我一直在为此奋斗了几天。我试图估计分段高斯函数的密度。谁能告诉我为什么现在我得到错误输出[0] = y0 ValueError:设置一个序列的数组元素
TypeError: output[0] = y0
ValueError:设置一个数组元素的序列。
它发生在这条线:
Zero_RG = integrate.romberg(gaussian(q,x,mu,sigma), Q1, Q2).`
下面是脚本:
import numpy as np
import sympy as sp
from sympy import *
from scipy import integrate
from sympy import Integral, log, exp, sqrt, pi
import matplotlib.pyplot as plt
from scipy.stats import norm, gaussian_kde
from quantecon import LAE
from sympy import symbols
var('Q1 Q2 x q sigma mu')
#q= symbols('q')
## == Define parameters == #
mu=80
sigma=20
b=0.2
Q=80
Q1=Q*(1-b)
Q2=Q*(1+b)
d = (sigma*np.sqrt(2*np.pi))
phi = norm()
n = 500
def p(x, y):
x, y = np.array(x, dtype=float), np.array(y, dtype=float)
Positive_RG = norm.pdf(x-y+Q1, mu, sigma)
print('Positive_R = ', Positive_RG)
Negative_RG = norm.pdf(x-y+Q2, mu, sigma)
print('Negative_RG = ', Negative_RG)
gaussian = lambda q,x,mu,sigma: 1/(sigma*np.sqrt(2*np.pi))*np.exp(-(x+q-mu)**2 /(2*sigma**2))
wrapped_gaussian = lambda q: gaussian(q, x, mu, sigma)
Zero_RG = integrate.romberg(wrapped_gaussian, Q1, Q2)
print('pdf',gaussian)
#Zero_RG = scipy.integrate.quad(norm.pdf(x + q, mu, sigma))
# Int_zerocase= lambda q: norm.pdf(x + q, u, sigma)
# Zero_RG = scipy.integrate.quad(Int_zerocase, Q1, Q2)
# print(Zero_RG)
if y>0.0 and x -y>=-Q1:
#print('printA', Positive_RG)
return Positive_RG
elif y<0.0 and x -y>=-Q2:
#print('printC', Negative_RG)
return Negative_RG
elif y==0.0 and x >=-Q1:
#print('printB', Zero_RG)
return Zero_RG
return 0.0
Z = phi.rvs(n)
X = np.empty(n)
for t in range(n-1):
X[t+1] = X[t] + Z[t]
#X[t+1] = np.abs(X[t]) + Z[t]
psi_est = LAE(p, X)
k_est = gaussian_kde(X)
fig, ax = plt.subplots(figsize=(10,7))
ys = np.linspace(-200.0, 200.0, 200)
ax.plot(ys, psi_est(ys), 'g-', lw=2, alpha=0.6, label='look ahead estimate')
ax.plot(ys, k_est(ys), 'k-', lw=2, alpha=0.6, label='kernel based estimate')
ax.legend(loc='upper left')
plt.show()
“integeate.romberg”函数的第一个参数必须是一个函数。你给它'gausian(...)',用参数调用lambda的结果。 – hpaulj
@hpaulj,仍然无法正常工作。我已经忍受了2天的尝试:-) .... –