Scipy optimize fmin ValueError: setting an array element with a sequence

Question

When using scipy.optimize's fmin I'm getting an error I don't understand:

ValueError: setting an array element with a sequence.

Here's a simple squared error example to demonstrate:

import numpy as np
from scipy.optimize import fmin

def cost_function(theta, X, y):    
    m = X.shape[0]
    error = X.dot(theta) - y 
    J = 1/(2*m) * error.T.dot(error)  
    return J

X = np.array([[1., 1.],
              [1., 2.],
              [1., 3.],
              [1., 4.]])

y = np.array([[2],[4],[6],[8]])   
initial_theta = np.ones((X.shape[1], 1)) * 0.01

# test cost_function
print cost_function(initial_theta, X, y)
# [[ 14.800675]] seems okay...

# but then error here...   
theta = fmin(cost_function, initial_theta, args=(X, y))

#Traceback (most recent call last):
#  File "C:\Users\me\test.py", line 21, in <module>
#    theta = fmin(cost_function, initial_theta, args=(X, y))
#  File "C:\Python27\lib\site-packages\scipy\optimize\optimize.py", line 278, in fmin
#    fsim[0] = func(x0)
#ValueError: setting an array element with a sequence.

I'd be grateful for any help to explain where I'm going wrong.

You're returning an array for the cost function, and fmin expects a single value. Either return J[0,0] or rewrite your cost function as J = 1/(2*m) * np.sum(error**2) — Joe Kington
– Joe Kington, Commented Mar 20, 2012 at 0:43

segasai · Accepted Answer · 2012-03-20 00:49:33Z

9

The reason is that the starting point (initial_theta) you gave to fmin is not a 1D array but a 2D array. So on a second iteration fmin passes a 1D array (that's how it supposed to work) and the result becomes non-scalar.

So you should refactor your cost function to accept 1d arrays as a first argument.

The simplest change is to make the code working is to flatten the initial_theta before passing to fmin and reshape theta inside cost_function to (X.shape[1],1) if you like.

edited Mar 20, 2012 at 0:49

answered Mar 20, 2012 at 0:43

segasai

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1 Comment

Kim Over a year ago

Thanks for the explanation. Yes, the number of dimensions was confusing me. So, I've now initially set both theta and y to one dim, and the cost function return value works out as a scalar without any further adjustment. Cheers.

YXD · Accepted Answer · 2012-03-20 00:41:48Z

7

cost_function should return a scalar, but your return value J is an array of some kind.

answered Mar 20, 2012 at 0:41

YXD

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