You have two options:

• If you no longer want the original shape, the easiest is just to assign a new shape to the array

  Copya.shape = (a.size//ncols, ncols)
  

  You can switch the a.size//ncols by -1 to compute the proper shape automatically. Make sure that a.shape[0]*a.shape[1]=a.size, else you'll run into some problem.

• You can get a new array with the np.reshape function, that works mostly like the version presented above

  Copynew = np.reshape(a, (-1, ncols))
  

  When it's possible, new will be just a view of the initial array a, meaning that the data are shared. In some cases, though, new array will be acopy instead. Note that np.reshape also accepts an optional keyword order that lets you switch from row-major C order to column-major Fortran order. np.reshape is the function version of the a.reshape method.

If you can't respect the requirement a.shape[0]*a.shape[1]=a.size, you're stuck with having to create a new array. You can use the np.resize function and mixing it with np.reshape, such as

Copy>>> a =np.arange(9)
>>> np.resize(a, 10).reshape(5,2)

A useful alternative answer to the first question, using the examples from tomeedee’s answer, would be to use numpy’s vstack and column_stack methods:

Given a matrix p,

>>> import numpy as np
>>> p = np.array([ [1,2] , [3,4] ])

an augmented matrix can be generated by:

>>> p = np.vstack( [ p , [5 , 6] ] )
>>> p = np.column_stack( [ p , [ 7 , 8 , 9 ] ] )
>>> p
array([[1, 2, 7],
       [3, 4, 8],
       [5, 6, 9]])

These methods may be convenient in practice than np.append() as they allow 1D arrays to be appended to a matrix without any modification, in contrast to the following scenario:

>>> p = np.array([ [ 1 , 2 ] , [ 3 , 4 ] , [ 5 , 6 ] ] )
>>> p = np.append( p , [ 7 , 8 , 9 ] , 1 )
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/lib/python2.6/dist-packages/numpy/lib/function_base.py", line 3234, in append
    return concatenate((arr, values), axis=axis)
ValueError: arrays must have same number of dimensions

In answer to the second question, a nice way to remove rows and columns is to use logical array indexing as follows:

Given a matrix p,

>>> p = np.arange( 20 ).reshape( ( 4 , 5 ) )

suppose we want to remove row 1 and column 2:

>>> r , c = 1 , 2
>>> p = p [ np.arange( p.shape[0] ) != r , : ] 
>>> p = p [ : , np.arange( p.shape[1] ) != c ]
>>> p
array([[ 0,  1,  3,  4],
       [10, 11, 13, 14],
       [15, 16, 18, 19]])

Note - for reformed Matlab users - if you wanted to do these in a one-liner you need to index twice:

>>> p = np.arange( 20 ).reshape( ( 4 , 5 ) )    
>>> p = p [ np.arange( p.shape[0] ) != r , : ] [ : , np.arange( p.shape[1] ) != c ]

This technique can also be extended to remove sets of rows and columns, so if we wanted to remove rows 0 & 2 and columns 1, 2 & 3 we could use numpy's setdiff1d function to generate the desired logical index:

>>> p = np.arange( 20 ).reshape( ( 4 , 5 ) )
>>> r = [ 0 , 2 ]
>>> c = [ 1 , 2 , 3 ]
>>> p = p [ np.setdiff1d( np.arange( p.shape[0] ), r ) , : ] 
>>> p = p [ : , np.setdiff1d( np.arange( p.shape[1] ) , c ) ]
>>> p
array([[ 5,  9],
       [15, 19]])