在二維陣列中創建黑白棋盤

Question

有沒有更好（也是更短）的如何創建類似棋盤棋盤的方法。對董事會的要求是：

• 板可以是不同的大小（以我的例子是3×3）董事會
• 左下方應始終是黑色
• 黑色方塊是由"B"介紹，白方是通過"W"

碼錶示，我有：

def isEven(number): 
    return number % 2 == 0 

board = [["B" for x in range(3)] for x in range(3)] 
if isEven(len(board)): 
    for rowIndex, row in enumerate(board): 
     if isEven(rowIndex + 1): 
      for squareIndex, square in enumerate(row): 
       if isEven(squareIndex + 1): 
        board[rowIndex][squareIndex] = "W" 
     else: 
      for squareIndex, square in enumerate(row): 
       if not isEven(squareIndex + 1): 
        board[rowIndex][squareIndex] = "W" 
else: 
    for rowIndex, row in enumerate(board): 
     if not isEven(rowIndex + 1): 
      for squareIndex, square in enumerate(row): 
       if isEven(squareIndex + 1): 
        board[rowIndex][squareIndex] = "W" 
     else: 
      for squareIndex, square in enumerate(row): 
       if not isEven(squareIndex + 1): 
        board[rowIndex][squareIndex] = "W" 

for row in board: 
    print row 

輸出：

['B', 'W', 'B'] 
['W', 'B', 'W'] 
['B', 'W', 'B'] 

Answer 1

如何：

>>> n = 3 
>>> board = [["BW"[(i+j+n%2+1) % 2] for i in range(n)] for j in range(n)] 
>>> print board 
[['B', 'W', 'B'], ['W', 'B', 'W'], ['B', 'W', 'B']] 
>>> n = 4 
>>> board = [["BW"[(i+j+n%2+1) % 2] for i in range(n)] for j in range(n)] 
>>> print board 
[['W', 'B', 'W', 'B'], ['B', 'W', 'B', 'W'], ['W', 'B', 'W', 'B'], ['B', 'W', 'B', 'W']] 

Answer 2

樣的黑客，但

print [["B" if abs(n - row) % 2 == 0 else "W" for n in xrange(3)] for row in xrange(3)][::-1] 

這似乎是要求蠕變什麼的=）

def make_board(n): 
    ''' returns an empty list for n <= 0 ''' 
    return [["B" if abs(c - r) % 2 == 0 else "W" for c in xrange(n)] for r in xrange(n)][::-1] 

Answer 3

for i in range(len(board)): 
    for j in range(len(board)): 
     if isEven(i + j + len(board)): 
      board[i][j] = "W" 

Answer 4

這一個正確設置左下角爲 'B'，總是：

def board(n): 
    def line(n, offset): 
     return [(i+offset) % 2 and 'W' or 'B' for i in range(n)] 
    return [line(n,i) for i in range(n+1,1,-1)] 

Answer 5

下面是一個itertools溶液：

from itertools import cycle 
N = 4 

colors = cycle(["W","B"]) 
row_A = [colors.next() for _ in xrange(N)] 
if not N%2: colors.next() 
row_B = [colors.next() for _ in xrange(N)] 

rows = cycle([row_A, row_B]) 
board = [rows.next() for _ in xrange(N)] 

對於N=4這給出

['W', 'B', 'W', 'B'] 
['B', 'W', 'B', 'W'] 
['W', 'B', 'W', 'B'] 
['B', 'W', 'B', 'W'] 

這應該是可擴展到多種顏色（例如，如果確保將每個新行和循環添加到行列表中，則該板將變爲「B」，「W」，「G」）。

Answer 6

殘酷地容易理解。此外，可以生成一個矩形板：

def gen_board(width, height): 
    bw = ['B', 'W'] 
    l = [[bw[(j + i) % 2] for j in range(width)] for i in range(height)] 
    # this is done to make sure B is always bottom left 
    # alternatively, you could do the printing in reverse order 
    l.reverse() 

    ## or, we could ensure B is always bottom left by adjusting the index 
    #offset = height%2 + 1 
    #l = [[bw[(j + i + offset) % 2] for j in range(width)] for i in range(height)] 
    return l 

def print_board(b): 
    for row in b: 
     print row 

試駕：

>>> print_board(gen_board(4, 3)) 
['B', 'W', 'B', 'W'] 
['W', 'B', 'W', 'B'] 
['B', 'W', 'B', 'W'] 

Answer 7

隨着一個線numpy的代碼，而無需循環：

import numpy as np 

chessbool = (np.arange(3)[:, None] + np.arange(3)) % 2 == 0 

並且輸出是：

array([[ True, False, True], 
     [False, True, False], 
     [ True, False, True]] 

要用填充陣列和B：

chessboard = np.where(chessbool,'B','W') 

，輸出是：

array([['B', 'W', 'B'], 
     ['W', 'B', 'W'], 
     ['B', 'W', 'B']])