FiPy 使用 python2 或 python3 在同一系统上评估不同的解决方案
FiPy evaluates different solutions on the same system using python2 or python3
我正在用偏微分方程组试验 FiPy，我得到了不同的结果 运行 Python2 或 Python3[=25 中的相同脚本=].更准确地说，我在 Python 2.7 中得到了我期望 运行 脚本的结果，而在 Python 3.8.
中得到了完全错误的结果
有人知道原因吗？我错过了什么吗？
我知道 FiPy 在 Python 2.7 中使用的一些求解器不支持 Python 3.x，但这足以解释不同的行为吗？
提前致谢。
注意：我想 post 图片，但我不能，因为我刚刚订阅了堆栈溢出。
代码
import fipy as fp
import fipy.tools as fpt
"""
Let's try to solve the reaction diffusion equation fith FiPy
A + B ---k---> C
"""
# define spatial domain (square)
nx = ny = 20
dx = dy = 1.
mesh = fp.Grid2D(dx=dx, dy=dy, nx=nx, ny=ny)
linear_dimension = (nx * ny,)
# define molecule as CellVariables
a = fp.CellVariable(name="molecule A",
mesh=mesh)
b = fp.CellVariable(name="molecule B",
mesh=mesh)
c = fp.CellVariable(name="molecule C",
mesh=mesh)
# define initial condition
def gauss_2d(mu_x, mu_y, sigma_x, sigma_y, x_1d, y_1d):
"""
Utility function to define initial conditions (see below). Provide a simil-gaussian
distribution (NOTICE: it's not an exact gaussian because I needed it to be 1 at the top)
"""
# initialize matrix
gauss_mat = fpt.numerix.empty((len(x_1d), len(y_1d)))
# define gaussian
gauss_x = fpt.numerix.exp((-1 / 2) * (((x_1d - mu_x) ** 2) / (sigma_x ** 2)))
gauss_y = fpt.numerix.exp((-1 / 2) * (((y_1d - mu_y) ** 2) / (sigma_y ** 2)))
# evaluate each point of the matrix
for i in range(0, len(x_1d)):
for j in range(0, len(y_1d)):
gauss_mat[i, j] = gauss_x[i] * gauss_y[j]
return gauss_mat
normal_distribution = gauss_2d(mu_x=nx / 2,
mu_y=ny / 2,
sigma_x=fpt.numerix.sqrt(10),
sigma_y=fpt.numerix.sqrt(10),
x_1d=fpt.numerix.arange(0, nx, dx),
y_1d=fpt.numerix.arange(0, ny, dy))
a_max = 100.
a0 = a_max * normal_distribution
a.setValue(a0.reshape(linear_dimension))
b_max = a_max / 2
b0 = b_max * normal_distribution
b.setValue(b0.reshape(linear_dimension))
c0 = 0.
c.setValue(c0)
# create viewer for the three molecules
vi = fp.Viewer(vars=(a, b, c), datamin=0., datamax=a_max)
vi.plot()
fp.input("Press enter to continue...")
# define the reaction term
k = 0.01
reaction_term = k * a * b
# define the equation for each molecule
D = 0.05
eq_a = fp.TransientTerm(var=a) == fp.DiffusionTerm(coeff=D, var=a) - reaction_term
eq_b = fp.TransientTerm(var=b) == fp.DiffusionTerm(coeff=D, var=b) - reaction_term
eq_c = fp.TransientTerm(var=c) == fp.DiffusionTerm(coeff=D, var=c) + reaction_term
# couple equations
sys = eq_a & eq_b & eq_c
# solve
dt = 0.25
steps = 400
for step in range(steps):
sys.solve(dt=dt)
vi.plot()
fp.input("Press enter to continue...")
tl;dr：您可能发现 Matplotlib 查看器的行为与求解器的行为不同。
- 当我使用 Matplotlib 调整错误和回归时，我看到 Py27 和 Py3k 的演变非常相似。在我们解决这个问题之前，您可以看看这些更改是否有帮助：
diff --git a/rxndiff.py b/rxndiff.py
index b41932a..ea30a95 100644
--- a/rxndiff.py
+++ b/rxndiff.py
@@ -1,5 +1,6 @@
import fipy as fp
import fipy.tools as fpt
+from matplotlib import pyplot as plt
"""
Let's try to solve the reaction diffusion equation fith FiPy
@@ -79,5 +80,8 @@ steps = 400
for step in range(steps):
sys.solve(dt=dt)
vi.plot()
+ for vw in vi.viewers:
+ vw.fig.canvas.draw_idle()
+ plt.show(block=False)
- 求解器不同（我用 Py27 得到 PySparse
LinearLUSolver，用 Py3k 得到 PETSc LinearGMRESSolver）。这可能很重要，特别是因为您没有扫除这个非线性问题，但我看不出有什么大的不同。
- 由于整数除法，Py27 的初始条件可能不是您想要的。我建议初始化：
mu_x=nx / 2.
mu_y=ny / 2.
sigma_x=fpt.numerix.sqrt(10.)
sigma_y=fpt.numerix.sqrt(10.)
normal_distribution = (fpt.numerix.exp((-1./2) * (mesh.x - mu_x) ** 2 / sigma_x ** 2)
* fpt.numerix.exp((-1./2) * (mesh.y - mu_y) ** 2 / sigma_y ** 2))
a_max = 100.
a0 = a_max * normal_distribution
a.setValue(a0)
b_max = a_max / 2
b0 = b_max * normal_distribution
b.setValue(b0)
c0 = 0.
c.setValue(c0)
我正在用偏微分方程组试验 FiPy，我得到了不同的结果 运行 Python2 或 Python3[=25 中的相同脚本=].更准确地说，我在 Python 2.7 中得到了我期望 运行 脚本的结果，而在 Python 3.8.
