How to plot the variables generated in a for loop and with " OverflowError: (34, 'Result too large') "
How to plot the variables generated in a for loop and with " OverflowError: (34, 'Result too large') "
我正在尝试用 Python3 中的四阶龙格 - 库塔法求解微分方程。该函数是故意使解趋于无穷大。
我的问题是有人要求我绘制迭代解，但在 for 循环中将迭代器和相应的解值存储在列表中时，我似乎无法理解如何处理无穷大，以及当我在循环外调用该列表时。下面是带有错误消息的完整代码和结果窗格。
import math
import numpy as np
import matplotlib.pyplot as plt
# Problem function #
def diffEq(x, y):
return (5*math.exp(y)) + (x)**2
# 4th order Runge - Kutta method #
def RK_fourth(x0,y0,x,h):
average_slope = 0
i = 0
x_axis = [] # For plotting #
y_axis = [] # For plotting #
print("Iterations of 4th Order Runge - Kutta Method")
for i in np.arange(x0,x,h, dtype=float):
k1 = diffEq(y0,i)
k2 = diffEq(y0 + (k1 * (h/2)), i + (h/2))
k3 = diffEq(y0 + (k2 * (h/2)), i + (h/2))
k4 = diffEq(y0 + (k3 * h), i + h)
average_slope = (1/6) * (k1 + (2* k2) + (2*k3) + k4)
print('average_slope:',average_slope)
y = y0 + (average_slope * h)
y0 = y
x_axis.append(i + h)
y_axis.append(y)
print("x:", i+h)
print("y:", y)
print(x_axis)
print(y_axis)
print(x_axis)
print(y_axis)
print("The solution using 4th order Runge - Kutta method is : ","%.4f"%y)
# main #
RK_fourth(0, 0 ,1, 0.1)
# Console #
Iterations of 4th Order Runge - Kutta Method
average_slope: 5.350250926709215
x: 0.1
y: 0.5350250926709216
[0.1]
[0.5350250926709216]
average_slope: 6.568756155777355
x: 0.2
y: 1.191900708248657
[0.1, 0.2]
[0.5350250926709216, 1.191900708248657]
average_slope: 9.107515760505036
x: 0.30000000000000004
y: 2.1026522842991606
[0.1, 0.2, 0.30000000000000004]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606]
average_slope: 14.993810753267782
x: 0.4
y: 3.602033359625939
[0.1, 0.2, 0.30000000000000004, 0.4]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939]
average_slope: 33.7279469045296
x: 0.5
y: 6.9748280500789
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789]
average_slope: 185.38472029098983
x: 0.6
y: 25.51330007917788
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788]
average_slope: 2568779.276837211
x: 0.7000000000000001
y: 256903.4409838003
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003]
average_slope: 1.4658111185680487e+68
x: 0.8
y: 1.4658111185680488e+67
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67]
---------------------------------------------------------------------------
OverflowError Traceback (most recent call last)
/var/folders/bt/kqf88mw53h55m9mj35rkwt6h0000gn/T/ipykernel_18686/1130051705.py in <module>
6 #ImprovedEuler(0.2, 0.1 ,0.3, 0.1)
7 #print("\n")
----> 8 RK_fourth(0, 0 ,1, 0.1)
9 #print("\n")
10 #GaussJordan(12,3,-5,1,1,5,3,28,3,7,13,76) # Enter the coefficients #
/var/folders/bt/kqf88mw53h55m9mj35rkwt6h0000gn/T/ipykernel_18686/303333894.py in RK_fourth(x0, y0, x, h)
74 k1 = diffEq(y0,i)
75 k2 = diffEq(y0 + (k1 * (h/2)), i + (h/2))
---> 76 k3 = diffEq(y0 + (k2 * (h/2)), i + (h/2))
77 k4 = diffEq(y0 + (k3 * h), i + h)
78 average_slope = (1/6) * (k1 + (2* k2) + (2*k3) + k4)
/var/folders/bt/kqf88mw53h55m9mj35rkwt6h0000gn/T/ipykernel_18686/303333894.py in diffEq(x, y)
12 def diffEq(x, y):
13 #Function for GVF
---> 14 return (5*math.exp(y)) + (x)**2
15
16
OverflowError: (34, 'Result too large')
请注意，我还没有调用 plt.plot(x_axis, y_axis)。为了验证 x_axis 和 y_axis 能够获取我在循环内和循环外分别调用它们的所有值。错误消息之前的最后打印来自循环内的 print() 语句。
如何解决这个问题，我对Python很陌生。如果您不清楚这个问题，我深表歉意，很乐意进一步澄清。
提前感谢您的帮助。
您可以使用 numpy 的 np.exp 而不是 math.exp 来处理无穷大：
def diffEq(x, y):
return (5*np.exp(y)) + (x)**2
