在Python中打印Bulirsch-Stoer算法的解决方案 [英] Printing the solution for Bulirsch-Stoer algorithm in Python
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问题描述
在下面的代码中，为什么我不断出现此错误:
In the following code, why do I keep getting this error:
TypeError:"NoneType"对象不可下标
TypeError: 'NoneType' object is not subscriptable
我的 PrintSoln(X，Y，freq)函数?我该如何解决?
for my PrintSoln(X, Y, freq) function? And how can I fix it?
我正在尝试解决IVP: 2 *(y'')+ y'+(y/0.45)= 9 ，初始条件为 y = 0且y'= 0 .'''
I am trying to solve the IVP: 2*(y'') + y' + (y/0.45) = 9,
with initial conditions y = 0 and y' = 0.'''
from numpy import zeros, float, sum
import math
def integrate_BulirschStoeir(F, x, y, xStop, tol):
def midpoint(F, x, y, xStop, nSteps):
h = (xStop - x)/ nSteps
y0 = y
y1 = y0 + h*F(x, y0)
for i in range(nSteps-1):
x = x + h
y2 = y0 + 2.0*h*F(x, y1)
y0 = y1
y1 = y2
return 0.5*(y1 + y0 + h*F(x, y2))
def richardson(r, k):
for j in range(k-1,0,-1):
const = (k/(k - 1.0))**(2.0*(k-j))
r[j] = (const*r[j+1] - r[j])/(const - 1.0)
return
kMax = 51
n = len(y)
r = zeros((kMax, n), dtype=float)
# Start with two integration steps
nSteps = 2
r[1] = midpoint(F, x, y, xStop, nSteps)
r_old = r[1].copy()
# Increase the number of integration points by 2 and refine result by Richardson extrapolation
for k in range(2,kMax):
nSteps = 2*k
r[k] = midpoint(F, x, y, xStop, nSteps)
richardson(r,k)
# Compute RMS change in solution
e = sqrt(sum((r[1] - r_old)**2)/n)
# Check for convergence
if e < tol:
return r[1]
r_old = r[1].copy()
print("Midpoint method did not converge")
# Bulirsch-Stoer Algorithm:-
''' X, Y = bulStoer(F, x, y, xStop, H, tol=1.0e-6).
Simplified Bulirsch-Stoer method for solving the
initial value problem {y}’ = {F(x,{y})}, where {y} = {y[0],y[1],...y[n-1]}
x, y = initial conditions
xStop = terminal value of x
H = increment of x at which results are stored
F = user-supplied function that returns the array F(x,y) = {y’[0],y’[1],...,y’[n-1]} '''
from numpy import array
def bulStoer(F, x, y, xStop, H, tol=1.0e-6):
X = []
Y = []
X.append(x)
Y.append(y)
while x < xStop:
H = min(H,xStop - x)
y = integrate_BulirschStoeir(F, x, y, x + H, tol) # Midpoint method
x = x + H
X.append(x)
Y.append(y)
return array(X), array(Y)
def printSoln(X, Y, freq):
def printHead(n):
print("\n x ")
for i in range (n):
print("y[",i,"]")
print
def printLine(x, y, n):
print("%13.4e"% x)
for i in range (n):
print("%13.4e"% y[i])
print
m = len(Y)
try: n = len(Y[0])
except TypeError: n = 1
if freq == 0: freq = m
printHead(n)
for i in range(0,m,freq):
printLine(X[i], Y[i], n)
if i != m - 1: printLine(X[m - 1], Y[m - 1], n)
# Code:-
from numpy import array, zeros
def F(x, y):
F = zeros(2)
F[0] = y[1]
F[1] =(-y[1] - (y[0]/0.45) + 9.0)/2.0
return F
x = 0.0
xStop = 10.0
H = 0.5
y = array([0.0, 0.0])
freq = 1
X, Y = bulStoer(F, x, y, xStop, H)
printSoln(X, Y, freq)
推荐答案
-
"def integration_BulirschStoeir(F，x，y，xStop，tol)"中存在缩进问题(从"kMax = 51"到函数结尾).因此，在"X，Y = bulStoer(F，x，y，xStop，H)"和Y bacame矩阵的第一行为[0,0]，其他行为None的情况下，Y中没有新数据.
