从给定的二维方差创建二维高斯随机场

Question

我一直在尝试使用我计算的方差创建一个物质斑点（高斯随机场）的2D地图。该方差是一个二维数组。我尝试过使用numpy.random.normal，因为它允许方差的2D输入，但它并没有真正创建一个带有我期望的输入参数趋势的映射。一个重要的输入常数lambda_c应显示为水滴的物理尺寸（直径）。但是，当我更改lambda_c时，水滴的大小根本不会改变。例如，如果我设置lambda_c=40 parsecs，则贴图需要直径为40 parsecs的blob。使用我的方差生成地图的MWE：

import numpy as np
import random
import matplotlib.pyplot as plt
from matplotlib.pyplot import show, plot
import scipy.integrate as integrate
from scipy.interpolate import RectBivariateSpline



n = 300
c = 3e8
G = 6.67e-11
M_sun = 1.989e30
pc = 3.086e16  # parsec
Dds = 1097.07889283e6*pc 
Ds = 1726.62069147e6*pc
Dd = 1259e6*pc

FOV_arcsec_original = 5.
FOV_arcmin = FOV_arcsec_original/60.   


pix2rad = ((FOV_arcmin/60.)/float(n))*np.pi/180.
rad2pix = 1./pix2rad

x_pix = np.linspace(-FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,n)
y_pix = np.linspace(-FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,FOV_arcsec_original/2/pix2rad/180.*np.pi/3600.,n)
X_pix,Y_pix = np.meshgrid(x_pix,y_pix)

conc = 10.
M = 1e13*M_sun
r_s = 18*1e3*pc

lambda_c = 40*pc  ### The important parameter that doesn't seem to manifest itself in the map when changed

rho_s = M/((4*np.pi*r_s**3)*(np.log(1+conc) - (conc/(1+conc)))) 
sigma_crit = (c**2*Ds)/(4*np.pi*G*Dd*Dds)
k_s = rho_s*r_s/sigma_crit
theta_s = r_s/Dd
Renorm = (4*G/c**2)*(Dds/(Dd*Ds))
#### Here I just interpolate and zoom into my field of view to get better resolutions
A = np.sqrt(X_pix**2 + Y_pix**2)*pix2rad/theta_s



A_1 = A[100:200,0:100]

n_x = n_y = 100

FOV_arcsec_x = FOV_arcsec_original*(100./300)
FOV_arcmin_x = FOV_arcsec_x/60.   
pix2rad_x = ((FOV_arcmin_x/60.)/float(n_x))*np.pi/180.
rad2pix_x = 1./pix2rad_x

FOV_arcsec_y = FOV_arcsec_original*(100./300)
FOV_arcmin_y = FOV_arcsec_y/60.   
pix2rad_y = ((FOV_arcmin_y/60.)/float(n_y))*np.pi/180.
rad2pix_y = 1./pix2rad_y

x1 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x)
y1 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y)
X1,Y1 = np.meshgrid(x1,y1)



n_x_2 = 500
n_y_2 = 500


x2 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x_2)
y2 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y_2)
X2,Y2 = np.meshgrid(x2,y2)

interp_spline = RectBivariateSpline(y1,x1,A_1)

A_2 = interp_spline(y2,x2)



A_3 = A_2[50:450,0:400]

n_x_3 = n_y_3 = 400

FOV_arcsec_x = FOV_arcsec_original*(100./300)*400./500.
FOV_arcmin_x = FOV_arcsec_x/60.   
pix2rad_x = ((FOV_arcmin_x/60.)/float(n_x_3))*np.pi/180.
rad2pix_x = 1./pix2rad_x

FOV_arcsec_y = FOV_arcsec_original*(100./300)*400./500.
FOV_arcmin_y = FOV_arcsec_y/60.   
pix2rad_y = ((FOV_arcmin_y/60.)/float(n_y_3))*np.pi/180.
rad2pix_y = 1./pix2rad_y

x3 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x_3)
y3 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y_3)
X3,Y3 = np.meshgrid(x3,y3)

n_x_4 = 1000
n_y_4 = 1000


x4 = np.linspace(-FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,FOV_arcsec_x/2/pix2rad_x/180.*np.pi/3600.,n_x_4)
y4 = np.linspace(-FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,FOV_arcsec_y/2/pix2rad_y/180.*np.pi/3600.,n_y_4)
X4,Y4 = np.meshgrid(x4,y4)

interp_spline = RectBivariateSpline(y3,x3,A_3)

A_4 = interp_spline(y4,x4)