中得到了完全错误的结果有人知道原因吗？我错过了什么吗？ 我知道 FiPy 在 Python 2.7 中使用的一些求解器不支持 Python 3.x，但这足以解释不同的行为吗？
提前致谢。
注意：我想 post 图片，但我不能，因为我刚刚订阅了堆栈溢出。
代码
import fipy as fp
import fipy.tools as fpt
"""
Let's try to solve the reaction diffusion equation fith FiPy
A + B ---k---> C
"""
# define spatial domain (square)
nx = ny = 20
dx = dy = 1.
mesh = fp.Grid2D(dx=dx, dy=dy, nx=nx, ny=ny)
linear_dimension = (nx * ny,)
# define molecule as CellVariables
a = fp.CellVariable(name="molecule A",
mesh=mesh)
b = fp.CellVariable(name="molecule B",
mesh=mesh)
c = fp.CellVariable(name="molecule C",
mesh=mesh)
# define initial condition
def gauss_2d(mu_x, mu_y, sigma_x, sigma_y, x_1d, y_1d):
"""
Utility function to define initial conditions (see below). Provide a simil-gaussian
distribution (NOTICE: it's not an exact gaussian because I needed it to be 1 at the top)
"""
# initialize matrix
gauss_mat = fpt.numerix.empty((len(x_1d), len(y_1d)))
# define gaussian
gauss_x = fpt.numerix.exp((-1 / 2) * (((x_1d - mu_x) ** 2) / (sigma_x ** 2)))
gauss_y = fpt.numerix.exp((-1 / 2) * (((y_1d - mu_y) ** 2) / (sigma_y ** 2)))
# evaluate each point of the matrix
for i in range(0, len(x_1d)):
for j in range(0, len(y_1d)):
gauss_mat[i, j] = gauss_x[i] * gauss_y[j]
return gauss_mat
normal_distribution = gauss_2d(mu_x=nx / 2,
mu_y=ny / 2,
sigma_x=fpt.numerix.sqrt(10),
sigma_y=fpt.numerix.sqrt(10),
x_1d=fpt.numerix.arange(0, nx, dx),
y_1d=fpt.numerix.arange(0, ny, dy))
a_max = 100.
a0 = a_max * normal_distribution
a.setValue(a0.reshape(linear_dimension))
b_max = a_max / 2
b0 = b_max * normal_distribution
b.setValue(b0.reshape(linear_dimension))
c0 = 0.
c.setValue(c0)
# create viewer for the three molecules
vi = fp.Viewer(vars=(a, b, c), datamin=0., datamax=a_max)
vi.plot()
fp.input("Press enter to continue...")
# define the reaction term
k = 0.01
reaction_term = k * a * b
# define the equation for each molecule
D = 0.05
eq_a = fp.TransientTerm(var=a) == fp.DiffusionTerm(coeff=D, var=a) - reaction_term
eq_b = fp.TransientTerm(var=b) == fp.DiffusionTerm(coeff=D, var=b) - reaction_term
eq_c = fp.TransientTerm(var=c) == fp.DiffusionTerm(coeff=D, var=c) + reaction_term
# couple equations
sys = eq_a & eq_b & eq_c
# solve
dt = 0.25
steps = 400
for step in range(steps):
sys.solve(dt=dt)
vi.plot()
fp.input("Press enter to continue...")
tl;dr：您可能发现 Matplotlib 查看器的行为与求解器的行为不同。
- 当我使用 Matplotlib 调整错误和回归时，我看到 Py27 和 Py3k 的演变非常相似。在我们解决这个问题之前，您可以看看这些更改是否有帮助：
diff --git a/rxndiff.py b/rxndiff.py index b41932a..ea30a95 100644 --- a/rxndiff.py +++ b/rxndiff.py @@ -1,5 +1,6 @@ import fipy as fp import fipy.tools as fpt +from matplotlib import pyplot as plt """ Let's try to solve the reaction diffusion equation fith FiPy @@ -79,5 +80,8 @@ steps = 400 for step in range(steps): sys.solve(dt=dt) vi.plot() + for vw in vi.viewers: + vw.fig.canvas.draw_idle() + plt.show(block=False) - 求解器不同（我用 Py27 得到 PySparse
LinearLUSolver，用 Py3k 得到 PETScLinearGMRESSolver）。这可能很重要，特别是因为您没有扫除这个非线性问题，但我看不出有什么大的不同。 - 由于整数除法，Py27 的初始条件可能不是您想要的。我建议初始化：
mu_x=nx / 2. mu_y=ny / 2. sigma_x=fpt.numerix.sqrt(10.) sigma_y=fpt.numerix.sqrt(10.) normal_distribution = (fpt.numerix.exp((-1./2) * (mesh.x - mu_x) ** 2 / sigma_x ** 2) * fpt.numerix.exp((-1./2) * (mesh.y - mu_y) ** 2 / sigma_y ** 2)) a_max = 100. a0 = a_max * normal_distribution a.setValue(a0) b_max = a_max / 2 b0 = b_max * normal_distribution b.setValue(b0) c0 = 0. c.setValue(c0)