scipy.optimize.newton gives TypeError: 'float' object is not callable

Question

Im new to python and I was writing this simply code for finding the roots of the function:

from scipy import optimize

x = eval(raw_input())                           #Initial guess
f = eval(raw_input())                           # function to be evaluated
F = eval(raw_input())                          #derivative of function f

round(optimize.newton(f, x, F, tol = 1.0e-9), 4)

But the interpreter returns: TypeError: 'float' object is not callable

I'm really not sure what im missing out from this code. Can someone help me out..thank you in advance

you are evaluating `raw_input`, likely becoming a float, for `f`. `newton` is expecting a function. — user2085282, Sep 22 '14 at 16:12

score 1 · Accepted Answer · answered Sep 22 '14 at 16:16

optimize.newton expects a reference to a callable object (for example a function). That does not mean that you give a function as a string like 'x*x' but you have to define one first, like:

def my_func (x):
    return x*x

Then you can plug my_func into optimize.newton (besides the other required parameters).

score 0 · Answer 2 · answered Sep 22 '14 at 16:17

0

This will depend on what you are inputting for f. If you enter something like

lambda x: x ** 2

then it will be interpreted as a function, for example

In [83]: f = eval('lambda x: x ** 2')

In [84]: f(5)
Out[84]: 25

answered Sep 22 '14 at 16:17

user2085282

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scipy.optimize.newton gives TypeError: 'float' object is not callable

2 Answers2