creating a spiral array in python?

Question

Me and my mate were trying to create a fun game in python where the elements entered in the array are accessed in a spiral manner. I have tried few methods like one given below (source).

def spiral(X, Y):
  x = y = 0
  dx = 0
  dy = -1
  for i in range(max(X, Y)**2):
    if (-X/2 < x <= X/2) and (-Y/2 < y <= Y/2):
        print (x, y)
        # DO STUFF...
    if x == y or (x < 0 and x == -y) or (x > 0 and x == 1-y):
        dx, dy = -dy, dx
    x, y = x+dx, y+dy

The above statement accesses the elements in spiral loop and prints them for a defined array AE. I would like to know how can I transform a given array AE to a spiral one

Answer 1

Introductory remarks

The question is closely related to a problem of printing an array in spiral order. In fact, if we already have a function which does it, then the problem in question is relatively simple.

There is a multitude of resources on how to produce a spiral matrix or how to loop or print an array in spiral order. Even so, I decided to write my own version, using numpy arrays. The idea is not original but use of numpy makes the code more concise.

The other reason is that most of examples of producing a spiral matrix I found (including the code in the question and in the other answers) deal only with square matrices of size n x n for odd n. Finding the start (or end) point in matrices of other sizes may be tricky. For example, for a 3x5 matrix it can't be the middle cell. The code below is general and the position of the starting (ending) point depends on the choice of the function spiral_xxx.

Code

The first function unwraps an array in spiral order clockwise:

import numpy as np

def spiral_cw(A):
    A = np.array(A)
    out = []
    while(A.size):
        out.append(A[0])        # take first row
        A = A[1:].T[::-1]       # cut off first row and rotate counterclockwise
    return np.concatenate(out)

We can write this function on eight different ways depending on where we start and how we rotate the matrix. I'll give another one, which is consistent (it will be evident later) with the matrix transformation in the image in the question. So, further on, I will be using this version:

def spiral_ccw(A):
    A = np.array(A)
    out = []
    while(A.size):
        out.append(A[0][::-1])    # first row reversed
        A = A[1:][::-1].T         # cut off first row and rotate clockwise
    return np.concatenate(out)

How it works:

A = np.arange(15).reshape(3,5)
print(A)
[[ 0  1  2  3  4]
 [ 5  6  7  8  9]
 [10 11 12 13 14]]

print(spiral_ccw(A))
[ 4  3  2  1  0  5 10 11 12 13 14  9  8  7  6]

Note that the end (or start) point is not the middle cell. This function works for all type of matrices but we will need a helper function that generates spiral indices:

def base_spiral(nrow, ncol):
    return spiral_ccw(np.arange(nrow*ncol).reshape(nrow,ncol))[::-1]

For example:

print(base_spiral(3,5))
[ 6  7  8  9 14 13 12 11 10  5  0  1  2  3  4]

Now come the two main functions. One transforms a matrix to a spiral form of the same dimensions, the other reverts the transformation:

def to_spiral(A):
    A = np.array(A)
    B = np.empty_like(A)
    B.flat[base_spiral(*A.shape)] = A.flat
    return B

def from_spiral(A):
    A = np.array(A)
    return A.flat[base_spiral(*A.shape)].reshape(A.shape)

Examples

Matrix 3 x 5:

A = np.arange(15).reshape(3,5)
print(A)
[[ 0  1  2  3  4]
 [ 5  6  7  8  9]
 [10 11 12 13 14]]

print(to_spiral(A))
[[10 11 12 13 14]
 [ 9  0  1  2  3]
 [ 8  7  6  5  4]]

print(from_spiral(to_spiral(A)))
[[ 0  1  2  3  4]
 [ 5  6  7  8  9]
 [10 11 12 13 14]]

Matrix from the question:

B = np.arange(1,26).reshape(5,5)
print(B)
[[ 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]]

print(to_spiral(B))
[[21 22 23 24 25]
 [20  7  8  9 10]
 [19  6  1  2 11]
 [18  5  4  3 12]
 [17 16 15 14 13]]

print(from_spiral(to_spiral(B)))
[[ 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]]

Remark

If you are going to work only with fixed size matrices, for example 5x5, then it's worth replacing base_spiral(*A.shape) in definitions of the functions with a fixed matrix of indices, say Ind (where Ind = base_spiral(5,5)).

Answer 2

You can build a spiral by starting near the center of the matrix and always turning right unless the element has been visited already:

#!/usr/bin/env python
NORTH, S, W, E = (0, -1), (0, 1), (-1, 0), (1, 0) # directions
turn_right = {NORTH: E, E: S, S: W, W: NORTH} # old -> new direction

def spiral(width, height):
    if width < 1 or height < 1:
        raise ValueError
    x, y = width // 2, height // 2 # start near the center
    dx, dy = NORTH # initial direction
    matrix = [[None] * width for _ in range(height)]
    count = 0
    while True:
        count += 1
        matrix[y][x] = count # visit
        # try to turn right
        new_dx, new_dy = turn_right[dx,dy]
        new_x, new_y = x + new_dx, y + new_dy
        if (0 <= new_x < width and 0 <= new_y < height and
            matrix[new_y][new_x] is None): # can turn right
            x, y = new_x, new_y
            dx, dy = new_dx, new_dy
        else: # try to move straight
            x, y = x + dx, y + dy
            if not (0 <= x < width and 0 <= y < height):
                return matrix # nowhere to go

def print_matrix(matrix):
    width = len(str(max(el for row in matrix for el in row if el is not None)))
    fmt = "{:0%dd}" % width
    for row in matrix:
        print(" ".join("_"*width if el is None else fmt.format(el) for el in row))

Example:

>>> print_matrix(spiral(5, 5))
21 22 23 24 25
20 07 08 09 10
19 06 01 02 11
18 05 04 03 12
17 16 15 14 13