Python - 'numpy.float64' object is not callable using minimize function for alpha optimization for Simple Exponential Smoothing

Question

I'm getting the TypeError: 'numpy.float64' object is not callable error for the following code:

import numpy as np
from scipy.optimize import minimize

def ses(data, alpha):
    fit=[]
    fit.append(alpha*data[1] + (1-alpha)*data[0])
    for i in range(2, len(data)):
        fit.append(data[i]*alpha + fit[i-2]*(1-alpha))
    return fit

def rmse(data, fit):
    se=[]
    for i in range(2,len(data)):
        se.append((data[i]-fit[i-2])*(data[i]-fit[i-2]))
    mse=np.mean(se)
    return np.sqrt(mse)


alpha=0.1555 # starting value
fit=ses(d[0], alpha)
error=rmse(d[0], fit)

result=minimize(error, alpha, (fit,), bounds=[(0,1)], method='SLSQP')

I've tried many alternatives and its just not working. Changed the lists to arrays and made the multiplications involve no exponentials (np.sqrt() as opposed to ()**0.5)

EDIT:

def ses(data, alpha):
    fit=[]
    fit.append(alpha*data[1] + (1-alpha)*data[0])
    for i in range(2, len(data)):
        fit.append(data[i]*alpha + fit[i-2]*(1-alpha))
    return fit

def rmse(data, alpha):
    fit=ses(data, alpha)
    se=[]
    for i in range(2,len(data)):
        print i, i-2
        se.append((data[i]-fit[i-2])*(data[i]-fit[i-2]))
    mse=np.mean(se)
    return np.sqrt(mse)


alpha=0.1555 # starting value
data=d[0]


result = minimize(rmse, alpha, (data,), bounds=[(0,1)], method='SLSQP')

Ok guys, thanks. Have edited to this and I have stopped the error, however now I am getting an index out of bounds error, which is strange as without the minimize line, the code runs perfectly fine.

EDIT 2:

There was a series of silly errors, most of which I didn't know were problems, but were solved by trial and error.

For some working code of optimized exponential smoothing:

def ses(data, alpha):
    'Simple exponential smoothing'

    fit=[]
    fit.append(data[0])
    fit.append(data[1]) ## pads first two
    fit.append(alpha*data[1] + (1-alpha)*data[0])

    for i in range(2, len(data)-1):
        fit.append(alpha*data[i] + (1-alpha)*fit[i])
    return fit


def rmse(alpha, data):
    fit=ses(data, alpha)
    se=[]
    for i in range(2,len(data)):
        se.append((data[i]-fit[i-2])*(data[i]-fit[i-2]))
    mse=np.mean(se)
    return np.sqrt(mse)

alpha=0.5
data = d[0]

result = minimize(rmse, alpha, (data,), bounds=[(0,1)], method='SLSQP')

I think you have the alpha and data arguments back-to-front in your function definitions. — ebarr
– ebarr, Commented Aug 4, 2014 at 13:46

ebarr · Accepted Answer · 2014-08-04 13:47:12Z

1

Its hard to tell exactly what the problem is here. I assume that minimize is actually Scipy's minimize.

If so the first argument should be a function. Instead, you are passing the output of the rmse function, which is a double precision number.

error=rmse(d[0], fit) # <--- returns a number

You should have:

result=minimize(<some function here>, alpha, (fit,), bounds=[(0,1)], method='SLSQP')

When minimize is called, it attempts to call error, thus throwing a TypeError: 'numpy.float64' object is not callable

There is a straightforward tutorial here that walks through exactly how to use minimize with the sequential least squares programming optimization algorithm.

I would hazard a guess that you actually want to be passing rmse as the first argument:

result=minimize(rmse, alpha, (fit,), bounds=[(0,1)], method='SLSQP')

After all, the rmse function is giving you the error value and that is what you are minimising in such an optimisation.

edited Aug 4, 2014 at 13:47

answered Aug 4, 2014 at 9:22

ebarr

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1 Comment

ebarr Over a year ago

Please read the documentation for minimize, your current code is clearly passing a number as the first argument.

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