结果：
Iterations of 4th Order Runge - Kutta Method
average_slope: 5.350250926709215
x: 0.1
y: 0.5350250926709216
[0.1]
[0.5350250926709216]
average_slope: 6.568756155777355
x: 0.2
y: 1.191900708248657
[0.1, 0.2]
[0.5350250926709216, 1.191900708248657]
average_slope: 9.107515760505036
x: 0.30000000000000004
y: 2.1026522842991606
[0.1, 0.2, 0.30000000000000004]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606]
average_slope: 14.993810753267782
x: 0.4
y: 3.602033359625939
[0.1, 0.2, 0.30000000000000004, 0.4]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939]
average_slope: 33.7279469045296
x: 0.5
y: 6.9748280500789
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789]
average_slope: 185.38472029098983
x: 0.6
y: 25.51330007917788
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788]
average_slope: 2568779.276837211
x: 0.7000000000000001
y: 256903.4409838003
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003]
average_slope: 1.4658111185680487e+68
x: 0.8
y: 1.4658111185680488e+67
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67]
average_slope: inf
x: 0.9
y: inf
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8, 0.9]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67, inf]
average_slope: inf
x: 1.0
y: inf
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8, 0.9, 1.0]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67, inf, inf]
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8, 0.9, 1.0]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67, inf, inf]
The solution using 4th order Runge - Kutta method is : inf
程序将输出运行时警告RuntimeWarning: overflow encountered in double_scalars...。您可以使用上下文管理器来抑制此警告：
import warnings
with warnings.catch_warnings():
warnings.filterwarnings("ignore", "overflow")
RK_fourth(0, 0 ,1, 0.1)
我正在尝试用 Python3 中的四阶龙格 - 库塔法求解微分方程。该函数是故意使解趋于无穷大。
我的问题是有人要求我绘制迭代解，但在 for 循环中将迭代器和相应的解值存储在列表中时，我似乎无法理解如何处理无穷大，以及当我在循环外调用该列表时。下面是带有错误消息的完整代码和结果窗格。
import math
import numpy as np
import matplotlib.pyplot as plt
# Problem function #
def diffEq(x, y):
return (5*math.exp(y)) + (x)**2
# 4th order Runge - Kutta method #
def RK_fourth(x0,y0,x,h):
average_slope = 0
i = 0
x_axis = [] # For plotting #
y_axis = [] # For plotting #
print("Iterations of 4th Order Runge - Kutta Method")
for i in np.arange(x0,x,h, dtype=float):
k1 = diffEq(y0,i)
k2 = diffEq(y0 + (k1 * (h/2)), i + (h/2))
k3 = diffEq(y0 + (k2 * (h/2)), i + (h/2))
k4 = diffEq(y0 + (k3 * h), i + h)
average_slope = (1/6) * (k1 + (2* k2) + (2*k3) + k4)
print('average_slope:',average_slope)
y = y0 + (average_slope * h)
y0 = y
x_axis.append(i + h)
y_axis.append(y)
print("x:", i+h)
print("y:", y)
print(x_axis)
print(y_axis)
print(x_axis)
print(y_axis)
print("The solution using 4th order Runge - Kutta method is : ","%.4f"%y)
# main #
RK_fourth(0, 0 ,1, 0.1)
# Console #
Iterations of 4th Order Runge - Kutta Method
average_slope: 5.350250926709215
x: 0.1
y: 0.5350250926709216
[0.1]
[0.5350250926709216]
average_slope: 6.568756155777355
x: 0.2
y: 1.191900708248657
[0.1, 0.2]
[0.5350250926709216, 1.191900708248657]
average_slope: 9.107515760505036
x: 0.30000000000000004
y: 2.1026522842991606
[0.1, 0.2, 0.30000000000000004]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606]
average_slope: 14.993810753267782
x: 0.4
y: 3.602033359625939
[0.1, 0.2, 0.30000000000000004, 0.4]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939]
average_slope: 33.7279469045296
x: 0.5
y: 6.9748280500789
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789]
