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This is my input:

row=6 col=9

6
9
s b k g s y w g f
r g y e q j j a s
s m s a s z s l e
u s q u e h s s s
g s f h s s e s g
x d r h g y s s s

This is my code:

r=int(input())
c=int(input())
n=min(r,c)
k=3
matrix=[list(map(str,input().split())) for i in range(r)]
t = []
for i in range(0,n-1,k):
  for j in range(0,n-1,k):
    t.append([matrix[i+ii][j+jj] for ii in range(k) for jj in range(k)])
print(t)

output got:

[['s', 'b', 'k', 'r', 'g', 'y', 's', 'm', 's'], ['g', 's', 'y', 'e', 'q', 'j', 'a', 's', 'z'], ['u', 's', 'q', 'g', 's', 'f', 'x', 'd', 'r'], ['u', 'e', 'h', 'h', 's', 's', 'h', 'g', 'y']]

this code not printing my last 3 columns why? Please help me

Comrade
  • 29
  • 2

5 Answers5

1
r=int(input())
c=int(input())
n=min(r,c)
k=3
matrix=[list(map(str,input().split())) for i in range(r)]
t = []
a=[]
for i in range(0,r,k):
  for j in range(0,c,k):
    t.append([matrix[i+ii][j+jj] for ii in range(k) for jj in range(k)])
print(t)

In for loop you used n variable in range. so it will end in the minimum of row and col .That's why you got error..Thank you

Hymdal
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0

If your matrix size is fixed and is divisible by 3 something like this would do the trick:

for i in range(3):
    for j in range(3):
        print(M[i:i+3,j:j+3])

It also works for bigger matrices but it will leave unused elements if it can't index three elements.

Basically I am indexing the matrix M in both axis. For mor info, please, check Numpy docs

Gonzalo Donoso
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0

If a be a NumPy array as:

a = np.array([["s", "b", "k", "g", "s", "y", "w", "g", "f"],
              ["r", "g", "y", "e", "q", "j", "j", "a", "s"],
              ['s', 'm', 's', 'a', 's', 'z', 's', 'l', 'e'],
              ['u', 's', 'q', 'u', 'e', 'h', 's', 's', 's'],
              ['g', 's', 'f', 'h', 's', 's', 'e', 's', 'g'],
              ['x', 'd', 'r', 'h', 'g', 'y', 's', 's', 's']])

Just by one-line NumPy, by little modifying this answer if k=3 --> (3*3):

a.reshape(a.shape[0] // 3, 3, -1, 3).swapaxes(1, 2).reshape(-1, 1, 3 * 3).squeeze()

# [['s' 'b' 'k' 'r' 'g' 'y' 's' 'm' 's']
#  ['g' 's' 'y' 'e' 'q' 'j' 'a' 's' 'z']
#  ['w' 'g' 'f' 'j' 'a' 's' 's' 'l' 'e']
#  ['u' 's' 'q' 'g' 's' 'f' 'x' 'd' 'r']
#  ['u' 'e' 'h' 'h' 's' 's' 'h' 'g' 'y']
#  ['s' 's' 's' 'e' 's' 'g' 's' 's' 's']]

NumPy solution is faster than loops, particularly in case of large array sizes.

Ali_Sh
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0

Сut a matrix into square pieces

If we transform the original list into a numpy.array then we can change its strides to get the desired result without touching the data:

import numpy as np
from numpy.lib.stride_tricks import as_strided

matrix = np.array(matrix)
size = matrix.itemsize
rows, cols, step = r, c, k
shape = rows//step, cols//step, step, step
strides = size*cols*step, size*step, size*cols, size

sliced = as_strided(matrix, shape, strides)

sliced is a view into matrix. With rows, cols, step = 6, 9, 3, this array exposes 2×3 squares of the shape 3×3, i.e. sliced.shape == (2, 3, 3, 3). To get a specific square we use the first two indices.

In order to get a list of lists we apply ravel.tolist to each square:

from itertools import chain
t = [x.ravel().tolist() for x in chain(*sliced)]

Another way to get the same is to use split:

# the matrix here is the original list of lists
t = [square.ravel().tolist() 
    for ribbon in np.vsplit(np.array(matrix), rows//step) 
    for square in np.hsplit(ribbon, cols//step)]
Vitalizzare
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0

If you are using numpy you can use a sliding window : https://numpy.org/devdocs/reference/generated/numpy.lib.stride_tricks.sliding_window_view.html

It is not fastest method but it is really convenient to use for small matrices! For example if you want to extract 3x3 submatrices from the following matrix a:

import numpy as np
from numpy.lib.stride_tricks import sliding_window_view

a = np.array(
    [
        ["s", "b", "k", "g", "s", "y", "w", "g", "f"],
        ["r", "g", "y", "e", "q", "j", "j", "a", "s"],
        ["s", "m", "s", "a", "s", "z", "s", "l", "e"],
        ["u", "s", "q", "u", "e", "h", "s", "s", "s"],
        ["g", "s", "f", "h", "s", "s", "e", "s", "g"],
        ["x", "d", "r", "h", "g", "y", "s", "s", "s"],
    ]
)
print(sliding_window_view(a, (3, 3)).reshape(-1, 3, 3))
tbeolet
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