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从odeint scipy python使用的函数中提取值

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我有以下脚本来使用odeint计算dRho .

P_r = 10e5
rho_r = 900
L = 750
H = 10
W = 150
A = H * W
V = A * L
fi = 0.17

k = 1.2e-13
c = 12.8e-9
mu = 2e-3

N = 50
dV = V/N
dx = L/N

P_in = P_r
rho_in = rho_r

P_w = 1e5    
rho_w = rho_r* np.exp(c*(P_w-P_r))

# init initial case
P = np.empty(N+1)*10e5
Q = np.ones(N+1)
out = np.empty(N+1)

P[0] = P_w
Q[0] = 0
out[0] = 0

def dRho(rho_y, t, N):

    P[1:N] = P_r + (1/c) * np.log(rho_y[1:N]/rho_r)
    P[N] = P_r + (1/c) * np.log(rho_y[N]/rho_r)


    Q[1:N] = (-A*k/mu)*((P[1-1:N-1] - P[1:N])/dx)
    Q[N] = (-A*k/mu)*((P[N]-P_r)/dx)


    out[1:N] = ((Q[1+1:N+1]*rho_y[1+1:N+1] - Q[1:N]*rho_y[1:N])/dV) 
    out[N] = 0

    return out

t0 = np.linspace(0,1e9, int(1e9/200))
rho0 = np.ones(N+1)*900
ti = time.time()
solve = odeint(dRho, rho0, t0, args=(N,))
plt.plot(t0,solve[:,1:len(rho0)], '-', label='dRho')
plt.legend(loc='upper right')
plt.show()

P和Q在函数dRho内计算,它们P作用并输入Q,P,Q和rho_y都作为out的输入 . 该函数返回“out” . 我可以毫无问题地进行策划,但是,我也有兴趣绘制P和Q.

我已经尝试了各种方法来实现这一点:在集成方法之后重新计算P和Q但这增加了脚本的运行时间 . 所以,由于计算是在dRho内完成的,我想知道是否以及如何从外部访问它来绘制它 .

我还尝试将p和Q与rho0一起作为odeint的输入,但是P和Q都在集成中被采用,导致函数返回时导致错误的结果 .

简化版:

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint
def dY(y, x):
    a = 0.001
    yin = 1
    C = 0.01
    N = 1
    dC = C/N
    b1 = 0
    y_diff = -np.copy(y)
    y_diff[0] += yin
    y_diff[1:] += y[:-1]
    print(y)
    return (a/dC)*y_diff+b1*dC

x = np.linspace(0,20,1000)
y0 = np.zeros(4)
res = odeint(dY, y0, x)
print(res)
plt.plot(x,res, '-')
plt.show()

在这个简化的例子中,我想创建一个额外的ydiff图 .

这是另一个简单的案例

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

def func(z,t):
    x, y=z
    xnew = x*2
    print(xnew)
    ynew = y*0.5
#     print y
    return [x, y]    

z0=[1,3]
t = np.linspace(0,10)
xx=odeint(func, z0, t)
plt.plot(t, xx[:,0],t,xx[:,1])
plt.show()

我有兴趣绘制所有xnew和ynew值 .

另一个例子:

xarr = np.ones(4)
def dY(y, x):
    a = 0.001
    yin = 1
    C = 0.01
    N = 1
    dC = C/N
    b1 = 0
    xarr[0] = 0.25
    xarr[1:] = 2 
    mult = xarr*2
    out = mult * y
    print(mult)
    return out

x = np.linspace(0,20,1000)
y0 = np.zeros(4)+1.25
res = odeint(dY, y0, x)
dif = np.array([dY(y,x) for y in res])
print(dif)
plt.plot(x,res, '-')
plt.show()

我想将多值映射到x

1 回答

  • 1

    以下可能是你想要的 . 您可以将中间值存储在列表中,然后绘制该列表 . 这也需要存储x值 .

    import numpy as np
    import matplotlib.pyplot as plt
    from scipy.integrate import odeint
    
    xs = []
    yd = []
    
    def dY(y, x):
        a = 0.001
        yin = 1
        C = 0.01
        N = 1
        dC = C/N
        b1 = 0
        y_diff = -np.copy(y)
        y_diff[0] += yin
        y_diff[1:] += y[:-1]
        xs.append(x)
        yd.append(y_diff)
        return (a/dC)*y_diff+b1*dC
    
    x = np.linspace(0,20,1000)
    y0 = np.zeros(4)
    res = odeint(dY, y0, x)
    
    plt.plot(x,res, '-')
    
    plt.gca().set_prop_cycle(plt.rcParams['axes.prop_cycle'])
    plt.plot(np.array(xs),np.array(yd), '-.')
    
    plt.show()
    

    enter image description here

    虚线是相同颜色的 res 解的相应 y_diff 值 .

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