迭代三个变量

Question

好的，这会很长。我正在尝试计算极地网格中解析解中的水位。它取决于 r 和 theta 以及这个变量 j。我要做的基本上是根据给定的方程计算网格中特定 r、theta 点的水位。这个方程有一部分对 j 的无限值求和。等式的一部分在这里Equation image

部分代码如下：

"""
Plot idealized solution for water levels in a quarter circle domain with 
constant bathymetry  
"""
import numpy as np
import matplotlib.pyplot as plt
#from pylab import meshgrid,cm,imshow,contour,clabel,colorbar,axis,title,show

# establish parameters
Ho = 300 #m
g = 9.81 #m/s2
r1 = 1000 #m
r2 = 10000 #m
rr = np.arange(r1,r2,10)

#radial size of domain
phi = np.pi/2
#theta is the angle in radians at a specific location within the domain
#theta = np.pi/4

theta = np.arange(0, phi, np.pi/360)#varies
Theta = theta[1:180]

zeta = [0] * len(rr) * len(Theta)
#converting from wind speed to wind shear stress
U = 10
Cd = (1/1000) * ((3/4) + (U/15))
Roair = 1.225 #kg/m3
Rowater = 997 #kg/m3

W = (Roair/Rowater) * Cd * (U**2)

#wind shear in m^2/s^2 in the 0 direction (W to E)
Wo = np.sqrt((W**2)/2)
#wind shear in m^2/s^2 in the phi direction 
Wphi = np.sqrt((W**2)/2)

zeta = np.zeros((len(rr), len(Theta)))
#determines the bathymetry
a_star = []
n = 0
kappa = (1-n)**(0.5)
for t in range(len(Theta)):
    a_star.append ( ( (np.sin(phi)) / (g*Ho*kappa* np.sin(kappa*Theta[t])) ) )

#first half of equation 19 that does not depend on j 
for r in range(len(rr)):
    for t in range(len(Theta)):
        zeta[r,t] = ( (a_star[t] * (rr[r]**(1-n)))*(Wo*np.cos(((1-n)**(0.5))*Theta[t]) + Wphi*np.cos(((1-n)**(0.5))*(Theta[t]-phi))) )  

#second half of equation 19 for j=0    
ajbj = []

for t in range(len(Theta)):
    j = 0
    Djo = np.sin(( ( (1-n)**(0.5) ) * phi ) ) / ( (1-n)**(0.5) * (phi) ) 
    Ejo = (np.sin(phi)) / (phi) 
    ajbj.append (  (r2**(1-n)) * (-a_star[t] * Djo))
    for r in range(len(rr)):
        zeta[r,t] = zeta[r,t] + (ajbj[t])*(Wo+Wphi) 

#second half of equation 19 for j=1,2,3 (summation) 

sj = []
tj = []
Dj = []
Ej = []
r1EogH = []
astarD = []
tjr1r2 = []
sjr2 = []
aj = []
bj = []


jj = [1,2,3]
for j in range(len(jj)):
    sj.append(- (n/2) + np.sqrt( ( (n/2)**2) + ( (jj[j]*np.pi / phi)**2) ) )
    tj.append (- (n/2) - np.sqrt( ( (n/2)**2) + ( (jj[j]*np.pi / phi)**2) ) )    
    Dj.append ( (2* ((-1)**jj[j]) * ((1-n)**(0.5)) * phi * np.sin( ((1-n)**(0.5)) * phi )) / ( (1-n) * (phi**2) - (jj[j]**2) * (np.pi**2) ) )
    Ej.append ( (2* ((-1)**jj[j]) * phi * np.sin(phi) ) / ( (phi**2) - (jj[j]**2) * (np.pi**2) ) )
    r1EogH.append ( ( (r1**(1-n)) * Ej[j] ) / ( g * Ho ) )        
    tjr1r2.append ( tj[j] * (r1**tj[j]) * (r2**sj[j]) )
    sjr2.append ( sj[j] * (r2**tj[j]) )
    for t in range(len(Theta)):
        #astarD.append ( a_star[t] * Dj[j] )
        aj.append (   ( a_star[t]*Dj[j] * (  ( tj[j] * (r1**tj[j]) * (r2**(1-n)) ) - ( (r2**tj[j]) * (r1**(1-n)) ) ) +  ( r1EogH[j] * r2**tj[j] ) ) / ( (sjr2[j] * (r1**sj[j])) - tjr1r2[j] ) )
        bj.append ( ( -1* a_star[t]*Dj[j] * (  ( sj[j] * (r1**sj[j]) * (r2**(1-n)) ) - ( (r2**sj[j]) * (r1**(1-n)) ) ) -  ( r1EogH[j] * r2**sj[j] ) ) / ( (sjr2[j] * (r2**sj[j])) - tjr1r2[j] ) ) 
        for r in range(len(rr)): 
            zeta[r,t] = zeta[r,t] + ( ( (aj[j] * rr[r]**(sj[j])) + (bj[j] * rr[r]**(tj[j])) ) * (Wo*np.cos( (jj[j]*np.pi*Theta[t])/phi)  +  Wphi*np.cos( (jj[j]*np.pi*(Theta[t]-phi))/phi)) )


x,y = np.meshgrid(Theta, rr)

