有没有办法检查同一个正方形上有没有两块，我试过用同样的方法来计算每种颜色的块数，但是没有用。我知道这本词典不能有重复的键。有没有办法检查重复的钥匙？

# count how many times each sqaure is mentioned (for ex: if g7 is mentioned twice it should detect it)

    countSquares.setdefault(m, 0)
        countSquares[m] = countSquares[m] + 1
    #check if more than one piece are in the same square
    for j in countSquares:
        if countSquares.get(j, 0) > 1:
            print('Multiple pieces cannot exist in the same square!!')
            check = False

另外，我想知道是否有什么方法可以使代码的行数更少或运行更快。我知道这个特定程序的速度不是问题，但我想知道是否有一种方法，使它更快，以防我写一个类似的，但更大的程序。
我试图解决的实践问题：
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，该函数接受一个dictionary参数并根据电路板是否有效返回true或false。一个有效的董事会将有正好一个黑国王和正好一个白国王。每个玩家最多只能有16个棋子，最多8个棋子，所有棋子必须在“1a”到“8h”的有效空间内；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，后跟“兵”、“骑士”、“主教”、“车”、“皇后”或“国王”。此函数应在错误导致棋盘不正确时进行检测。
提前谢谢

import pprint
chessBoard = {'a1': 'wrook', 'b1': 'wknight', 'c1': 'wbishop', 'd1': 'wqueen', 'e1': 'wking', 'f1': 'wbishop', 'g1': 'wknight', 'h1': 'wrook',
              'a2': 'wpawn', 'b2': 'wpawn', 'c2': 'wpawn', 'd2': 'wpawn', 'e2': 'wpawn', 'f2': 'wpawn', 'g2': 'wpawn', 'h2': 'wpawn',
              'a8': 'brook', 'b8': 'bknight', 'c8': 'bbishop', 'd8': 'bqueen', 'e8': 'bking', 'f8': 'bbishop', 'g8': 'bknight', 'h8': 'brook',
              'a7': 'bpawn', 'b7': 'bpawn', 'c7': 'bpawn', 'd7': 'bpawn', 'e7': 'bpawn', 'f7': 'bpawn', 'g7': 'bpawn', 'g7': 'bpawn'}
validBoard = ['a1', 'a2', 'a3', 'a4', 'a5', 'a6', 'a7', 'a8',
              'b1', 'b2', 'b3', 'b4', 'b5', 'b6', 'b7', 'b8',
              'c1', 'c2', 'c3', 'c4', 'c5', 'c6', 'c7', 'c8',
              'd1', 'd2', 'd3', 'd4', 'd5', 'd6', 'd7', 'd8',
              'e1', 'e2', 'e3', 'e4', 'e5', 'e6', 'e7', 'e8',
              'f1', 'f2', 'f3', 'f4', 'f5', 'f6', 'f7', 'f8',
              'g1', 'g2', 'g3', 'g4', 'g5', 'g6', 'g7', 'g8',
              'h1', 'h2', 'h3', 'h4', 'h5', 'h6', 'h7', 'h8']
whitePieces = ['wking', 'wqueen', 'wbishop', 'wknight', 'wrook', 'wpawn']
blackPieces = ['bking', 'bqueen', 'bbishop', 'bknight', 'brook', 'bpawn']
pieces = {'white': whitePieces, 'black': blackPieces}
count = {}
countSquares = {}

def isValidChessBoard(board):
    check = True
    for m, n in board.items():
        # Checks if the places occupied are a valid place on the chess board
        if m not in validBoard:
            print('{} is an invalid place'.format(m))
            check = False
        #keep count of every type of piece
        count.setdefault(n, 0)
        count[n] = count[n] + 1
        #count how many times each sqaure is mentioned (for ex: if g7 is mentioned twice it should detect it)
        countSquares.setdefault(m, 0)
        countSquares[m] = countSquares[m] + 1
    #check if more than one piece are in the same square
    for j in countSquares:
        if countSquares.get(j, 0) > 1:
            print('Multiple pieces cannot exist in the same square!!')
            check = False
    #for loop that checks for black pieces and white pieces
    for v in pieces.keys():
        num = 0
        for k in pieces[v]:
            num = num + count.get(k, 0) #keeps count of total pieces per colour
            # conditions to check for, for each colour
            if k == 'bking' or k == 'wking':
                if count.get(k, 0) == 1:
                    pass
                else:
                    print('invalid number of kings for {}'.format(v))
                    check = False
            if k == 'bpawn' or k == 'wpawn':
                if count.get(k, 0) <= 8:
                    pass
                else:
                    print('invalid number of pawns for {}'.format(v))
                    check = False
        print('{} has: {} pieces'.format(v, num))
        #check for total number of pieces per colour
        if num <= 16:
            pass
        else:
            print('invalid total number of pieces for {}'.format(v))
            check = False
    pprint.pprint(countSquares)

    if check:
        return True
    else:
        return False

print(isValidChessBoard(chessBoard))