如何在Python中有效地重用代码

Question

我正在做一个python项目，其中的一部分涉及在游戏中排列棋盘状态。这个游戏，Sneaky雕像，类似于连接四个，你可以连续获得四个棋子。我对董事会的天真估计是你连续有多少件（1,2,3或4）。棋盘是一个三角形，所以你可以将棋子连成一排，或者沿任一方向对角。这是我现在使用的函数来查找片段。

def score(player): 

    player_x = sorted(player, key=lambda statue: statue.x) #player's pieces sorted by x coordinate 
    player_y = sorted(player, key=lambda statue: statue.y) 
    max_score = [0] 

    count = 1 
    #pieces are in a horizontal line if they share a y coord and have sequential x coords 
    for cur_s, next_s in zip(player_x, player_x[1:]): 
     if cur_s.x + 1 == next_s.x and cur_s.y == next_s.y: 
      count += 1 
     else: 
      max_score.append(count) 
      count = 1 
    max_score.append(count) 

    count = 1 
    #pieces are diagonal if they share an x and have sequental y's 
    for cur_s, next_s in zip(player_y, player_y[1:]): 
     if cur_s.y + 1 == next_s.y and cur_s.x == next_s.x: 
      count += 1 
     else: 
      max_score.append(count) 
      count = 1 
    max_score.append(count) 

    count = 1 
    #they are diagonal if both x's and y's are sequential 
    for cur_s, next_s in zip(player_y, player_y[1:]): 
     if cur_s.y + 1 == next_s.y and cur_s.x + 1 == next_s.x: 
      count += 1                                        
     else: 
      max_score.append(count) 
      count = 1 
    max_score.append(count) 

    return max(max_score) 

据我可以告诉它的工作，但我基本上是在重复自己的三倍。我的问题是，我写这个函数的最pythonic方式是什么，这样我就可以更少地重复自己了？

Answer 1

这可能不是最好的，但第一眼我看你能所有的循环组合成有三个参数的单一功能：

def score(player): 

    player_x = sorted(player, key=lambda statue: statue.x) #player's pieces sorted by x coordinate 
    player_y = sorted(player, key=lambda statue: statue.y) 
    max_score = [0] 

    def take_count(player, x_offset, y_offset): 
     count = 1 
     for cur_s, next_s in zip(player, player[1:]): 
      if cur_s.x + x_offset == next_s.x and cur_s.y + y_offset == next_s.y: 
       count += 1 
      else: 
       max_score.append(count) 
       count = 1 
     max_score.append(count) 

    #pieces are in a horizontal line if they share a y coord and have sequential x coords 
    take_count(player_x, 1, 0) 

    #pieces are diagonal if they share an x and have sequental y's 
    take_count(player_y, 0, 1) 

    #they are diagonal if both x's and y's are sequential 
    take_count(player_y, 1, 1) 

    return max(max_score)