average_slope: 185.38472029098983
x: 0.6
y: 25.51330007917788
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788]
average_slope: 2568779.276837211
x: 0.7000000000000001
y: 256903.4409838003
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003]
average_slope: 1.4658111185680487e+68
x: 0.8
y: 1.4658111185680488e+67
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67]
---------------------------------------------------------------------------
OverflowError Traceback (most recent call last)
/var/folders/bt/kqf88mw53h55m9mj35rkwt6h0000gn/T/ipykernel_18686/1130051705.py in <module>
6 #ImprovedEuler(0.2, 0.1 ,0.3, 0.1)
7 #print("\n")
----> 8 RK_fourth(0, 0 ,1, 0.1)
9 #print("\n")
10 #GaussJordan(12,3,-5,1,1,5,3,28,3,7,13,76) # Enter the coefficients #
/var/folders/bt/kqf88mw53h55m9mj35rkwt6h0000gn/T/ipykernel_18686/303333894.py in RK_fourth(x0, y0, x, h)
74 k1 = diffEq(y0,i)
75 k2 = diffEq(y0 + (k1 * (h/2)), i + (h/2))
---> 76 k3 = diffEq(y0 + (k2 * (h/2)), i + (h/2))
77 k4 = diffEq(y0 + (k3 * h), i + h)
78 average_slope = (1/6) * (k1 + (2* k2) + (2*k3) + k4)
/var/folders/bt/kqf88mw53h55m9mj35rkwt6h0000gn/T/ipykernel_18686/303333894.py in diffEq(x, y)
12 def diffEq(x, y):
13 #Function for GVF
---> 14 return (5*math.exp(y)) + (x)**2
15
16
OverflowError: (34, 'Result too large')
请注意，我还没有调用 plt.plot(x_axis, y_axis)。为了验证 x_axis 和 y_axis 能够获取我在循环内和循环外分别调用它们的所有值。错误消息之前的最后打印来自循环内的 print() 语句。
如何解决这个问题，我对Python很陌生。如果您不清楚这个问题，我深表歉意，很乐意进一步澄清。
提前感谢您的帮助。
您可以使用 numpy 的 np.exp 而不是 math.exp 来处理无穷大：
def diffEq(x, y):
return (5*np.exp(y)) + (x)**2
结果：
Iterations of 4th Order Runge - Kutta Method
average_slope: 5.350250926709215
x: 0.1
y: 0.5350250926709216
[0.1]
[0.5350250926709216]
average_slope: 6.568756155777355
x: 0.2
y: 1.191900708248657
[0.1, 0.2]
[0.5350250926709216, 1.191900708248657]
average_slope: 9.107515760505036
x: 0.30000000000000004
y: 2.1026522842991606
[0.1, 0.2, 0.30000000000000004]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606]
average_slope: 14.993810753267782
x: 0.4
y: 3.602033359625939
[0.1, 0.2, 0.30000000000000004, 0.4]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939]
average_slope: 33.7279469045296
x: 0.5
y: 6.9748280500789
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789]
average_slope: 185.38472029098983
x: 0.6
y: 25.51330007917788
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788]
average_slope: 2568779.276837211
x: 0.7000000000000001
y: 256903.4409838003
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003]
average_slope: 1.4658111185680487e+68
x: 0.8
y: 1.4658111185680488e+67
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67]
average_slope: inf
x: 0.9
y: inf
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8, 0.9]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67, inf]
average_slope: inf
x: 1.0
y: inf
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8, 0.9, 1.0]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67, inf, inf]
[0.1, 0.2, 0.30000000000000004, 0.4, 0.5, 0.6, 0.7000000000000001, 0.8, 0.9, 1.0]
[0.5350250926709216, 1.191900708248657, 2.1026522842991606, 3.602033359625939, 6.9748280500789, 25.51330007917788, 256903.4409838003, 1.4658111185680488e+67, inf, inf]
The solution using 4th order Runge - Kutta method is : inf
程序将输出运行时警告RuntimeWarning: overflow encountered in double_scalars...。您可以使用上下文管理器来抑制此警告：
import warnings
with warnings.catch_warnings():
warnings.filterwarnings("ignore", "overflow")
RK_fourth(0, 0 ,1, 0.1)