X = Theta
Y = rr

fig = plt.figure()
ax = fig.add_subplot(111, polar='True')
ax.pcolormesh(X, Y, zeta) #X,Y & data2D must all be same dimensions

ax.set_thetamin(0)
ax.set_thetamax(90)

plot = ax.pcolor(zeta)
fig.colorbar(plot)

plt.show()

Solution for above 我得到的 zeta 值非常大。它们应该更类似于以下 theta PI/4 解决方案中的值。我知道方程式很复杂。我想要的是网格中的一个点说 r=1000, theta= pi/4 计算 j=1, j=2 和 j=3 的 zeta 值，然后将它们加在一起，然后做同样的事情网格中每个点的东西。我想知道我是否只需要以不同的方式构造我的循环？还是不使用 .append 功能？有人有什么建议吗？

为 THETA PI/4 完成

"""
Plot idealized solution for water levels in a quarter circle domain with 
constant bathymetry  
"""
import numpy as np
import matplotlib.pyplot as plt
#from pylab import meshgrid,cm,imshow,contour,clabel,colorbar,axis,title,show

# establish parameters
Ho = 300 #m
g = 9.81 #m/s2
r1 = 1000 #m
r2 = 10000 #m
rr = np.arange(r1,r2,10)
zeta = [0] * len(rr)
#radial size of domain
phi = np.pi/2
#theta is the angle in radians at a specific location within the domain
theta = np.pi/4

#converting from wind speed to wind shear stress
U = 10
Cd = (1/1000) * ((3/4) + (U/15))
Roair = 1.225 #kg/m3
Rowater = 997 #kg/m3

W = (Roair/Rowater) * Cd * (U**2)

#wind shear in m^2/s^2 in the 0 direction (W to E)
Wo = np.sqrt((W**2)/2)
#wind shear in m^2/s^2 in the phi direction 
Wphi = np.sqrt((W**2)/2)

#determines the bathymetry
n = 0
kappa = (1-n)**(0.5)
a_star = ( (np.sin(phi)) / (g*Ho*kappa* np.sin(kappa*theta)) )

#first half of equation 19 that does not depend on j 
for r in range(len(rr)):
    zeta[r] = ( a_star * (rr[r]**(1-n))*(Wo*np.cos(((1-n)**(0.5))*theta) + Wphi*np.cos(((1-n)**(0.5))*(theta-phi))) ) 

#plt.xlabel("rr")
#plt.ylabel("zeta")
###
#plt.plot(rr,zeta, label = 'LHS eq 19') 

#second half of equation 19 for j=0    
for r in range(len(rr)):
    j = 0
    Djo = np.sin(( ( (1-n)**(0.5) ) * phi ) ) / ( (1-n)**(0.5) * (phi) ) 
    Ejo = (np.sin(phi)) / (phi) 
    ajbj =  (r2**(1-n)) * (-a_star * Djo) 
    zeta[r] = zeta[r] + (ajbj)*(Wo+Wphi) 

#plt.xlabel("rr")
#plt.ylabel("zeta")
###
#plt.plot(rr,zeta, label = 'j=0') 

#second half of equation 19 for j=1,2,3 (summation) 

sj = []
tj = []
Dj = []
Ej = []
r1EogH = []
astarD = []
tjr1r2 = []
sjr2 = []
aj = []
bj = []

