我還沒有使用Fsolve自己,但根據它的文檔它需要一個可調用的函數。 像這樣的東西可以處理多個變量數量未知的函數。請記住,這裏的參數必須正確排序,但每個函數都只需要一個列表。
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)
對於Fsolve,你可以嘗試這樣的事情(未經測試):
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, ...)
我打了SciPy的,但我沒有成功與活力的部分。你基本上需要事先爲你的方程式定義函數,這不是我想要的。然後我在R.中研究了BBsolve,似乎解決了我尋求的問題。 – pouria3 2012-04-10 06:03:37