,1000)
for wire in sorted(wires):
    wire.get_R20()
    wire.get_RT(vector_temp)
    line, = plt.plot(wire.vector_T,wire.RT)
    leg.append(line)
plt.legend(leg,sorted(wires))

问题是我在图例中没有得到正确的名称，而是对对象的引用:

如果我添加一个name属性

def __init__(self,name, ro, L,d,alpha):
    self.name = name

我可以追加名字

leg = []

names= []
vector_temp = np.linspace(20,1000,1000)
for wire in sorted(wires):
    wire.get_R20()
    wire.get_RT(vector_temp)
    line, = plt.plot(wire.vector_T,wire.RT)
    leg.append(line)
    names.append(wire.name)
plt.legend(leg,names,loc='best') 

但我想知道是否有更简单的方法直接使用电线列表中的对象名称来解决这个问题:

kantal = Heating_wire(1.45,0.25,0.3,4e-5)
nicromo = Heating_wire(1.18,0.25,0.3,0.0004)
ferroniquel = Heating_wire(0.86,0.25,0.3,9.3e-4)

wires = [kantal,nicromo,ferroniquel]

Answer 1

就这样去做，没有重复:

wires = [
    Heating_wire("kantal", 1.45,0.25,0.3,4e-5),
    Heating_wire("nicromo", 1.18,0.25,0.3,0.0004),
    Heating_wire("ferroniquel", 0.86,0.25,0.3,9.3e-4)
]

要回答您的问题，不，对象不能访问它们被赋予的名称。