jj = [1,2,3]
for j in range(len(jj)):
    sj.append(- (n/2) + np.sqrt( ( (n/2)**2) + ( (jj[j]*np.pi / phi)**2) ) )
    tj.append (- (n/2) - np.sqrt( ( (n/2)**2) + ( (jj[j]*np.pi / phi)**2) ) )    
    Dj.append ( (2* ((-1)**jj[j]) * ((1-n)**(0.5)) * phi * np.sin( ((1-n)**(0.5)) * phi )) / ( (1-n) * (phi**2) - (jj[j]**2) * (np.pi**2) ) )
    Ej.append ( (2* ((-1)**jj[j]) * phi * np.sin(phi) ) / ( (phi**2) - (jj[j]**2) * (np.pi**2) ) )
    r1EogH.append ( ( (r1**(1-n)) * Ej[j] ) / ( g * Ho ) )
    astarD.append ( a_star * Dj[j] )
    tjr1r2.append ( tj[j] * (r1**tj[j]) * (r2**sj[j]) )
    sjr2.append ( sj[j] * (r2**tj[j]) )
    aj.append (   ( astarD[j] * (  ( tj[j] * (r1**tj[j]) * (r2**(1-n)) ) - ( (r2**tj[j]) * (r1**(1-n)) ) ) +  ( r1EogH[j] * r2**tj[j] ) ) / ( sjr2[j] * (r1**sj[j]) - tjr1r2[j] ) )
    bj.append (- ( astarD[j] * (  ( sj[j] * (r1**sj[j]) * (r2**(1-n)) ) - ( (r2**sj[j]) * (r1**(1-n)) ) ) -  ( r1EogH[j] * r2**sj[j] ) ) / ( sjr2[j] * (r2**sj[j]) - tjr1r2[j] ) ) 
    for r in range(len(rr)):
        zeta[r] = zeta[r] + ( ( (aj[j] * rr[r]**(sj[j])) + (bj[j] * rr[r]**(tj[j])) ) * (Wo*np.cos( (jj[j]*np.pi*theta)/phi)  +  Wphi*np.cos( (jj[j]*np.pi*(theta-phi))/phi)) )

plt.xlabel("rr")
plt.ylabel("zeta")
plt.title("Wind In at 45 degrees")
#
plt.plot(rr,zeta, label = 'Ho=100m') 

plt.legend(loc='upper right')

Solution for theta 45degrees

Answer 1

很遗憾，我不是数学天才。在我尝试以某种方式重构它之前，我尝试运行您的代码以查看它得到了什么输出。但是，存在错误，因为在提供的 sn-p 中未定义来自 sj.append() 的 sj 之类的内容。我不知道如何阅读您图片中提供的方程式（数学课程是多年前的，自大学以来就没有使用过）。

至于建议，在没有完全理解你的问题的情况下，我能给你的最好的就是考虑制作函数。这使您/其他人可以更清楚地看到代码。另一个优点是当某些东西发生变化时（似乎不太可能用方程，但也许你想在未来的某个时候改变你的输入类型），这些变化会以更小的部分进行。

def calculate_j(point_or_other_necessary_input):
  """ do the maths for j """
  return some_equation_or_value_for_j

def calculate_t(point_or_other_necessary_input):
  """ do the maths for t """
  return some_equation_or_value_for_t

def calculate_r(point_or_other_necessary_input):
  """ do the maths for r """
  return some_equation_or_value_for_r

def process_grid(zeta_grid):
  """ process each point in the provided grid"""
  for point in zeta_grid:
    # or whatever makes sense to combine the values/equations
    calculated_sum = calculate_j(point) + calculate_t(point) + calculate_r(point)
    # store value in some reasonable way, probably another grid
  return calculated_points

print(process_grid(zeta))

经过进一步评估，并从您的编辑中获得更多信息，我不得不问您是否真的打算这样做：

jj = [1,2,3]
for j in range(len(jj)):

我猜测，但不确定，您希望 j 是 1，然后是 2，然后是 3，不是吗？目前你真的得到了0、1、2，因为你正在迭代列表的长度，而不是列表中的项目。这在您使用 jj[j] 获取 jj 值的第一部分并不重要，但在您使用 j 而不是 jj[j] 的 r 部分中，例如：

zeta[r] = zeta[r] + ( ( (aj[j] * rr[r]**(sj[j])) + (bj[j] * rr[r]**(tj[j])) )

如前所述，您的预期行为可能是只使用 0,1,2 而不是 1,2,3。 至于为什么这些值可能太大，您是否打算计算r 代表所有 rr 3 次？您将它嵌套在第一个 for 循环中，这意味着它在 for j in ... 循环的每次迭代中执行。因此它会多次添加到自身 (zeta[r] = zeta[r] + ...)。 （再一次，不是数学家）但不是运行 len(rr)^3 次或 3^len(rr) 次吗？比可能预期的多一些倍。

Answer 2

建议编写一个函数 zeta (r, t, j)，您可以使用一些手动计算的值进行测试：

def zeta(r, theta, j):
   #your complex calculations here

assert zeta(4000, pi/4, 2) == 0.00043 #This is a test to be sure your formulae 
                                      # are right (eyeballed from your graph)

然后遍历 rr 和 theta 来填充数组：

zeta_out = np.zeros((len(rr), len(Theta)))

for r in range(len(rr)):
    for t in range(len(theta)):
        zeta_out[r, t] = np.mean(zeta(rr[r], theta[t], j) for j in (1,2,3))