There is this issue about lists in python I have encountered which is probably about the code I wrote so some insight will be appreciated.
Problem Wk.6.1.3: A bigger bunch of zeros and code goes wrong is :
def zeroVector(n):
a=list(range(n))
b=[]
k=0
while k

Question

There is this issue about lists in python I have encountered which is probably about the code I wrote so some insight will be appreciated.
 Problem Wk.6.1.3: A bigger bunch of zeros and code goes wrong is :
 def zeroVector(n):
    a=list(range(n))
    b=[]
    k=0
    while k<n:
        b.append(0)
        k+=1
    for e in a:
        a[e]=0
        print(a)
    return a
def zeroArray(m,n):
    foo=zeroVector(m)
    print(foo)
    for f in foo:
        foo[f]=zeroVector(n)
        print(foo)
    return foo 

which has this return for zeroArray(3,4):
[[0, 0, 0, 0], 0, 0]

anonymous · Answer

Are you trying to generate a two dimensional list of zeros? I'm not sure I understand your purpose, but if so there are a few ways to do it.

For a list:

    def zeroArray(m, n):
        return [[0]*n]*m

Alternatively, if you want a numpy array:

    import numpy as np
    def zeroArray(m,n):
        return np.zeros(m*n).reshape(m,n)

anonymous · Answer

Also, I see what you did wrong in your code, if you do want to do it that way.

You say:
    foo=zeroVector(m)
    print(foo)
    for f in foo:
        foo[f]=zeroVector(n)
        print(foo)

However, look at what foo is. It is zeroVector(m), which outputs a vector full of zeros that is of length m. So you are looping over values of zero m times.

Essentially you are doing foo[0]=[0,0....,0] a lot of times.

Your for statement should be 'for f in range(len(foo)):'

anonymous · Answer

Thank you a lot.Apart from the first statement does it nicely, now I know a bit more about what numpy does that I didnt bother when it refused to install (have both python 2.7 and 3.3).However from my ambigious question this remains still; if foo is a list and "for ... in" statement iterates on any list's elements why should use  range(len(foo)).
 And once again you've earned best credits for your great help already

anonymous · Answer

Imagine a list ">>> alphabet = ['a', 'b', 'c']"
When you do 
    for x in alphabet:
        print x
What you are doing is iterating over the elements in alphabet. So the output would be:
    'a'
    'b'
    'c'

So in the original statement, you say ">>>for f in foo:". But foo is a list of zeros. So you are repeatedly giving f a value of zero. You then loop through foo[f]. So what are you doing? You are assigning things to the 0th index of foo and then assigning something else to foo[0]. You are saying foo[0] = [0,0,0...,0] a lot of times in a row.

range(n) creates a list object that goes from 0 -> n-1. [0, 1, 2, 3, ..., n-1]. You are trying to access the f index of foo. Now you are saying
    foo[0] = [0,0,0...,0]
    foo[1] = [0,0,0...,0]
    ...
    foo[n-1] = [0,0,0...,0]
Which is what you were trying to do.

anonymous · Answer

So if I understand correctly iteration element f (f in foo) has the value of 0 (more elaborately any item's value by an index number from 0 to top which is not shown).I got it now.I began to dig through some other lessons as well and YOU have been a great help.
 Yet when everything looked settled a new issue arose;if we use the code below
 
   def zeroArray(m, n):
        return [[0]*n]*m
 
and try to assign 2nd element of the 2nd element of constructed list say like:

   z=zeroArray(3,4)  

by

  z[2][2]=1 

value of z becomes;

   [[0, 0, 1, 0], [0, 0, 1, 0], [0, 0, 1, 0]]

you are already a charm even if you cant tell me if any way exists to make such assignment