This can be accomplished using a simple helper function in pure python:

def mirror(seq):    output = list(seq[::-1])    output.extend(seq[1:])    return outputinputs = [   ['a', 'b', 'c'],   ['d', 'e', 'f'],   ['g', 'h', 'i'],]print(mirror([mirror(sublist) for sublist in inputs]))

Obviously, once the mirrored list is created, you can use it to create a numpy array or whatever...

Usenumpy.lib.pad with'reflect'

m = [['a', 'b', 'c'],      ['d', 'e', 'f'],      ['g', 'h', 'i']]n=np.lib.pad(m,((2,0),(2,0)),'reflect')nOut[8]: array([['i', 'h', 'g', 'h', 'i'],       ['f', 'e', 'd', 'e', 'f'],       ['c', 'b', 'a', 'b', 'c'],       ['f', 'e', 'd', 'e', 'f'],       ['i', 'h', 'g', 'h', 'i']],       dtype='<U1')

import numpy as npX= [[1, 2, 3],    [4, 5, 6],    [7, 8, 9]]A = np.asanyarray(X)B= np.flipud(A)C= np.concatenate((B, A[1:]), axis=0)D = C[:,1:]F = np.fliplr(C)E = np.concatenate((F, D), axis=1)print(E)

I have added step by step transformation.flipud andflipud refrence

output

[[9 8 7 8 9] [6 5 4 5 6] [3 2 1 2 3] [6 5 4 5 6] [9 8 7 8 9]]

This is taggednumpy so I'll assume your matrix is a 2d array

In [937]: A=np.arange(9).reshape(3,3)In [938]: AOut[938]: array([[0, 1, 2],       [3, 4, 5],       [6, 7, 8]])

flipping it on the rows:

In [939]: A[::-1,:]Out[939]: array([[6, 7, 8],       [3, 4, 5],       [0, 1, 2]])

concatenating vertically

In [940]: np.concatenate((A[::-1,:],A), axis=0)Out[940]: array([[6, 7, 8],       [3, 4, 5],       [0, 1, 2],       [0, 1, 2],       [3, 4, 5],       [6, 7, 8]])

removing the duplicate first line

In [941]: np.concatenate((A[::-1,:],A[1:,:]), axis=0)Out[941]: array([[6, 7, 8],       [3, 4, 5],       [0, 1, 2],       [3, 4, 5],       [6, 7, 8]])

Do you think you can do the same with a horizontal (column) reversal and concatenate (axis=1)?

And here is a non-numpy solution:

a = [['a', 'b', 'c'], ['d', 'e', 'f'], ['g', 'h', 'i']]b = list(reversed(a[1:])) + a # vertical mirrorc = list(zip(*b)) # transposed = list(reversed(c[1:])) + c # another vertical mirrore = list(zip(*d)) # transpose again

Suppose you have

from numpy import array, concatenatem = array([[1, 2, 3],           [4, 5, 6],           [7, 8, 9]])

You can invert this along the first (vertical) axis via

>>> m[::-1, ...]array([[7, 8, 9],       [4, 5, 6],       [1, 2, 3]])

where::-1 selects the rows from last to first in steps of-1.

To omit the last row, explicitly ask for the selection to stop immediately before0:

>>> m[:0:-1, ...]array([[7, 8, 9],       [4, 5, 6]])

This can then be concatenated along the first axis

p = concatenate([m[:0:-1, ...], m], axis=0)

to form:

>>> parray([[7, 8, 9],       [4, 5, 6],       [1, 2, 3],       [4, 5, 6],       [7, 8, 9]])

This can be repeated along the other axis too:

q = concatenate([p[..., :0:-1], p], axis=1)

yielding

>>> qarray([[9, 8, 7, 8, 9],       [6, 5, 4, 5, 6],       [3, 2, 1, 2, 3],       [6, 5, 4, 5, 6],       [9, 8, 7, 8, 9]])

m = [['a', 'b', 'c'],      ['d', 'e', 'f'],      ['g', 'h', 'i']]m_m = [[m[abs(i)][abs(j)]       for j in range(-len(m)+1, len(m))]       for i in range(-len(m)+1, len(m))]

Or using numpy

m = array([['a', 'b', 'c'],            ['d', 'e', 'f'],            ['g', 'h', 'i']])m_m = m.T[meshgrid(*2*[abs(arange(-len(m) + 1, len(m)))])]

• python
• arrays
• python-3.x
• numpy
• matrix