TypeError:model()只需3个参数(给定5个参数)

时间:2016-03-29 04:46:38

标签: python scipy odeint

odeint ,以下设置正常;

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint

v0 = 10.0
k1 = 0.5
k2 = 0.35

def model(x,t):
    dx0 = v0 - k1*x[0]
    dx1 = k1*x[0] - k2*x[1]
    return [dx0, dx1]
time = linspace(0.0,20.0,100)
xinit = array([0.0,0.0])
x = odeint(model,xinit,time)
plt.plot(time, x[:,0], time, x[:,1])

但是当我想定义一个参数化模型然后将参数传递给model()函数时,当 odeint 调用它时我遇到了这个错误:TypeError:model()只需3个参数(给出5) 。这有什么不对?传递参数的正确设置是什么?

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

def model (x,t,p):
    dot_x = np.zeros(2)
    v0 = p[0]
    k1 = p[1]
    k2 = p[2]
    dot_x[0] = v0 - k1*x[0]
    dot_x[1] = k1*x[0] - k2*x[1]
    return dot_x

p = (10,0.5,.35)
xinit = [0.0,0.0] 
time = linspace(0.0,20.0,100)

x = odeint(model,xinit,time,p)

plt.plot(time, x[:,0], time, x[:,1])

1 个答案:

答案 0 :(得分:0)

感谢Thiru的评论,这是解决方案:

import numpy as np
from scipy.integrate import odeint
import matplotlib.pyplot as plt

def model (x,t,*p):
    dot_x = np.zeros(2)
    v0 = p[0]
    k1 = p[1]
    k2 = p[2]
    dot_x[0] = v0 - k1*x[0]
    dot_x[1] = k1*x[0] - k2*x[1]
    return dot_x

p = (10,0.5,.35)
xinit = [0.0,0.0] 
time = linspace(0.0,20.0,100)

x = odeint(model,xinit,time,p)

plt.plot(time, x[:,0], time, x[:,1])