############### Function to calculate variance

variance = np.zeros((len(A_4),len(A_4)))



def variance_fluctuations(x):
    for i in xrange(len(x)):
        for j in xrange(len(x)):
            if x[j][i] < 1.:
                variance[j][i] = (k_s**2)*(lambda_c/r_s)*((np.pi/x[j][i]) - (1./(x[j][i]**2 -1)**3.)*(((6.*x[j][i]**4. - 17.*x[j][i]**2. + 26)/3.)+ (((2.*x[j][i]**6. - 7.*x[j][i]**4. + 8.*x[j][i]**2. - 8)*np.arccosh(1./x[j][i]))/(np.sqrt(1-x[j][i]**2.)))))
            elif x[j][i] > 1.:
                variance[j][i] = (k_s**2)*(lambda_c/r_s)*((np.pi/x[j][i]) - (1./(x[j][i]**2 -1)**3.)*(((6.*x[j][i]**4. - 17.*x[j][i]**2. + 26)/3.)+ (((2.*x[j][i]**6. - 7.*x[j][i]**4. + 8.*x[j][i]**2. - 8)*np.arccos(1./x[j][i]))/(np.sqrt(x[j][i]**2.-1)))))



variance_fluctuations(A_4)

#### Creating the map 

mean = 0

delta_kappa = np.random.normal(0,variance,A_4.shape)  

xfinal = np.linspace(-FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,1000)
yfinal = np.linspace(-FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,FOV_arcsec_x*np.pi/180./3600.*Dd/pc/2,1000)

Xfinal, Yfinal = np.meshgrid(xfinal,yfinal)
plt.contourf(Xfinal,Yfinal,delta_kappa,100)
plt.show()

地图看起来是这样的，斑点的密度向右增加。但是，无论我使用lambda_c=40*pc还是lambda_c=400*pc，水滴的大小都不会改变，贴图看起来几乎相同

我想知道np.random.normal函数是否真的没有达到我期望的效果？我觉得地图的像素比例和样本的绘制方式与斑点的大小没有任何联系。也许有一个更好的方法来创建地图使用方差，如果有任何见解将不胜感激

我希望映射看起来像这样，blob大小根据我的方差的输入参数而变化：

Answer 1

ThunderFlash，请尝试以下代码绘制地图：

# function to produce blobs:
from scipy.stats import multivariate_normal

def blob (positions, mean=(0,0),  var=1):
    cov = [[var,0],[0,var]]
    return multivariate_normal(mean, cov).pdf(positions)
    
"""
now prepare for blobs generation.
note that I use less dense grid to pick blobs centers (regulated by `step`)
this makes blobs more pronounced and saves calculation time.

use this part instead of your code section below comment #### Creating the map 
"""
delta_kappa = np.random.normal(0,variance,A_4.shape) # same
    
step = 10 # 
dk2 = delta_kappa[::step,::step] # taking every 10th element
x2, y2 = xfinal[::step],yfinal[::step]
field = np.dstack((Xfinal,Yfinal)) 
print (field.shape, dk2.shape, x2.shape, y2.shape)

>> (1000, 1000, 2), (100, 100), (100,), (100,)

result = np.zeros(field.shape[:2]) 

for x in range (len(x2)):
    for y in range (len(y2)):
        res2 = blob(field, mean = (x2[x], y2[y]), var=10000)*dk2[x,y]
        result += res2

# the cycle above took over 20 minutes on Ryzen 2700X. It could be accelerated by vectorization presumably.

plt.contourf(Xfinal,Yfinal,result,100)
plt.show()

您可能希望使用blob（）中的var参数平滑图像，并使用step使其更压缩。 这是我用你的代码得到的图像（不知何故轴被翻转，顶部有更密集的区域）：

Answer 2

这是天文学和宇宙学中一个备受关注的问题

您可以使用lenstool:https://lenstools.readthedocs.io/en/latest/examples/gaussian_random_field.html

您也可以在这里尝试：

https://andrewwalker.github.io/statefultransitions/post/gaussian-fields

更不用说：

https://github.com/bsciolla/gaussian-random-fields

我不是在这里复制代码，因为所有的功劳都归于上述作者。然而，他们都是通过谷歌搜索出来的：/

最简单的可能是python模块FyeldGenerator，它显然就是为这个目的而设计的：

https://github.com/cphyc/FyeldGenerator

因此（改编自github示例）：

pip install FyeldGenerator

from FyeldGenerator import generate_field

from matplotlib import use
use('Agg')

import matplotlib.pyplot as plt
import numpy as np



plt.figure()

# Helper that generates power-law power spectrum
def Pkgen(n):
    def Pk(k):
        return np.power(k, -n)

    return Pk

# Draw samples from a normal distribution
def distrib(shape):
    a = np.random.normal(loc=0, scale=1, size=shape)
    b = np.random.normal(loc=0, scale=1, size=shape)
    return a + 1j * b


shape = (512, 512)

field = generate_field(distrib, Pkgen(2), shape)

plt.imshow(field, cmap='jet')

plt.savefig('field.png',dpi=400)
plt.close())

这使得：

在我看来很简单：）

PS:FoV意味着望远镜对高斯随机场的观测：）

Answer 3

一种完全不同且更快的方法可能只是用高斯滤波器模糊delta_kappa阵列。尝试调整sigma参数以改变blob大小

from scipy.ndimage.filters import gaussian_filter
dk_gf = gaussian_filter(delta_kappa, sigma=20)
Xfinal, Yfinal = np.meshgrid(xfinal,yfinal)
plt.contourf(Xfinal,Yfinal,dk_ma,100, cmap='jet')
plt.show();

这是带有sigma=20的图像

这是带有sigma=2.5的图像