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Question

我在这里有一个python类方法的怪胎。我认为做这件事的唯一方法是一个巨大而丑陋的if/elif/else块。你们有没有想出任何想法来使这个更好？

对于上下文来说，这是pygame的一个网格制作库的一部分，它是一个函数，它接受网格中的一个瓦片并返回周围的瓦片。如果“水平”设置为false，则它仅返回与瓦片垂直相邻的瓦片，反之亦然。

def getSurroundingTiles(self, tile, horizontal = True, vertical = True): 
    index = list(self.getTiles()).index(tile) 
    maxtile = self.sqrtnum - 1 # Offset for 0 indexing 

    i = int(math.floor(index/self.sqrtnum)) 
    j = int(index % self.sqrtnum) 

    surroundingTiles = [] 

    if i == 0 and j == 0: 
     #Top left corner 
     if horizontal: 
      surroundingTiles.extend((self[i + 1][j], self[i][j + 1])) 
     if vertical: 
      surroundingTiles.append(self[i + 1][j + 1]) 
    elif i >= maxtile and j == 0: 
     #Top right corner 
     if horizontal: 
      surroundingTiles.extend((self[i - 1][j], self[i][j + 1])) 
     if vertical: 
      surroundingTiles.append(self[i - 1][j + 1]) 
    elif i == 0 and j >= maxtile: 
     #Bottom left corner 
     if horizontal: 
      surroundingTiles.extend((self[i + 1][j], self[i][j - 1])) 
     if vertical: 
      surroundingTiles.append(self[i + 1][j - 1]) 
    elif i >= maxtile and j >= maxtile: 
     #Bottom right corner 
     if horizontal: 
      surroundingTiles.extend((self[i - 1][j], self[i][j - 1])) 
     if vertical: 
      surroundingTiles.append(self[i - 1][j - 1]) 

    elif i == 0: 
     #Top border 
     if horizontal: 
      surroundingTiles.extend((self[i + 1][j], self[i][j + 1], 
            self[i][j - 1])) 
     if vertical: 
      surroundingTiles.extend((self[i + 1][j + 1], 
            self[i + 1][j - 1])) 
    elif i >= maxtile: 
     #Bottom border 
     if horizontal: 
      surroundingTiles.extend((self[i - 1][j], self[i][j + 1], 
            self[i][j - 1])) 
     if vertical: 
      surroundingTiles.extend((self[i - 1][j + 1], 
            self[i - 1][j - 1])) 
    elif j == 0: 
     #Left border 
     if horizontal: 
      surroundingTiles.extend((self[i + 1][j], self[i - 1][j], 
            self[i][j + 1])) 
     if vertical: 
      surroundingTiles.extend((self[i + 1][j + 1], 
            self[i - 1][j + 1])) 
    elif j >= maxtile: 
     #Right border 
     if horizontal: 
      surroundingTiles.extend((self[i + 1][j], self[i - 1][j], 
            self[i][j - 1])) 
     if vertical: 
      surroundingTiles.extend((self[i + 1][j - 1], 
            self[i - 1][j - 1])) 

    else: 
     if horizontal: 
      surroundingTiles.extend((self[i + 1][j], self[i - 1][j], 
            self[i][j + 1], self[i][j - 1])) 
     if vertical: 
      surroundingTiles.extend((self[i + 1][j + 1], self[i + 1][j - 1], 
            self[i - 1][j + 1], self[i - 1][j - 1])) 

    return surroundingTiles 

Answer 1

尝试是这样的：（未测试）

# indices 0 - 3 are for horizontal, 4 - 7 are for vertical 
dij = [(0, 1), (1, 0), (0, -1), (-1, 0), 
     (1, 1), (1, -1), (-1, 1), (-1, -1)] 

def getSurroundingTiles(self, tile, horizontal = True, vertical = True): 
    index = list(self.getTiles()).index(tile) 
    maxtile = self.sqrtnum - 1 # Offset for 0 indexing 

    i = int(math.floor(index/self.sqrtnum)) 
    j = int(index % self.sqrtnum) 

    surroundingTiles = [] 

    startat = 0 if horizontal else 4 
    stopat = 8 if vertical else 4 

    for di, dj in dij[startat:stopat]: 
     if 0 <= i + di <= maxtile and 0 <= j + dj <= maxtile: 
      surroundingTiles.append(self[i + di][j + dj]) 

    return surroundingTiles 

它计算邻居指数相对使用单位的步骤，使您不必下明确键入出来。它也一次测试所有out of bounds个案。

我假设你的意思是horizontal和vertical，就像horizontally and vertically adjacent和diagonally adjacent，正如代码所示。

Answer 2

以@ irrelephant的想法更进一步（它是合乎逻辑的结论或归谬法，你决定）：

d = (
    ((0, 1), (1, 0), (0,-1), (-1, 0)), 
    ((1, 1), (1,-1), (-1, 1), (1,-1)) 
) 

def getSurroundingTiles(self, tile, horizontal = True, vertical = True): 
    index = list(self.getTiles()).index(tile) 
    maxtile = self.sqrtnum - 1 # Offset for 0 indexing 

    fhv = (horizontal, vertical) 
    ij = (int(math.floor(index/self.sqrtnum)), 
      int(index % self.sqrtnum)) 

    surroundingTiles = [] 
    for ihv in range(2): 
     if fhv[ihv]: 
      for k in range(4): 
       n = [sum(p) for p in zip(ij, d[ihv][k])] 
       if all([0 <= n[i] <= maxtile for i in range(2)]): 
        surroundingTiles.append(self[n[0]][n[1]]) 

    return surroundingTiles 

请注意：此代码是未经测试。