我正在尝试使用 Python 来求解一个包含 6 个非线性方程的系统。有 9 个变量，其中 3 个是预先确定的(留下 6 个方程组和 6 个未知数)。问题是，它可能是任何 3，我无法事先知道。
这是方程式(如果您感兴趣的话)。
c11*c12 + c21*c22 + c31*c32 = 0
c11*c13 + c21*c23 + c31*c33 = 0
c12*c13 + c22*c23 + c32*c33 = 0
c11*c21 + c12*c22 + c13*c23 = 0
c11*c31 + c12*c32 + c13*c33 = 0
c21*c31 + c22*c32 + c23*c33 = 0
注意:这是我认为最快/最容易解决的方法。另一种可能的表达方式是:
|c11 c21 c31|
A = |c12 c22 c32|
|c13 c23 c33|
|c11 c12 c13|
B = |c21 c22 c23|
|c31 c32 c33|
|1 0 0|
A*B = |0 1 0|
|0 0 1|
我的问题是:是否可以将其中 3 个设置为固定参数，并让 scipy.optimize.fsolve(或更合适的模块？)求解其余参数？
请您参考如下方法:
所以，我自己找到了一个有效的解决方案。不确定它是否是最好的解决方案，但它的功能。
为了回答我的问题，scipy.optimize.fsolve 接受一个参数 args =(这里有额外的参数)。我把预定的参数放在这里。调用函数时，首先解析 args 并将 3 个预定值放在适当的位置。
其余 6 个变量在列表中传递，并迭代以填补剩余的空白。由于参数不变，每个变量始终位于矩阵中的相同位置。
使用此方法，可以预先确定任意 3 个矩阵元素，fsolve 将尝试确定余数。
fsolve 的调用语句如下所示:
paramSolve1, infodict, ier, mesg = scipy.optimize.fsolve(func,(i,i,i,i,i,i),args = (knownVals[0],knownVals[1],knownVals[2]), full_output = True, warning = False)
knwonVals 是一个预先确定的参数列表，i 是一个起始猜测值(所有 6 个缺失的参数得到相同的起始猜测值)。 full_output 允许返回可选输出，并且 warning = False 关闭未找到解决方案时出现的警告消息。有关更多信息，请查看 http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fsolve.html
对于那些感兴趣的人，问题的完整代码如下。
import scipy
from scipy.optimize import fsolve
def func(params, *args):
c = propMatrix(createMatrix(args), params)
ans =(scipy.dot(c[:, 0],c[:, 1]), scipy.dot(c[:, 1],c[:, 2]), scipy.dot(c[:, 0],c[:, 2]),scipy.dot(c[:, 0],c[:, 0])-1,scipy.dot(c[:, 1],c[:, 1])-1,scipy.dot(c[:, 2],c[:, 2])-1)
return ans
def createMatrix(knownVals):
c = [['____', '____', '____'],['____', '____', '____'], ['____', '____', '____']]
for element in knownVals:
x, y, val = element
c[y][x] = float(val)
return c
def propMatrix(c, params):
for p in params:
assign = True
for x in range(3):
for y in range(3):
if c[x][y]=='____' and assign:
c[x][y] = float(p)
assign = False
return scipy.array(c)
def test(c):
v1 = c[:, 0]
v2 = c[:, 1]
v3 = c[:, 2]
h1 = c[0, :]
h2 = c[1, :]
h3 = c[2, :]
ans = (scipy.dot(v1,v1)-1, scipy.dot(v1,v2), scipy.dot(v1, v3), scipy.dot(v2, v2)-1, scipy.dot(v2, v3), scipy.dot(v3,v3)-1, scipy.dot(h1,h1)-1, scipy.dot(h1,h2), scipy.dot(h1, h3), scipy.dot(h2, h2)-1, scipy.dot(h2, h3), scipy.dot(h3,h3)-1)
return ans
def getInput():
knownVals = []
print """\n\nThis module analytically solves for the rotation matrix\n
First, enter 3 known values of the matrix:\n
x
1 2 3
1 | c11 c12 c13 |
y 2 | c21 c22 c23 |
3 | c31 c32 c33 |\n\n"""
for i in range(3):
invalid = True
print "Point Number %i:"%(i)
while invalid:
x = int(raw_input("\tx-coordinate:"))-1
if x>2 or x<0:
print "\tInvalid x-coordinate."
else:
invalid = False
invalid = True
while invalid:
y = int(raw_input("\ty-coordinate:"))-1
if y>2 or y<0:
print "\tInvalid y-coordinate."
else:
invalid = False
invalid = True
while invalid:
val = float(raw_input("\tValue:"))
if val>1 or val<-1:
print "\tInvalid value. Must be -1 <= value <= 1"
else:
invalid = False
knownVals.append((x, y, val))
c = createMatrix(knownVals)
print "Input Matrix:\n\n", scipy.array(c)
choice = raw_input("\nIs this correct (y/n)? ")
if choice == "y":
return knownVals
elif choice == "n":
return getInput()
def Main():
solution = False
knownVals = getInput()
for i in (-1,-.5,0,.5,1):
paramSolve1, infodict, ier, mesg = scipy.optimize.fsolve(func,(i,i,i,i,i,i),args = (knownVals[0],knownVals[1],knownVals[2]), full_output = True, warning = False)
if ier == 1:
print "\nInitial value: %r"%(i)
print propMatrix(createMatrix(knownVals),paramSolve1)
solution = True
if not solution:
print "Could not find a valid solution"
scipy.set_printoptions(precision = 4, suppress = True)
Main()
