类型错误 'float'对象在使用 for 循环时不可调用

Question

# Prototype of N-R for a system of two non-linear equations
#evaluating  functions of two variables
# f(x,y)=1.6 * x ** 2 + 3.6 * x * y - 7.8 * x - 2.6 * y + 5.2
# g(x,y)=0.9 * y ** 2 + 3.1 * x **2 - 6.2 * x + 6.2 * y
# x = 0.5
# y =0.4
from math import *
eq1 = raw_input('Enter the equation 1: ')
eq2 = raw_input('Enter the equation 2: ')
x0 = float(input('Enter x: '))
y0 = float(input('Enter y: '))
def f(x,y):
    return eval(eq1)
def g(x,y):
    return eval(eq2)
Ea_X = 1
x = x0
y = y0
for n in range(1, 8):
    a = (f(x + 1e-06, y) - f(x,y)) / 1e-06   #in this one start the trouble
    b = (f(x, y + 1e-06) - f(x,y)) / 1e-06
    c = 0 - f(x,y)
    d = (g(x + 1e-06, y) - g(x,y)) / 1e-06
    eE = (g(x, y + 1e-06) - g(x,y)) / 1e-06
    f = 0 - g(x,y)

    print "f(x, y)= ", eq1
    print "g(x, y)= ", eq2
    print """x   y """
    print x, y
    print """a   b   c   d   e   f """
    print a, b, c, d, e, f
    print """
    a * x + b * y = c
    d * x + e * y = f
    """
    print a," * x  +  ",b," * y  =  ",c
    print d," * x  +  ",eE," * y  =  ",f
    _Sy = (c - a * f / d) / (b - a * eE / d)
    _Sx = (f / d) - (eE / d) * _Sy
    Ea_X = (_Sx ** 2 + _Sy ** 2)**0.5

    x = x + _Sx
    y = y + _Sy
    print "Sx = ", _Sx
    print "Sy = ", _Sy
    print "x = ", x
    print "y = ", y
    print "|X_1 - X_0| = ", Ea_X

我一直在测试两个非线性方程的牛顿-拉普森方法，原型代码有效，但后来我正在考虑使其更多有用，因为原型是关于 2 个方程的输入和第一个猜测，最好实现一个 for 循环而不是像 6 或 10 那样启动过程来只解决一个在我正在使用的众多方程中

# Prototype of N-R for a system of two non-linear equations
# f(x,y)=1.6 * x ** 2 + 3.6 * x * y - 7.8 * x - 2.6 * y + 5.2
# g(x,y)=0.9 * y ** 2 + 3.1 * x **2 - 6.2 * x + 6.2 * y
# x = 0.5
# y =0.4

# evaluating  functions of two variables
from math import *

eq1 = raw_input('Enter the equation 1: ')
eq2 = raw_input('Enter the equation 2: ')
x0 = float(input('Enter x: '))
y0 = float(input('Enter y: '))
def f(x,y):
    return eval(eq1)
def g(x,y):
    return eval(eq2)
Ea_X = 1
x = x0
y = y0
a = (f(x + 1e-06, y) - f(x,y)) / 1e-06
b = (f(x, y + 1e-06) - f(x,y)) / 1e-06
c = 0 - f(x,y)
d = (g(x + 1e-06, y) - g(x,y)) / 1e-06
eE = (g(x, y + 1e-06) - g(x,y)) / 1e-06
f = 0 - g(x,y)

print "f(x, y)= ", eq1
print "g(x, y)= ", eq2
print """x   y """
print x, y
print """a   b   c   d   e   f """
print a, b, c, d, e, f
print """
a * x + b * y = c
d * x + e * y = f
"""
print a," * x  +  ",b," * y  =  ",c
print d," * x  +  ",eE," * y  =  ",f
_Sy = (c - a * f / d) / (b - a * eE / d)
_Sx = (f / d) - (eE / d) * _Sy
Ea_X = (_Sx ** 2 + _Sy ** 2)**0.5

x = x + _Sx
y = y + _Sy
print "Sx = ", _Sx
print "Sy = ", _Sy
print "x = ", x
print "y = ", y
print "|X_1 - X_0| = ", Ea_X

Answer 1

在行中

f = 0 - g(x,y)

为名称 f 分配一个编号。由于函数和其他变量在 Python 中共享一个命名空间（函数只是一个可调用的对象，绑定到任何变量），这使得未来的迭代失败。为在上行中分配的值选择另一个名称。

Answer 2

这是问题所在：

f = 0 - g(x,y)

您正在将f从函数重新绑定到float。