自动化枯燥的东西-第5章-国际象棋字典验证器

Question

第一个帖子，新的编程和乐趣！欢迎所有关于这个帖子的反馈和我的问题。

我正在通过自动化乏味的东西和攻击第一章问题国际象棋字典验证器。

在本章中，我们使用字典值{'1h':'bking','6c':'wqueen','2g':'bbishop','5h':'bqueen','3e':'wking'}来表示棋盘。编写一个名为isValidChessBoard（）的函数，它接受一个字典参数，并根据板是否有效返回True或False。

一个有效的板将有一个黑王和一个白王。每个玩家最多只能拥有16枚棋子，最多8枚棋子，所有棋子必须在'1a'到'8H'的有效空间上；也就是说，一块不能在空间'9z'上。作品名称以“W”或“B”开头，代表白色或黑色，后面是“Pawn”、“Knight”、“Bishop”、“Rook”、“Queen”或“King”。当错误导致不正确的棋盘时，这个函数应该检测。

我的问题和代码：

1. 通过这些for循环+多个if语句来评估字典键/值是“最佳实践”吗？好像有很多代码。如果在for循环中与另一个if语句后面进行更改以包含一些elif会导致问题。
2. 第23行如果i[0]=='b':出错，因为空字符串值的象棋空格在i[0]处没有字符。表达/评估空值的最佳方法是什么？如果是“”，是否应该在计算值==“”的循环中添加前导条件，然后添加“continue”？
3. 为什么我不能将第15行折叠为11，这样，如果board.values():中没有'bking'或'wking'就有一个语句:？如果我这样做，陈述结果是真的；然而，字典包含这两个值，所以它不应该计算为False并保持代码运行吗？

def isValidChessBoard(board):
    while True:
        blackPieces = 0
        whitePieces = 0
        wpawn = 0
        bpawn = 0
        letterAxis = ('a','b','c','d','e','f','g','h')
        pieceColour = ('b','w')
        pieceType = ('pawn','knight','bishop','rook','queen','king')

        #one black king and one white king
        if 'bking' not in board.values():
            print('KingError')
            return False
            break
        if 'wking' not in board.values():
            print('KingError')
            return False
            break

        #each player has <= 16 pieces
        for i in board.values():
            if i[0] == 'b':
                blackPieces+=1
            if i[0] == 'w':
                whitePieces+=1
            if whitePieces >= 17:
                print('TotalPieceError')
                return False
                break
            if blackPieces >= 17:
                print('TotalPieceError')
                return False
                break

        #each player has <= 8 pawns
        for i in board.values():
            if i == 'wpawn':
                wpawn+=1
            elif i == 'bpawn':
                bpawn+=1
            if wpawn or bpawn >= 9:
                print('PawnError')
                return False
                break

        #all pieces must be on valid space from '1a' to '8h'
        for i in board.keys():
            if int(i[0]) >= 9:
                print('SpacesError')
                return False
                break
            if i[1] not in letterAxis:
                print('yAxisError')
                return False
                break

        #piece names begin with 'w' or 'b'
        for i in board.values():
            if i[0] not in pieceColour:
                print('WhiteOrBlackError')
                return False
                break

        #piece names must follow with 'pawn', 'knight', 'bishop', 'rook', 'queen', 'king'
        for i in board.values():
            if i[1:] not in pieceType:
                print('PieceTypeError')
                return False
        return 'This board checks out'

board = {'1a': 'bking','2a': 'bqueen','3a': 'brook','4a': 'brook',
'5a': 'bknight','6a': 'bknight','7a':'bbishop','8a': 'bbishop',
'1b': 'bpawn','2b': 'bpawn','3b': 'bpawn','4b':'bpawn',
'5b': 'bpawn','6b': 'bpawn','7b': 'bpawn','8b': 'bpawn',
'1c': 'wking','2c': 'wqueen','3c': 'wrook','4c': 'wrook',
'5c': 'wbishop','6c': 'wbishop','7c': 'wknight','8c':'wknight',
'1e': 'wpawn','2e': 'wpawn','3e': 'wpawn','4e': 'wpawn',
'5e': 'wpawn','6e': 'wpawn','7e': 'wpawn','8e': 'wpawn',
'1f': '','2f': '','3f': '','4f': '','5f': '','6f': '','7f': '','8f': '',
'1g': '','2g': '','3g': '','4g': '','5g': '','6g': '','7g': '','8g': '',
'1h': '','2h': '','3h': '','4h': '','5h': '','6h': '','7h': '','8h': '',}

print(isValidChessBoard(board))

Error:
Traceback (most recent call last):
line 23, in isValidChessBoard
    if i[0] == 'b':
IndexError: string index out of range




Answer 1

编辑：当你定义有效的计数时，它没有考虑到你可能会得到一个棋子到棋盘的另一端，最终得到两个皇后，尽管我不知道你如何修复这一点，而不导致程序接受不确定的答案（例如在游戏开始时得到两个皇后）。如果您能在此提供任何帮助，我们将不胜感激！

既然你问是否有其他选择，我提出以下几点。

步骤

null

def isValidChessBoard(board):
      """Validate counts and location of pieces on board"""
    
      # Define pieces and colors
      pieces = ['king','queen','rook', 'knight','bishop', 'pawn']
      colors = ['b', 'w']
      # Set of all chess pieces
      all_pieces = set(color+piece for piece in pieces for color in colors)
    
      # Define valid range for count of chess pieces by type (low, high) tuples
      valid_counts = {'king': (1, 1),
                  'queen': (0, 1),
                  'rook': (0, 2),
                  'bishop': (0, 2),
                  'knight': (0, 2),
                  'pawn': (0, 8)}
    
      # Get count of pieces on the board
      piece_cnt = {}
      for v in board.values():
        if v in all_pieces:
          piece_cnt.setdefault(v, 0)
          piece_cnt[v] += 1
    
      # Check if there are a valid number of pieces
      for piece in all_pieces:
        cnt = piece_cnt.get(piece, 0)
        lo, hi = valid_counts[piece[1:]]
        if not lo <= cnt <= hi:   # Count needs to be between lo and hi
          if lo != hi:
            print(f"There should between {lo} and {hi} {piece} but there are {cnt}")
          else:
            print(f"There should be {lo} {piece} but there are {cnt})")
          return False
    
      # Check if locations are valid
      for location in board.keys():
        row = int(location[:1])
        column = location[1:]
        if not ((1 <= row <= 8) and ('a' <= column <= "h")):
          print(f"Invaid to have {board[location]} at postion {location}")
          return False

      # Check if all pieces have valid names
      for loc, piece in board.items():
        if piece:
          if not piece in all_pieces:
            print(f"{piece} is not a valid chess piece at postion {loc}")
            return False

      return True