How can I slice 2D array in python without Numpy module?

Question

What I wanna process is slice 2D array partially without numpy module like following example with numpy.

and I want to know Time Complexity of Slicing Lists in python basic function

import numpy as np    
A = np.array([ [1,2,3,4,5,6,7,8] for i in range(8)])
n = len(A[0])
x = int(n/2)

TEMP = [[None]*2 for i in range(2)]

for w in range(2):
    for q in range(2):
        TEMP[w][q] = A[w*x:w*x+x,q*x:q*x+x]

for w in range(2):
    for q in range(2):
        print(TEMP[w][q])

here is the result that i wanna get

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

Process finished with exit code 0

Salah Eddine Lahniche · Accepted Answer · 2017-03-21 15:05:37Z

1

For the first question:

A = [ [1,2,3,4,5,6,7,8] for i in range(8)]
n = len(A[0])
x = int(n/2)

TEMP = [[None]*2 for i in range(2)]

for w in range(2):
    for q in range(2):
        TEMP[w][q] = [item[q * x:(q * x) + x] for item in A[w * x:(w * x) + x]]

for w in range(2):
    for q in range(2):
        print("{i}, {j}: {item}".format(i=w, j=q, item=repr(TEMP[w][q])))

answered Mar 21, 2017 at 15:05

Salah Eddine Lahniche

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1 Comment

Seung Jin Lee Over a year ago

Thanks a lot for your answer

Asser Yehia · Accepted Answer · 2020-04-15 22:47:08Z

Numpy makes the vertical slicing of 2 dimensional array easier. However, you can achieve the same results without it. Imagine if we have the following 2D list:

arr1=[[1,1,1,0,0,0],[0,1,0,0,0,0],[1,1,1,0,0,0],[0,0,2,4,4,0],[0,0,0,2,0,0],[0,0,1,2,4,0]]

which is represented as the following matrix:

[[1, 1, 1, 0, 0, 0],
 [0, 1, 0, 0, 0, 0],
 [1, 1, 1, 0, 0, 0],
 [0, 0, 2, 4, 4, 0],
 [0, 0, 0, 2, 0, 0],
 [0, 0, 1, 2, 4, 0]]

Let's say that you want to slice this to form the pattern below :

This can be achieved without numpy by the following :

for x in range (0,4):
        for y in range (0,4):
            # Here we traverse through the 2D array vertically
            temp_matrix= arr1[x][y:y+3],arr1[x+1][y:y+3],arr1[x+2][y:y+3]
             print (temp_matrix)

If we use numpy instead we can re-write the following code :

arr=np.array(arr1)
rows = arr.shape[0]
cols = arr.shape[1]
for x in range (0,rows-2):
    for y in range (0, cols-2):
        print(arr[x:x+3,y:y+3])

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