get the column from 2d array to calculate the normalization and cross product in python

It depends on how you set up your array to start with. I like to use numpy arrays as I find the indexing easier to get my head around. I think the below code is what you are after. As you always have 3 colulmns it doesnt matter how long A is, you can just slice it into 3 columns.

import numpy as np
A=np.array([[-0.00022939, -0.04265404, 0.00022939], 
            [-0.00022939, -0.04265404, 0.00022939], 
            [0., -0.2096513, 0.],
            [0.00026388, 0.00465183, 0.00026388]])
for idx in range(3):
    b = A[:, idx]
    print b # call your function here

EDIT:: Full implementation showing the code & the output

import numpy as np
def vectors(b): 
    b = b/np.sqrt(np.sum(b**2.,axis=0))
    b = b/np.linalg.norm(b)     
    z = np.array([0.,0.,1.])
    n1 = np.cross(z,b,axis=0)
    n1 = n1/np.linalg.norm(n1) ## normalize n
    return [n1] 
A=np.array([[-0.00022939, -0.04265404, 0.00022939],
            [ 0.,         -0.2096513,  0.        ],
            [ 0.00026388, 0.00026388,  0.00026388]])
for idx in range(3):
    b = A[:, idx]
    n1 = vectors(b)
    print 'idx', idx, '\nb ', b, '\nn1 ', n1, '\n'
Output:
    idx 0 
    b  [-0.00022939  0.          0.00026388] 
    n1  [array([ 0., -1.,  0.])] 
    idx 1 
    b  [-0.04265404 -0.2096513   0.00026388] 
    n1  [array([ 0.9799247 , -0.19936794,  0.        ])] 
    idx 2 
    b  [ 0.00022939  0.          0.00026388] 
    n1  [array([ 0.,  1.,  0.])] 

You can try this:

A=[[1,2,3],[4,5,6],[7,8,9]]
def getColumn(m):
    res=[]
    for x in A:
        res.append(x[m])
    return res
def countSomething(x):
    # counting code here
    print x
def looper(n): # n is the second dimension size
    for x in xrange(0,n):
        countSomething(getColumn(x))
looper(3)