Python坐标变换下的montecarlo积分
我试图用蒙特卡罗方法计算积分,其中被积函数经历了从柱坐标到笛卡尔坐标的转换。被积函数本身非常简单,可以使用Python坐标变换下的montecarlo积分,python,montecarlo,Python,Montecarlo,我试图用蒙特卡罗方法计算积分,其中被积函数经历了从柱坐标到笛卡尔坐标的转换。被积函数本身非常简单,可以使用scipy.integrate.quad进行计算,但我需要它位于笛卡尔坐标系中,以便以后用于特定用途 这就是被积函数:rho*k_-r**2*(k_-r**2*kn(0,k_-r*rho)**2+k_-n**2*kn(1,k_-r*rho)**2)d rho 这里的kn(i,rho)是第二类贝塞尔函数的修正 使用quad求解它会得到以下结果: from scipy.special impor
scipy.integrate.quad
进行计算,但我需要它位于笛卡尔坐标系中,以便以后用于特定用途
这就是被积函数:rho*k_-r**2*(k_-r**2*kn(0,k_-r*rho)**2+k_-n**2*kn(1,k_-r*rho)**2)d rho
这里的kn(i,rho)
是第二类贝塞尔函数的修正
使用quad
求解它会得到以下结果:
from scipy.special import kn
from scipy.integrate import quad
import random
k_r = 6.2e-2
k_n = k_r/1.05
C_factor = 2*np.pi*1e5
lmax,lmin = 80,50
def integration_polar():
def K_int(rho):
return rho*k_r**2*(k_r**2*kn(0,k_r*rho)**2+k_n**2*kn(1,k_r*rho)**2)
rho = np.linspace(lmin,lmax,200)
I,_ = quad(K_int,lmin,lmax)
Gamma = I*C_factor
print("expected=",Gamma)
输出:Expected=7.641648442007296
现在,使用蒙特卡罗法(点击或未点击法)的相同积分给出了几乎相同的结果:
def integration_polar_MC():
random.seed(1)
n = 100000
def K_int(rho):
return rho*k_r**2*(k_r**2*kn(0,k_r*rho)**2+k_n**2*kn(1,k_r*rho)**2)
def sampler():
x = random.uniform(lmin,lmax)
y = random.uniform(0,c_lim)
return x,y
c_lim = 2*K_int(50) #Upper limit of integrand
sum_I = 0
for i in range(n):
x,y = sampler()
func_Int = K_int(x)
if y>func_Int:
I = 0
elif y<=func_Int:
I = 1
sum_I += I
Gamma = C_factor*(lmax-lmin)*c_lim*sum_I/n
print("MC_integral_polar:",Gamma)
以下是我的尝试:
def integration_cartesian_MCtry():
random.seed(1)
lmin,lmax = -100,100
n = 100000
def K_int(x,y):
rho = np.sqrt(x**2+y**2)
if rho>=50 and rho<=80:
return k_r**2*(k_r**2*kn(0,k_r*rho)**2+k_n**2*kn(1,k_r*rho)**2)
else:
return 0
def sampler():
x = random.uniform(lmin,lmax)
y = random.uniform(lmin,lmax)
z = random.uniform(0,c_lim)
return x,y,z
c_lim = K_int(50,0)
sum_I = 0
for i in range(n):
x,y,z = sampler()
func_Int = K_int(x,y)
if z>func_Int:
I = 0
elif z<=func_Int:
I = 1
sum_I += I
Gamma = C_factor*(lmax-lmin)**2*c_lim*sum_I/n
print("MC_integral_cartesian:",Gamma)
def集成\u笛卡尔\u MCtry():
随机种子(1)
lmin,lmax=-100100
n=100000
定义K_int(x,y):
rho=np.sqrt(x**2+y**2)
如果rho>=50且rhofunc_Int:
I=0
Elifz正如我所说,问题在于雅可比矩阵。如果是polar,则需要对其进行积分
f(ρ)*ρ*dρ*dφ
对dφ进行解析积分(f(ρ)与φ无关),得到2π
在笛卡尔坐标的情况下,没有解析积分,所以它超过dx*dy,没有因子
2π。Python 3.9.1、Windows 10 x64的代码对此进行了说明,它给出了几乎相同的答案
import numpy as np
from scipy.special import kn
k_r = 6.2e-2
k_n = k_r/1.05
C_factor = 2*np.pi*1e5
lmin = 50
lmax = 80
def integration_polar_MC(rng, n):
def K_int(rho):
if rho>=50 and rho<=80:
return rho*k_r**2*(k_r**2*kn(0, k_r*rho)**2 + k_n**2*kn(1, k_r*rho)**2)
return 0.0
def sampler():
x = rng.uniform(lmin, lmax)
y = rng.uniform(0.0, c_lim)
return x,y
