我尝试为囚徒困境中的学习算法创建相同类型的不同对象，但 python 混淆了它们

Question

显然，我想不是 Python 让他们感到困惑。我只是想不出我的错误的来源。

这是我的代码：

import numpy as np

class EW_learner:
    #p is the probability distribution (from which we chose action C or D)
    #pp are the potential payoffs (if 1 action was taken consistently)
    #s is my real payoff or score (i.e. negated total prison years)
    ##pp[0] and action = 0 is cooperate (C) => (0=C)
    ##pp[1] and action = 1 is defect (D) => (1=D)
    #ps is a list of the probability distribution at each round
    #ss is a list of the scores at each round
    def __init__(self, lr, p=[0.5,0.5], pp=[0,0], s=0, ps=[], ss=[]):
        self.lr = lr
        self.p = p
        self.pp = pp
        self.s = s
        self.ps = ps
        self.ss = ss

    #Return an action (C or D) => (0 or 1)
    def action(self):
        return int(np.random.choice(2, 1, p=self.p))

    def update(self, my_act, adv_act):
        if (my_act == 0) and (adv_act == 0):
            self.s -= 3
            self.pp[0] -= 3
            self.pp[1] -= 5
        elif (my_act == 1) and (adv_act == 0):
            self.s -= 5
            self.pp[0] -= 3
            self.pp[1] -= 5
        elif (my_act == 0) and (adv_act == 1):
            #self.s -= 0
            #self.pp[0] -= 0
            self.pp[1] -= 1
        elif (my_act == 1) and (adv_act == 1):
            self.s -= 1
            #self.pp[0] -= 0
            self.pp[1] -= 1

        self.p[0] = np.power(1.0+self.lr, self.pp[0])
        self.p[1] = 1 - self.p[0]

    def collect_data(self):
        (self.ps).append(self.p)
        (self.ss).append(self.s)

def play(p1, p2, n_rounds):
    for r in range(n_rounds):
        act1 = p1.action()
        act2 = p2.action()
        p1.update(act1, act2)
        p2.update(act2, act1)
        p1.collect_data()
        p2.collect_data()
    print('P1 Score: ' + str(p1.s) + ', P2 Score: ' + str(p2.s))
    print('P1 ProbDist: ' + str(p1.p) + ', P2 ProbDist: ' +  str(p2.p))
    return p1.ss, p2.ss, p1.ps, p2.ps

lucas = EW_learner(0.1)
paula = EW_learner(0.9)
sim = play(lucas, paula, 10)

当我调用 sim[0] 时，它基本上应该输出 p1.ss 这是卢卡斯在每一轮的得分，它输出这个：

in [85]: sim[0]

out[85]: 
[-5,
 0,
 -6,
 -1,
 -6,
 -6,
 -7,
 -7,
 -8,
 -8,
 -9,
 -9,
 -10,
 -10,
 -11,
 -11,
 -12,
 -12,
 -13,
 -13]

长度为 20（这没有意义，因为有 10 个回合）。此外，此输出似乎与 sim[1] 的输出完全相同，应该是 paula 在每一轮的得分。出于某种原因，他们合并到相同的列表中，交替使用卢卡斯的分数和保拉的分数。我知道我可以使用 %2 轻松拆分列表，但我更愿意了解错误。

Answer 1

这样做：

import numpy as np

class EW_learner():
    #p is the probability distribution (from which we chose action C or D)
    #pp are the potential payoffs (if 1 action was taken consistently)
    #s is my real payoff or score (i.e. negated total prison years)
    ##pp[0] and action = 0 is cooperate (C) => (0=C)
    ##pp[1] and action = 1 is defect (D) => (1=D)
    #ps is a list of the probability distribution at each round
    #ss is a list of the scores at each round
    def __init__(self, lr, p=[0.5,0.5], pp=[0,0], s=0):
        self.lr = lr
        self.p = p
        self.pp = pp
        self.s = s
        self.ps = []
        self.ss = []

sim[0] 结果：

[0, -5, -6, -7, -8, -9, -10, -11, -12, -13]

我尝试为囚徒困境中的学习算法创建相同类型的不同对象，但 python 混淆了它们

I try to create different objects of the same type for Learning Algorithms in Prisoner's Dilemma but python confuses them

python

class

machine-learning

object

supervised-learning