错误'integrand() 需要 3 个位置参数，但给出了 4 个'，但我不知道如何。使用 scipy 集成 nquad

Question

如果这只是一个快速回答错误修复问题，我深表歉意；希望相反，我有一个可以从中学习的概念错误。

我正在尝试对积分进行优化，因此我的代码中有很多函数。当我运行它时，我得到一个错误

TypeError: integrand() 接受 3 个位置参数，但给出了 4 个

对应我的代码第53行：

return 积分.nquad(integrand, [bounds_a(), bounds_x()], args = (t_est))[0]

但是根据我对 nquad 上的the documentation 的理解，我在这一行中写的内容完全对应于 3 个参数：a 和 x，隐含在给出它们的限制，以及 t_est，一个向量。不喜欢 t_est 是向量这一事实吗？

代码：

#!/usr/bin/env python3

from scipy.optimize import minimize
import numpy as np
from scipy import integrate





#0)Number of covariates:
p=5



#2)Obtain data y (use a mdified algorithm that takes as input the covariate distributions))
def g(z):
    g = np.power((1+np.exp(-z)),-1)
    return g

#betatrue for the values of y and how they depend on the coavriates x
betatrue = [1.3,3.5,2.6,0.9,2.1,1.1]

#)Z distribution
means=0
sigmas=1

#)Function as integral
def integrand(a, x, t_est):
    t0 = t_est[0]
    t1 = t_est[1]
    Z = np.power(a,2)/np.sqrt(2*np.pi*np.power(sigmas,2)) * np.exp(-np.power((a-t0-t1*np.sum(x)),2)/(2*sigmas*sigmas))
    xint = np.append([1.0],x)
    h = g(xint.dot(betatrue))
    px = np.divide(np.exp(-np.divide(np.power((x-means),2),(2*np.power(sigmas,2)))), np.sqrt(2*np.pi*np.power(sigmas)))
    Px = np.multiply(px)
    YV = Px*h
    return YV * Z

#Integration limits


def bounds_x():
    return [-np.inf,np.inf]

def bounds_a():
    return [0,np.inf]


#integration function. If the control is set to 0, gets the integrand. If control is set to 1
#gets the jacobian. Control set to 2, gets the Hessian.
def integration(t_est):
        return integrate.nquad(integrand, [bounds_a(), bounds_x()], args = (t_est))[0]

def integration_jac(t_est):

    def jacobian_integrand(a,x,t_est):
        t0=t_est[0]
        t1=t_est[1]
        j0 = (t0 + t1 * np.sum(x))/np.power(sigmas,2)
        j1 = np.sum(x)*(t0 + t1 * np.sum(x))/np.power(sigmas,2)
        jacintegrands = [integrand(a,x,t_est) * j0, integrand(a,x,t_est) * j1]
        return jacintegrands

    jac0 = integrate.nquad(jacobian_integrand[0], [bounds_a(), bounds_x()], args = (t_est))[0]
    jac1 = integrate.nquad(jacobian_integrand[1], [bounds_a(), bounds_x()], args = (t_est))[0]
    output = [jac0,jac1]
    return output



def integration_hessian(t_est):
    t0=t_est[0]
    t1=t_est[1]

    def hessian_integrand(a,x,t_est):

        h00 = (1 + np.power((t0+t1*np.sum(x)),2))/sigmas*sigmas
        h01 = np.sum(x)*( + np.power((t0+t1*np.sum(x)),2))/sigmas*sigmas
        h11 = np.power(np.sum(x),2)*( + np.power((t0+t1*np.sum(x)),2))/sigmas*sigmas
        Hint = [[integrand(a,x,t_est)*h00,integrand(a,x,t_est)*h01],[integrand(a,x,t_est)*h01,integrand(a,x,t_est)*h11]]
        return Hint

    entry00 = integrand.nquad(hessian_integrand(0,0), [bounds_a(), bounds_x()], args = (t_est))[0]
    entry01 = integrand.nquad(hessian_integrand(0,1), [bounds_a(), bounds_x()], args = (t_est))[0]
    entry11 = integrand.nquad(hessian_integrand(1,1), [bounds_a(), bounds_x()], args = (t_est))[0]
    output = [[entry00,entry01],[entry01,entry11]]

    return output


##And then a maximisation here!
t_est = [0,0]
res = minimize(integration, t_est, method='Newton-CG', jac=integration_jac, hess=integration_hessian, options={'xtol': 1e-8, 'disp': True})

Answer 1

添加尾随逗号以使 args 成为单元素元组：args = (t_est,)