Streamplot 2D domain in python

[mat_2]

Hello. I got one question, when i want to streamplot this, i got eror: "raise ValueError("The columns of 'y' must be equal")", i can't figure it out. :/
Thank you for your time.

this is entire code:

import matplotlib.pyplot as plt
import numpy as np

# discretization of domain
x_start, x_end = 0. , 10.
y_start , y_end = 0. , 8.
i = np.linspace(x_start, x_end, 80)
j = np.linspace(y_start, y_end, 50)
X,Y = np.meshgrid(i,j)
Y[:,:40] += 1

Dx = x_end/(np.size(i)-1)
Dy = y_end/(np.size(j)-1)
beta = Dx/Dy
nx = np.size(i) #number of points on x axis
ny = np.size(j) #number of points on y axis

#matrix
nk = nx*ny
A = np.zeros([nk, nk])
B = np.zeros([nk,1])                    

# solving
for i in range (nx):
    for j in range (ny):
        k = j*nx + i
        if i == 0:                       # left edge
            k1 = j*nx + 1
            k0 = j*nx + 0
            A[k,k1] = 1
            A[k,k0] = -1
            B[k] = Dx
        elif i == nx-1:                  # right edge
            k_1 = j*nx + (nx-1)
            k_2 = j*nx + (nx-2)
            A[k,k_2] = -1
            A[k,k_1] = 1
            B[k] = Dx
        elif j == 0:                     # bottom
            K1 = nx + i
            K0 = i
            A[k,K0] = -1
            A[k,K1] = 1
            B[k] = 0
        elif j == ny-1:                  # topper
            K_1 = (ny-1)*nx + i
            K0 = (ny-2)*nx + i
            A[k,K_1] = 1
            A[k,K0] = -1
            B[k] = 0        
        else:                           # middle
            k_top = (j+1)*nx + i        # fi i,j+1
            k_down = (j-1)*nx + i       # fi i,j-1
            k_left = j*nx + (i-1)       # fi i-1,j
            k_right = j*nx + (i+1)      # fi i+1,j
            A[k,k_top] = beta**2
            A[k,k_down] = beta**2
            A[k,k_left] = 1
            A[k,k_right] = 1
            A[k,k] = -2*(1+beta**2)
            B[k] = 0

sol_vec = np.linalg.solve(A,B)
fi = np.reshape(sol_vec,(ny,nx))

# u and v
u = np.zeros([ny,nx])
v = np.zeros([ny,nx])

# velocity u i v
for i in range(nx):
    for j in range (ny):
        if i == 0:                                 # left edge               
            v[j,i] = 0
            u[j,i] = (fi[j,i+1] - fi[j,i])/Dx
        elif i == nx-1:                             # right edge
            v[j,i] = 0
            u[j,i] = (fi[j,i] - fi[j,i-1])/Dx
        elif j == 0:                                # bottom, 'v' 
            v[j,i] = 0
        elif j == ny-1:                             # topper, 'v' 
            v[j,i] = 0
        else:                                       # middle               
            u[j,i] = (fi[j,i+1] - fi[j,i-1])/(2*Dx)
            v[j,i] = (fi[j+1,i] - fi[j-1,i])/(2*Dy)

for i in range (1,nx-1):                                    
    u[0,i] = (fi[j,i+1] - fi[j,i-1])/(2*Dx)         # 'u'         
    u[-1,i] = (fi[j,i+1] - fi[j,i-1])/(2*Dx)        

u[15,40] = 0
v[15,40] = 0

plt.figure()                                       
plt.quiver(X,Y,u,v)

plt.streamplot(X,Y,u,v,linewidth=1.5,arrowsize=1)

plt.show()

[rproffitt]

The last time I used a streaming plotter, yes, you can only feed it values along the y that match since the chart scrolls along and y must be same or the next value and then match until it's time for the chart to scroll. Nothing seems odd about that.