Python: 2D to 1D list in clockwise direction?

Question

I am a bit newbie to Python and I am looking for a function that will convert any n x n 2D list to a 1D list in clockwise direction.

For example:

When n = 3

list = [[2, 3, 5],[ 8, 7, 1],[ 0, 4, 6]]

or

list = [[2, 3, 5]
       ,[8, 7, 1]
       ,[0, 4, 6]]

would become

result = [2, 3, 5, 1, 6, 4, 0, 8, 7]

and when n = 5

list = [[2, 3, 5, 9, 10],[ 8, 7, 1, 11, 13],[ 0, 4, 6, 21, 22], [12, 19, 17, 18, 25], [14, 15, 16, 23, 24]]

or

list = [[  2,  3,  5,  9, 10]
       ,[  8,  7,  1, 11, 13]
       ,[  0,  4,  6, 21, 22]
       ,[ 12, 19, 17, 18, 25]
      , [ 14, 15, 16, 23, 24]]

would become

result = [2, 3, 5, 9, 10, 13, 22, 25, 24, 23, 16, 15, 14, 12, 0, 8, 7, 1, 11, 21, 18, 17, 19, 4, 6]

How can I efficiently do that for any value of nxn ??

Are you looking for something that will work for lists of any nxn? or just 3x3 — Jack Evans, Jun 07 '17 at 12:13
Could you give additional examples for clockwise rotation in differently sized lists? — Jack Evans, Jun 07 '17 at 12:17
Possible duplicate of [Print two-dimensional array in spiral order](https://stackoverflow.com/questions/726756/print-two-dimensional-array-in-spiral-order) — Jack Evans, Jun 07 '17 at 12:45

score 1 · Accepted Answer · answered Jun 07 '17 at 12:49

1

Adapted from Print two-dimensional array in spiral order

import itertools

arr = [[2,  3,  5,  9, 10],
       [8,  7,  1, 11, 13],
       [0,  4,  6, 21, 22],
       [12, 19, 17, 18, 25],
       [14, 15, 16, 23, 24]]


def transpose_and_yield_top(arr):
    while arr:
        yield arr[0]
        arr = list(zip(*arr[1:]))[::-1]


rotated = list(itertools.chain(*transpose_and_yield_top(arr)))

answered Jun 07 '17 at 12:49

Jack Evans

1,697
3
17
33

Appreciated. Thank you :') – Mohammad Zain Abbas Jun 07 '17 at 13:12

Damjan Temelkovski · Answer 2 · 2017-06-07T13:07:07.493

This sounds like an exam question - is it?!

This would be my solution using recursion and Python's list manipulation operators:

def clockwise(input_list, output_list):
    list_size = len(input_list[0])
    if list_size == 1:
        output_list.append(input_list[0][0])
    else:
        for i in range(list_size):
            output_list.append(input_list[0][i])

        for i in range(list_size)[1:]:
            output_list.append(input_list[i][list_size - 1])

        for i in reversed(range(list_size)[:-1]):    
            output_list.append(input_list[list_size - 1][i])

        for i in reversed(range(list_size)[1:-1]):    
            output_list.append(input_list[i][0])

        new_list = list()
        for i in range(list_size - 2):
            new_list.append(input_list[i + 1][1:-1])

        return clockwise(new_list, output_list)

l = [[2, 3, 5, 9, 10],[ 8, 7, 1, 11, 13],[ 0, 4, 6, 21, 22], [12, 19, 17, 18, 25], [14, 15, 16, 23, 24]]
output_list = []
clockwise(l, output_list)

print output_list

No, It's one of my project thing; needed it for Local binary pattern algo. Thank you so much for the help though. :') — Mohammad Zain Abbas, Jun 07 '17 at 13:17

score 1 · Answer 3 · answered Jun 07 '17 at 13:01

This also works:

from math import floor

lists = [[  2,  3,  5,  9, 10]
       ,[  8,  7,  1, 11, 13]
       ,[  0,  4,  6, 21, 22]
       ,[ 12, 19, 17, 18, 25]
      , [ 14, 15, 16, 23, 24]]

n = len(lists) # assume each list also has n-length

output_list = []

idx = 0
while idx <= floor(n/2):

    if len(output_list) == n*n:
        break

    # across ->
    print("Across ->")
    for item in lists[idx][idx:n-idx]:
        output_list.append(item)
    print(output_list)

    if len(output_list) == n*n:
        break

    # down
    print("Down")
    for _idx in range(idx+1, n-idx-1):
        output_list.append(lists[_idx][n-idx-1])
    print(output_list)

    if len(output_list) == n*n:
        break

    # across <-
    print("Across <-")
    for _idx in range(n-idx-1, idx-1, -1):
        output_list.append(lists[n-idx-1][_idx])
    print(output_list)

    if len(output_list) == n*n:
        break

    # up
    print("Up")
    for _idx in range(n-idx-2, idx, -1):
        output_list.append(lists[_idx][idx])
    print(output_list)

    idx += 1


print("\nOutput: ")
print(output_list)

It could be cleaned up in a few ways but the logic is there. — J. Darnell, Jun 07 '17 at 13:03

stamaimer · Answer 4 · 2017-06-07T13:32:52.483

import math
import numpy as np

def print_wall(input, result):

    n = input.shape[0]

    for i in range(n):  # print top outer
        result.append(input[0][i])

    for i in range(n - 1):  # print right outer
        result.append(input[i + 1][n - 1])

    for i in range(n - 1):  # print bottom outer
        result.append(input[n - 1][n - 2 - i])

    for i in range(n - 2):  # print left outer
        result.append(input[n - 2 - i][0])

def clock_wise(input):

    n = input.shape[0]

    result = list()

    for i in range(math.ceil(n / 2)):  # from the outer to the inner

        print_wall(input[i: n - i, i: n - i], result)

    print(result)

This method print from the outer of the array to the inner of the array. For example, The array is:

1  2  3  4  5
6  7  8  9  10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

At first, print the outer of the 5x5 array in the clock wise direction, got 1 2 3 4 5 10 15 20 25 24 23 22 21 16 11 6;

Then, deal with the inner array(3x3):

7  8  9
12 13 14
17 18 19

print the outer of the 3x3 array in the clock wise direction, got 7 8 9 14 19 18 17 12;

At last, deal with 13.

score 0 · Answer 5 · answered Jun 07 '17 at 12:07

0

lists = [[2, 3, 5],[ 8, 7, 1],[ 0, 4, 6]]
output_list = [item for list in lists for item in list]

answered Jun 07 '17 at 12:07

J. Darnell

519
7
15

See, that's the point. You've only converted 2D list into 1D list (missing the point of **clockwise conversion**) – Mohammad Zain Abbas Jun 07 '17 at 12:11

Python: 2D to 1D list in clockwise direction?

5 Answers5