python - 我怎样才能将我的两个 for 循环重新组合成一个，两者之间只有微小的差异？

Question

我有以下代码，唯一的区别是列表中的 j 和 I 位置。有什么方法可以通过编写一个函数或类似的东西来让它变得更好，因为我不太明白吗？

for i in range(dimension):
    for j in range(dimension - 4):
        k = 1
        while k < 5 and gameboard[i][j+k] == gameboard[i][j]:
            k += 1
        if k == 5:
            winner = gameboard[i][j]

for i in range(dimension):
    for j in range(dimension - 4):
        k = 1
        while k < 5 and gameboard[j+k][i] == gameboard[j][i]:
            k += 1
        if k == 5:
            winner = gameboard[j][i]

Answer 1

您可以通过插入一个额外的嵌套循环来合并这两个循环，该循环处理 i 和 j 的排列以及相应的维度增量:

for i in range(dimension):
    for j in range(dimension - 4):
        for i,j,di,dj in [ (i,j,0,1), (j,i,1,0) ]:
            k = 1
            while k < 5 and gameboard[i+k*di][j+k*dj] == gameboard[i][j]:
                k += 1
            if k == 5:
                winner = gameboard[i][j]

所有方向的通用化

或者，您可以创建一个函数，告诉您在给定方向上是否有胜利，并在循环中使用它。

def directionWinner(board,i,j,di,dj):
    player,pi,pj = board[i][j],i,j
    # if player == empty: return
    for _ in range(4):
        pi,pj = pi+di, pj+dj
        if pi not in range(len(board)): return
        if pj not in range(len(board)): return
        if board[pi][pj] != player:     return  
    return player

然后用它来检查所有方向:

for pos in range(dimensions*dimensions):
    i,j = divmod(pos,dimensions)
    for direction in [ (0,1),(1,0),(1,1),(1,-1) ]:
        winner = directionWinner(gameboard,i,j,*direction)
        if winner is not None: break
    else: continue; break

方向由每一步的垂直和水平坐标 (delta) 的增加/减少表示。所以 [ (0,1),(1,0),(1,1),(1,-1) ] 给你 "down", "across", "diagonal 1", diagonal分别为 2"。

只检查最后一步

同样的想法可以用来检查特定位置的获胜者(例如检查最后一步是否获胜):

# count how may consecutive in a given direction (and its inverse)
def countDir(board,i,j,di,dj,inverted=False):
    player,pi,pj = board[i][j],i,j
    count  = 0
    while pi in range(len(board) and pj in range(len(board)):
          if board[pi][pj] == player: count += 1
          else: break
          pi, pj = pi+di, pj+dj 
    if not inverted: 
       count += countDir(board,i,j,-di,-dj,True)-1
    return count

 def winnerAt(board,i,j):
     for direction in [ (0,1),(1,0),(1,1),(1,-1) ]:
         if countDir(board,i,j,*direction)>=5:
             return board[i,j]

然后，在i,j位置下棋后，可以立即知道这步棋是否获胜:

if winnerAt(gameboard,i,j) is not None:
    print(gameboard[i][j],"wins !!!")