c_lim = 2*K_int(50) # Upper limit of integrand
sum_I = 0
for i in range(n):
x,y = sampler()
func_Int = K_int(x)
I = 1
if y>func_Int:
I = 0
sum_I += I
Gamma = C_factor*(lmax-lmin)*c_lim*sum_I/n
return Gamma
def integration_cartesian_MC(rng, n):
def K_int(x,y):
rho = np.hypot(x, y)
if rho>=50 and rho<=80:
return k_r**2*(k_r**2*kn(0,k_r*rho)**2+k_n**2*kn(1,k_r*rho)**2)
return 0.0
def sampler():
x = rng.uniform(lmin,lmax)
y = rng.uniform(lmin,lmax)
z = rng.uniform(0,c_lim)
return x,y,z
lmin,lmax = -100,100
c_lim = K_int(50, 0)
sum_I = 0
for i in range(n):
x,y,z = sampler()
func_Int = K_int(x,y)
I = 1
if z>func_Int:
I = 0
sum_I += I
Gamma = C_factor*(lmax-lmin)**2*c_lim*sum_I/n
return Gamma/(2.0*np.pi) # to compensate for 2π in the constant
rng = np.random.default_rng()
q = integration_polar_MC(rng, 100000)
print("MC_integral_polar:", q)
q = integration_cartesian_MC(rng, 100000)
print("MC_integral_cart:", q)
将numpy导入为np
来自scipy.special进口kn
k_r=6.2e-2
k_n=k_r/1.05
C_系数=2*np.pi*1e5
lmin=50
lmax=80
def集成\u polar\u MC(rng,n):
定义K_int(rho):
如果rho>=50且rhofunc_Int:
I=0
和I+=I
Gamma=C_系数*(lmax lmin)*C_lim*总和I/n
返回伽马
def集成\u笛卡尔\u MC(rng,n):
定义K_int(x,y):
rho=np.hypot(x,y)
如果rho>=50且rhofunc_Int:
I=0
和I+=I
Gamma=C_系数*(lmax-lmin)**2*C_-lim*总和I/n
返回伽马/(2.0*np.pi)#以补偿常数中的2π
rng=np.random.default\u rng()
q=集成度、极性、最大连续性(rng,100000)
打印(“MC_积分_极坐标:”,q)
q=积分笛卡尔矩阵(rng,100000)
打印(“MC_积分车:”,q)
Jacobian?这将导致不同的采样密度谢谢!显然,我已经在常数中加入了2*pi,并且完全忘记了它来自于圆上的dñ积分。
import numpy as np
from scipy.special import kn
k_r = 6.2e-2
k_n = k_r/1.05
C_factor = 2*np.pi*1e5
lmin = 50
lmax = 80
def integration_polar_MC(rng, n):
def K_int(rho):
if rho>=50 and rho<=80:
return rho*k_r**2*(k_r**2*kn(0, k_r*rho)**2 + k_n**2*kn(1, k_r*rho)**2)
return 0.0
def sampler():
x = rng.uniform(lmin, lmax)
y = rng.uniform(0.0, c_lim)
return x,y
c_lim = 2*K_int(50) # Upper limit of integrand
sum_I = 0
for i in range(n):
x,y = sampler()
func_Int = K_int(x)
I = 1
if y>func_Int:
I = 0
sum_I += I
Gamma = C_factor*(lmax-lmin)*c_lim*sum_I/n
return Gamma
def integration_cartesian_MC(rng, n):
def K_int(x,y):
rho = np.hypot(x, y)
if rho>=50 and rho<=80:
return k_r**2*(k_r**2*kn(0,k_r*rho)**2+k_n**2*kn(1,k_r*rho)**2)
return 0.0
def sampler():
x = rng.uniform(lmin,lmax)
y = rng.uniform(lmin,lmax)
z = rng.uniform(0,c_lim)
return x,y,z
lmin,lmax = -100,100
c_lim = K_int(50, 0)
sum_I = 0
for i in range(n):
x,y,z = sampler()
func_Int = K_int(x,y)
I = 1
if z>func_Int:
I = 0
sum_I += I
Gamma = C_factor*(lmax-lmin)**2*c_lim*sum_I/n
return Gamma/(2.0*np.pi) # to compensate for 2π in the constant
rng = np.random.default_rng()
q = integration_polar_MC(rng, 100000)
print("MC_integral_polar:", q)
q = integration_cartesian_MC(rng, 100000)
print("MC_integral_cart:", q)