Unable to print variables in Python when using def function

Question

I am trying to implement a simple neural net. I want to print the initial pattern, weights, activation. I then want it to print the learning process (i.e. every pattern it goes through as it learns). I am as yet unable to do this - it returns the initial and final pattern (whn I put print p in appropriate places), but nothing else. Hints and tips appreciated - I'm a complete newbie to Python!

#!/usr/bin/python
import random

p = [ [1, 1, 1, 1, 1],
      [1, 1, 1, 1, 1],
      [0, 0, 0, 0, 0],
      [1, 1, 1, 1, 1],
      [1, 1, 1, 1, 1] ] # pattern I want the net to learn
n = 5
alpha = 0.01
activation = []   # unit activations
weights = []      # weights
output = []       # output



def initWeights(n):  # set weights to zero, n is the number of units
    global weights 
    weights = [[[0]*n]*n]   # initialised to zero


def initNetwork(p):  # initialises units to activation
    global activation
    activation = p

def updateNetwork(k): # pick unit at random and update k times
    for l in range(k):
        unit = random.randint(0,n-1)
        activation[unit] = 0
        for i in range(n):
            activation[unit] += output[i] * weights[unit][i]
        output[unit] = 1 if activation[unit] > 0 else -1

def learn(p):
    for i in range(n):
        for j in range(n):
            weights += alpha * p[i] * p[j]

One thing to check is the PEP-8 coding standard for python. Generally, everyone uses 4 spaces for indentation of python code, along with a number of other conventions: python.org/dev/peps/pep-0008 Quick summary here wwd.ca/blog/2009/07/09/pep-8-cheatsheet — mcpeterson
– mcpeterson, Commented Oct 18, 2010 at 18:57

salezica · Accepted Answer · 2010-10-18 18:33:05Z

7

You have a problem with the line:

weights = [[[0]*n]*n]

When you use*, you multiply object references. You are using the same n-len array of zeroes every time. This will cause:

>>> weights[0][1][0] = 8
>>> weights
[[[8, 0, 0], [8, 0, 0], [8, 0, 0]]]

The first item of all the sublists is 8, because they are one and the same list. You stored the same reference multiple times, and so modifying the n-th item on any of them will alter all of them.

answered Oct 18, 2010 at 18:33

salezica

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5 Comments

jlv Over a year ago

What's the best way to generate a null array then? List comprehensions?

Jochen Ritzel Over a year ago

[[0 for _ in range(n)] for _ in range(n)]

salezica Over a year ago

A null list is not the same as a list of null lists. You can use the technique above to generate a single null list, it's multiplying that list where the problem arises. I'd go for list comprehension with xrange.

salezica Over a year ago

Didn't see THC's answer. That's right, only you should use xrange if n is big.

Daenyth Over a year ago

More generally, using * on any mutable object will create the same issue of shared mutability

mouad · Accepted Answer · 2010-10-18 19:30:48Z

0

this the line is where you get : "IndexError: list index out of range"

output[unit] = 1 if activation[unit] > 0 else -1

because output = [] , you should do output.append() or ...

answered Oct 18, 2010 at 19:30

mouad

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