Scipy 最小化结果高于初始猜测值

Question

我正在尝试使用非线性最小二乘法联合估计模型的参数，以最小化实际估计值和基于模型的估计值之间的平方差之和。然而，结果值高于我猜测的 SSE。猜测值SSE为2951687，优化参数SSE为4281096。

版本：python 3.7.6、numpy 1.19.2、scipy 1.5.2

    import numpy as np
    import pandas as pd
    from scipy.optimize import minimize
    
    ###################### importing the excel file ######################
    df = pd.read_csv('data2.csv')
    
    ###################### Setting up variables and arrays ######################
    a = df.loc[:,'C(ADD)'].values #measured added customers
    l = df.loc[:,'C(Loss)'].values #measured lost customers
    m = df.loc[:,'m'].values #number of months
    mkt = df.loc[:,'Marketing Expense'].values #maketing dollars     in each month
    e = 5596 #end measured value, Calculated from the cac/total marketing spend over the time period
    n = len(df) #creates a variable of the length of the dataframe
    
    ###################### Defining equations ######################
    
    g0 = np.zeros(n) #guess values
    g0[0] = 0.0001
    g0[1] = 0.006
    g0[2] = 96755.00
    g0[3] = 1.7
    g0[4] = 0.6
    g0[5] = 0.1
    g0[6] = 0.006
    g0[7] = 1.7
    g0[8] = 0.6
    
    def addhat(g): #Add predict values
        pNT = g[0]
        r = g[1]
        alpha = g[2]
        c = g[3]
        Bm = g[4]
        ah = np.empty(len(df)) #an empty array for the add hat values
        b = np.empty(n) #an empty array for the B(m,m') values
    
        b[0] = np.exp(np.log(mkt[0])*Bm)
        ah[0] = 400000*((1-pNT) * (1 - (alpha/(alpha + b[0]))**r))
    
        for i in range(1, n):
            b[i] = b[i-1] + (m[i]**c - m[i-1]**c)*np.exp(np.log(mkt[i])*Bm)
            ah[i] = 400000*((1-pNT) * (1 - (alpha/(alpha + (b[i])))**r))
        return ah
    
    print('add pred values: ' + str(addhat(g0)))
    
    def rethat(g): #Retention percentage
        rr = g[5]
        alphar = g[6]
        cr = g[7]
        Bmr = g[8]
    
        k = np.empty(n) #an empty array for exponent section of the formula
        w = np.empty(n) #an empty array for the retention values
    
        #The value of b(t)r when i = 0
        k[0] = np.exp(np.log(mkt[0])*Bmr)
    
        w[0] = 1 - (alphar/(alphar + k[0]))**rr
    
        # the value of B(t) for all other values of q
        for i in range(1, n):
            k[i] = k[i-1] + (m[i]**cr - m[i-1]**cr)*np.exp(np.log(mkt[i])*Bmr)
            w[i] = 1 - (alphar/(alphar + (k[i])))**rr
        return w
    
    def endpred(g): #predicting the end hat values
        eh = np.empty(n) #an empty array for the end hat values
    
        eh[0] = 213
        for i in range(1, n):
            eh[i] = (eh[i-1] * rethat(g)[i]) + addhat(g)[i]
        return eh
        
    endhat = sum(endpred(g0))
    
    def losshat(g):
        lh = np.empty(n) #an empty array for the loss hat values
    
        lh[0] = 0
        for i in range(1, n):
            lh[i] = endpred(g)[i-1] - (endpred(g)[i] - addhat(g)[i])
        return lh
    
    ###################### Sum of square errors ######################
    
    def objective(g):
        sse = sum((addhat(g)-a)**2 + (losshat(g)-l)**2) + (endhat-e)**2
        return sse
    print("SSE Initial: " + str(objective(g0)))
    ###################### Constraints ######################
    def constraint1(g): #c is greater than 1
        return g[3] - 1
    
    def constraint2(g): #cr is greater than 1
        return g[7] - 1
    
    def constraint3(g): #pNT is greater than 0
        return g[0]
    
    con1 = {'type': 'ineq', 'fun': constraint1}
    con2 = {'type': 'ineq', 'fun': constraint2}
    con3 = {'type': 'ineq', 'fun': constraint3}
    cons = [con1, con2, con3]
    
    ###################### Optimize ######################
    s = minimize(objective, g0, method='SLSQP', constraints = cons)
    g = s.x
    
    print(g)
    print("SSE Final: " + str(objective(g)))

生成的 SSE 值为 4,281,096.9，其值为： 3.48133574e+02, 6.84452015e+02, 9.67550032e+04, 2.22008198e+00, -3.28153006e+03, -1.91454144e+02, 2.20947909e+02, 1.70207912e+00, -1.24649708e+01

我使用的初始猜测值与实际结果值非常接近（我正在检查我的代码，遇到一个我知道结果的问题）。结果应为 0.0001001361, 0.006035783, 96,755.64542, 1.78204741, 0.636357403, 0.152, 0.0065432195, 1.73490796, 0.62625507，其 SSE 为 912,278。

Link 到data2.csv。

再次感谢您的帮助

Answer 1

您似乎将同一变量 l 用于 2 个不同的目的。 它首先使用 CSV 文件中的固定值进行初始化，但也用作函数 rethat 中的内部变量。并且也用在目标函数中——这意味着每次优化时，也会改变目标函数。 这看起来不太好......