>>> test[:,0]
array([1, 3, 5])

this command gives you a row vector, if you just want to loop over it, it's fine, but if you want to hstack with some other array with dimension 3xN, you will have

ValueError: all the input arrays must have same number of dimensions

while

>>> test[:,[0]]
array([[1],
       [3],
       [5]])

gives you a column vector, so that you can do concatenate or hstack operation.

e.g.

>>> np.hstack((test, test[:,[0]]))
array([[1, 2, 1],
       [3, 4, 3],
       [5, 6, 5]])

This should give you a start

>>> for col in range(arr.shape[1]):
    some_function(arr[:,col])


[1 2 3 4]
[99 14 12 43]
[2 5 7 1]

This is a trick that it is slightly more general than the other useful answers here:

def normalize(array, imin = -1, imax = 1):
    """I = Imin + (Imax-Imin)*(D-Dmin)/(Dmax-Dmin)"""

    dmin = array.min()
    dmax = array.max()

    array[...] = imin + (imax - imin)*(array - dmin)/(dmax - dmin)

Here we are assigning values to the view array[...] rather than assigning these values to some new local variable within the scope of the function.

x = np.arange(5, dtype='float')
print x
normalize(x)
print x

>>> [0. 1. 2. 3. 4.]
>>> [-1.  -0.5  0.   0.5  1. ]

EDIT:

It's slower; it allocates a new array. But it may be valuable if you are doing something more complicated where builtin in-place operations are cumbersome or don't suffice.

def normalize2(array, imin=-1, imax=1):
    dmin = array.min()
    dmax = array.max()

    array -= dmin;
    array *= (imax - imin)
    array /= (dmax-dmin)
    array += imin

A = np.random.randn(200**3).reshape([200] * 3)
%timeit -n5 -r5 normalize(A)
%timeit -n5 -r5 normalize2(A)

>> 47.6 ms ± 678 µs per loop (mean ± std. dev. of 5 runs, 5 loops each)
>> 26.1 ms ± 866 µs per loop (mean ± std. dev. of 5 runs, 5 loops each)