There was an indentation problem in "def integrate_BulirschStoeir(F, x, y, xStop, tol)" (from "kMax = 51" to the end of function). So, there was no new data in Y after "X, Y = bulStoer(F, x, y, xStop, H)" and Y bacame a matrix with [0,0] as first row and None as other rows.
此外，在"def printLine"中的打印(在打印功能中未添加"end ="")导致列向量，并添加了"end ="(")以使结果更易读.从numpy导入
Also the print in "def printLine" without addition of " end=' ' " in print-function led to a column-vector and " end=' ' " was added to make results more readable.
from numpy import zeros, float, sum, sqrt
import math
def integrate_BulirschStoeir(F, x, y, xStop, tol):
def midpoint(F, x, y, xStop, nSteps):
h = (xStop - x)/ nSteps
y0 = y
y1 = y0 + h*F(x, y0)
for i in range(nSteps-1):
x = x + h
y2 = y0 + 2.0*h*F(x, y1)
y0 = y1
y1 = y2
return 0.5*(y1 + y0 + h*F(x, y2))
def richardson(r, k):
for j in range(k-1,0,-1):
const = (k/(k - 1.0))**(2.0*(k-j))
r[j] = (const*r[j+1] - r[j])/(const - 1.0)
return
kMax = 51 ## Indentation problem from here
n = len(y)
r = zeros((kMax, n), dtype=float)
# Start with two integration steps
nSteps = 2
r[1] = midpoint(F, x, y, xStop, nSteps)
r_old = r[1].copy()
# Increase the number of integration points by 2 and refine result by Richardson extrapolation
for k in range(2,kMax):
nSteps = 2*k
r[k] = midpoint(F, x, y, xStop, nSteps)
richardson(r,k)
# Compute RMS change in solution
e = sqrt(sum((r[1] - r_old)**2)/n)
# Check for convergence
if e < tol:
return r[1]
r_old = r[1].copy()
print("Midpoint method did not converge")
# Bulirsch-Stoer Algorithm:-
''' X, Y = bulStoer(F, x, y, xStop, H, tol=1.0e-6).
Simplified Bulirsch-Stoer method for solving the
initial value problem {y}’ = {F(x,{y})}, where {y} = {y[0],y[1],...y[n-1]}
x, y = initial conditions
xStop = terminal value of x
H = increment of x at which results are stored
F = user-supplied function that returns the array F(x,y) = {y’[0],y’[1],...,y’[n-1]} '''
from numpy import array
def bulStoer(F, x, y, xStop, H, tol=1.0e-6):
X = []
Y = []
X.append(x)
Y.append(y)
while x < xStop:
H = min(H,xStop - x)
y = integrate_BulirschStoeir(F, x, y, x + H, tol) # Midpoint method
x = x + H
X.append(x)
Y.append(y)
return array(X), array(Y)
def printSoln(X, Y, freq):
def printHead(n):
print("\n x ",end=" ") ## end=" " here
for i in range(n):
print(" y[",i,"] ",end=" ") ## end=" " here
print()
def printLine(x, y, n):
print('{:13.4e}'.format(x), end=' ') ## end=" " here
for i in range(n):
print('{:13.4e}'.format(y[i]), end=' ') ## end=" " here
print()
m = Y.shape[0]
try: n = Y.shape[1]
except Exception: n = 1
if freq == 0: freq = m
printHead(n)
for i in range(0,m,freq):
printLine(X[i], Y[i], n)
if i != m - 1: printLine(X[m - 1], Y[m - 1], n)
from numpy import array, zeros
def F(x, y):
F = zeros(2)
F[0] = y[1]
F[1] =(-y[1] - (y[0]/0.45) + 9.0)/2.0
return F
x = 0.0
xStop = 10.0
H = 0.5
y = array([0.0, 0.0])
freq = 1
X, Y = bulStoer(F, x, y, xStop, H)
printSoln(X, Y, freq)
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