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FsolveinScipy似乎是合适的人选,我只需要动态传递方程的帮助。我提前感谢任何想法。

通过动态,我的意思是方程的数量从一次运行到另一次运行不同,例如我有一种情况:

alpha*x + (1-alpha)*x*y - y = 0
beta*x  + (1- beta)*x*z - z = 0
A*x + B*y + C*z = D

我还有另一种情况:

alpha*x + (1-alpha)*x*y - y = 0
beta*x  + (1- beta)*x*z - z = 0
gama*x  + (1 -gama)*x*w - w =0
A*x + B*y + C*z + D*w = E

alpha, beta, A, B, C,DE都是常数。x, y, z,w是变量。

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1 回答 1

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我自己没有使用过 Fsolve,但根据它的文档,它需要一个可调用的函数。像这样的东西可以处理具有未知数量变量的多个函数。请记住,此处的参数必须正确排序,但每个函数只需要一个列表。

def f1(argList):
    x = argList[0]
    return x**2
def f2(argList):
    x = argList[0]
    y = argList[1]
    return (x+y)**2
def f3(argList):
    x = argList[0]
    return x/3

fs = [f1,f2,f3]
args = [3,5]
for f in fs:
    print f(args)

对于 Fsolve,您可以尝试这样的事情(未经测试):

def func1(argList, constList):
    x = argList[0]
    y = argList[1]
    alpha = constList[0]
    return alpha*x + (1-alpha)*x*y - y
def func2(argList, constList):
    x = argList[0]
    y = argList[1]
    z = argList[2]
    beta = constList[1]
    return beta*x  + (1- beta)*x*z - z
def func3(argList, constList):
    x = argList[0]
    w = argList[1] ## or, if you want to pass the exact same list to each function, make w argList[4]
    gamma = constList[2]
    return gama*x  + (1 -gama)*x*w - w
def func4(argList, constList):

    return A*x + B*y + C*z + D*w -E ## note that I moved E to the left hand side


functions = []
functions.append((func1, argList1, constList1, args01))
# args here can be tailored to fit your  function structure
# Just make sure to align it with the way you call your function:
# args = [argList, constLit]
# args0 can be null.
functions.append((func1, argList2, constList2, args02))
functions.append((func1, argList3, constList3, args03))
functions.append((func1, argList4, constList4, args04))

for func,argList, constList, args0 in functions: ## argList is the (Vector) variable you're solving for.
    Fsolve(func = func, x0 = ..., args = constList, ...)
于 2012-04-05T16:57:24.433 回答