根据2d列表中的内容编辑2d列表

Question

对于家庭作业，我必须做一个基于文本的康威的生活游戏在蟒蛇。它的规则是这样。

任何活细胞如果少于两个活的邻居就会死

任何有两个或三个活邻居的活细胞都会活到下一代。

任何有三个以上活邻居的活细胞都会死。

任何有三个活邻居的死细胞都会变成活细胞。

我把棋盘放在一个2d的列表中，我很难找到如何循环列表中的每个元素(游戏中的每个单元，死或活，'0‘或'X')，并根据周围的单元格检查条件。

启动板的一个例子是

00000
00000
000X0
000X0
000X0

首先，我尝试遍历每个元素，并检查当前单元格右侧的单元格是否为X。

这是我做的代码，试着这样做。下一次迭代的函数应该是返回板和康威生命游戏的下一次迭代，但是现在，为了解决问题，我只想让它检查右边的单元格，如果右边的单元格是X的话。

def nextIteration(board):
    newBoardTemp = board[:]
    newBoard = [board[:] for board in newBoardTemp]
    column = 0
    for row in newBoard:
        column = 0
        for item in range(len(row)):
             column += 1
             item2 = row[column + 1]
             if item == item2:
                 board[item][column] = 'X'


def printBoard(board):
   for row in board:
       for item in row:
           print(item, end="")
       print()


def main():
    rows = input("Please enter number of rows: ")
    columns = input("Please enter number of columns: ")
    print()
    cellRow = 0
    cellRows = []
    cellColumns = []
    total = 0
    #the cellRow and cellColumn will contain all of the inputted rows           
    #and columns connected by the index value                                   
    while cellRow != "q":
        cellRow = input("Please enter the row of a cell to turn on (or q to exi\
t): ")
        if cellRow != "q":
                cellColumn = input("Please enter a column for that cell: ")
                cellRows.append(cellRow)
                cellColumns.append(cellColumn)
                total = total + 1
                print()
        else:
            print()
    board = []
    #boardTemp will hold a list that contains one row of the entire board       
    boardTemp = []
    for i in range(int(rows)):
        boardTemp.append("0")
    for i in range(int(columns)):
        board.append(boardTemp[:])
    for i in range(total):
        temp = i
        board[int(cellRows[temp - 1])][int(cellColumns[temp - 1])] = "X"
    iterations = input("How many iterations should I run? ")
 print()
    print("Starting Board:")
    print()
    printBoard(board)
    for i in range(int(iterations)):
        iterationNumber = 1
        nextIteration(board)
        print("Iteration", iterationNumber,":")
        printBoard(board)

main()

这是我在前面展示的启动板上遇到的错误。

File "hw7.py", line 9, in nextIteration
    item2 = row[column + 1]
IndexError: list index out of range

Answer 1

for item in range(len(row)):
    column += 1
    item2 = row[column + 1]
    if item == item2:
        board[item][column] = 'X'

在这个循环的第四次迭代中，您将item2设置为row5，这超出了范围。如果要比较两个相邻的单元格，则应在以后增加列(在第一个迭代项==行和item2 == row2中)。您还应该更早地停止循环。以下应有助于纠正您的错误：

for item in range(len(row) - 1):
    item2 = row[column + 1]
    if item == item2:
        board[item][column] = 'X'
    column += 1

Answer 2

你想用边缘做什么？一种可能是把表面当作一个圆环，也就是甜甜圈.如果你不介意，请跟我来

nr = 5 ;  nc = 4
# here I define a List Of Lists with content unrelated to Conway's Life
# just to have an easy to follow example
lol = [["%d"%(r*10+c+11,) for c in range(nc)] for r in range(nr)]
print(lol)
# [['11', '12', '13', '14'],
#  ['21', '22', '23', '24'],
#  ['31', '32', '33', '34'],
#  ['41', '42', '43', '44'],
#  ['51', '52', '53', '54']]

# now, we build an AUGmented List Of Lists, prepending each row with
# its last element and appending the last one, prepending the first
# augmented row with the last row and appending the first to the last
aug_lol = [[lol[r%nr][c%nc] for c in range(-1,nc+1)] for r in range(-1,nr+1)]
print(aug_lol)
# [['54', '51', '52', '53', '54', '51'],
#  ['14', '11', '12', '13', '14', '11'],
#  ['24', '21', '22', '23', '24', '21'],
#  ['34', '31', '32', '33', '34', '31'],
#  ['44', '41', '42', '43', '44', '41'],
#  ['54', '51', '52', '53', '54', '51'],
#  ['14', '11', '12', '13', '14', '11']]

使用列表的aug_lol列表，您可以以最自然的方式查询元素11 (或其他什么)的邻居，

def neighborhood(alol, r, c):
    return [alol[nr][nc] for nc in range(c,c+3)
                         for nr in range(r,r+3)
                         if (nr!=(r+1) or nc!=(c+1))]

去数那些活着的邻居，

ln = sum(c=="X" for c in neighborhood(...))

要打印板，以下内容更有效

print('\n'.join(''.join(ch for ch in r) for r in board))