如何獲取代碼以生成一組值而不是一個值？

Question

嗨，我編寫了以下代碼來生成菲涅耳衍射圖：

import math
import cmath
import numpy as np
import matplotlib.pyplot as plt

lamda=0.00000005
k=(2*math.pi)/lamda
z=0.03
h=6.67e-34
c=3e8
e0=8.85e-12
E0=h*c/lamda
xp1=-1e-6
xp2=1e-6
t=1
N=100
y=0 
yp=1
yp1=-1e-6
yp2=1e-6
xp=0



def expfuncX(x,xp): #gives the x function to be integrated

    j=cmath.sqrt(-1)
    g=(k/2*z)*((x-xp)**2)    

    return cmath.cos(g)+cmath.sin(g)*j


def X(xp1,xp2,x,xp,f,N): #integrates the x function

    h=(xp2-xp1)/N
    ff=0
    xp=xp1
    for i in np.arange(1, N/2 +1): #summing odd order func terms

        ff+=4*f(x,xp)
        xp+=2*h

    xp=xp1+2*h
    for i in np.arange(2,N/2): #summing even order func terms

        ff+=2*f(x,xp)
        xp+=2*h

    integral= (h/3)*(ff+f(x, xp1)+f(x, xp2))    

    return integral




def expfuncXY(y,yp):  #gives the 2d func to be integrated

    j=cmath.sqrt(-1)
    g=(k/2*z)*((y-yp)**2)    

    return X(xp1,xp2,x,xp,expfuncX,N)*cmath.cos(g)+cmath.sin(g)*j    


def simpsonXY(yp1,yp2,y,yp,f,N): #integrates 2d function

    h=(yp2-yp1)/N
    ff=0
    yp=yp1
    for i in np.arange(1, N/2 +1): #summing odd order func terms

        ff+=4*f(y,yp)
        yp+=2*h

    yp=yp1+2*h
    for i in np.arange(2,N/2): #summing even order func terms

        ff+=2*f(y,yp)
        yp+=2*h

    integral= ((E0*k)/(2*(math.pi)*z))*(h/3)*(ff+f(y, yp1)+f(y, yp2))    

    return integral

print(simpsonXY(-1e-6,1e-6,1,0,expfuncXY,100))




NumPoints = 200
delta = 4.0*np.pi / (NumPoints - 1)
intensity = np.zeros( (NumPoints,NumPoints) )
for i in range(NumPoints):
    x = i * delta
for j in range(NumPoints):
    y = j * delta
intensity[i,j] =e0*c*((abs(simpsonXY(-1e-6,1e-6,1,0,expfuncXY,100))**2))
plt.imshow(intensity)
plt.show()

print(intensity)

但是代碼產生了這個！

我如何獲得它，以便它打印出數組中其余的值，而不只是一個？ 我想您需要定義一個xvalue范圍並將它們存儲到函數中的x中，但是我不太確定如何去做。

謝謝

Answer 1

您的縮進有點古怪。 你有：

intensity = np.zeros( (NumPoints,NumPoints) )
for i in range(NumPoints):
    x = i * delta
for j in range(NumPoints):
    y = j * delta
intensity[i,j] =e0*c*((abs(simpsonXY(-1e-6,1e-6,1,0,expfuncXY,100))**2))

當我認為您真正想要的是：

intensity = np.zeros( (NumPoints,NumPoints) )
for i in range(NumPoints):
    x = i * delta
    for j in range(NumPoints):
        y = j * delta
        intensity[i,j] =e0*c*((abs(simpsonXY(-1e-6,1e-6,1,0,expfuncXY,100))**2))

否則（在代碼的第一段），您僅設置一個強度，即強度[NumPoints-1，NumPoints-1]

Answer 2

使用meshgrid ，這只是numpy.meshgrid中的示例

import numpy as np
from matplotlib import pyplot as plt


x = np.arange(-5, 5, 0.1)
y = np.arange(-5, 5, 0.1)
xx, yy = np.meshgrid(x, y, sparse=True)
z = np.sin(xx**2 + yy**2) / (xx**2 + yy**2)
h = plt.contourf(x,y,z)