康威的人生游戏没有正确计算邻居

Question

我正在使用Python编写标准的Conway的“生命游戏”程序。 在遍历数组时尝试计算邻居时遇到问题。 我创建了打印语句，该语句打印连续的if语句以及每个语句的count值。

这是我的代码：（我在整个代码中的＃内都有问题）

import random

numrows = 10
numcols = 10
def rnd():
    rn = random.randint(0,1)
    return rn

def initial():
    grid = []
    count = 0
    for x in range(numrows):
        grid.append([])
        for y in range(numcols):
            rand=random.randrange(1,3)
            if(rand == 1):
                grid[x].append('O')
            else:
                grid[x].append('-')
    for x in grid:    
        print(*x, sep=' ',end="\n") #this prints the random 2d array 

    print("")
    print("")

    answer = 'y'
    newgrid = []
    count = 0

    while(answer == 'y'): # I believe I am going through, checking neighbors
                          # and moving onto the next index inside these for 
                          #loops below
        for r in range(0,numrows):
            grid.append([])
            for c in range(0,numcols):

                if(r-1 > -1 and c-1 > -1): #I use this to check out of bound
                    if(newgrid[r-1][c-1] == 'O'):#if top left location is O
                        count = count + 1        #should get count += 1
                    else:
                        count = count
                print("top left check complete")
                print(count)
                if(r-1 > -1):
                    if(newgrid[r-1][c] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("top mid check complete")
                print(count)
                if(r-1 > -1 and c+1 < numcols):
                    if(newgrid[r-1][c+1] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("top right check complete")
                print(count)
                if(c-1 > -1 and r-1 > -1):
                    if(newgrid[r][c-1] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("mid left check complete")
                print(count)
                if(r-1 > -1 and c+1 < numcols): 
                    if(newgrid[r][c+1] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("mid right check complete")
                print(count)
                if(r+1 < numrows and c-1 > -1):
                    if(newgrid[r+1][c-1] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("bot left check complete")
                print(count)
                if(r+1 < numrows and c-1 > -1):
                    if(newgrid[r+1][c] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("bot mid check complete")
                print(count)
                if(r+1 < numrows and c+1 < numcols):
                    if(newgrid[r+1][c+1] == 'O'):
                        count = count + 1
                    else:
                        count = count
                print("bot right check complete")
                print(count)

# I am not sure about the formatting of the code below, how do I know that
# the newgrid[r][c] location is changing? should it be according to the for-
# loop above? Or should it get it's own? If so, how could I construct it as
# to not interfere with the other loops and items of them?


            if(newgrid[r][c] == '-' and count == 3):
                newgrid[r][c] ='O'

            elif(newgrid[r][c] == 'O' and count < 2):
                newgrid[r][c] = '-'

            elif(newgrid[r][c] == 'O' and (count == 2 or count == 3)):
                newgrid[r][c] = 'O'

            elif(newgrid[r][c] == 'O' and count > 3):
                newgrid[r][c] = '-'

# I'm also confused how to go about printing out the 'new' grid after each
# element has been evaluated and changed. I do however know that after the
# new grid prints, that I need to assign it to the old grid, so that it can
# be the 'new' default grid. How do I do this? 


        for z in newgrid:
            print(*z, sep=' ',end="\n")


        answer = input("Continue? y or n( lower case only): ")

        newgrid = grid

        if(answer != 'y'):
            print(" Hope you had a great life! Goodbye!")

initial()

这是当前的输出和错误消息：

 >>> initial() - O - - O - - O - - - O - - O - - - OO - O - - O - OO - O O - - O - - OOOO O - OO - - - OO - O - O - O - O - O - O - OOOO - - O - - - - - OOO - - O OO - O - - O - - - - - OOO - O - - - top left check complete 0 top mid check complete 0 top right check complete 0 mid left check complete 0 mid right check complete 0 bot left check complete 0 bot mid check complete 0 Traceback (most recent call last): File "<pyshell#68>", line 1, in <module> initial() File "C:\\Users\\Ted\\Desktop\\combined.py", line 86, in initial if(newgrid[r+1][c+1] == 'O'): IndexError: list index out of range 

当我遍历随机数组以查看邻居是什么时，在检查bot右邻居的同时，直到移至[0] [1]为止，它似乎都可以正常工作。

同样，中右邻居应该还活着就算+1。 但是，即使继承了if语句，计数仍为0？

问题1：我怎么可能知道条件是否足以满足数组所有面的[r] [c]的每个实例？

问题2：我目前的检查范围是否最适合我的情况？ 有没有一种方法可以在我检查值之前进行“全部检查”？

在这一点上，我已经不知所措。 预先感谢您抽出宝贵时间来回答我的问题

Answer 1

您收到该索引错误，因为您的newgrid仅包含一个空行。 然后您在newgrid而不是在grid对邻居进行测试（如Blckknght在评论中所述）。 我进行了一些修复，但是可以做很多事情来改进此代码。 看起来它现在正在运行，但是很难确定何时使用随机Life表单。 :)我建议给您的程序某种方式来使用已知的生活模式，例如眨眼和滑翔伞，以查看它们的行为是否正确。

确保newgrid有效的最简单方法是从grid复制它。 如果我们只是做newgrid = grid ，那只会使newgrid成为grid对象的另一个名称。 为了正确复制列表列表，我们需要复制每个内部列表。 我的新代码使用了copy_grid函数。

我已经修复了计算邻居的本节中的if测试中的一些小错误，并且简化了从邻居计数更新单元格的逻辑。 我还压缩了构成随机网格的代码，并添加了一个简单的函数，该函数可以从字符串中读取Life模式并从中构建网格。 这使我们可以使用Glider测试代码。 我还添加了一个使空白网格的函数。 尽管我在测试中使用过该程序，但该程序目前未使用该功能，我想这是一个有用的示例。 :)

import random

# Set a seed so that we get the same random numbers each time we run the program
# This makes it easier to test the program during development
random.seed(42)

numrows = 10
numcols = 10

glider = '''\
----------
--O-------
---O------
-OOO------
----------
----------
----------
----------
----------
----------
'''

# Make an empty grid
def empty_grid():
    return [['-' for y in range(numcols)]
        for x in range(numrows)]

# Make a random grid
def random_grid():
    return [[random.choice('O-') for y in range(numcols)]
        for x in range(numrows)]

# Make a grid from a pattern string
def pattern_grid(pattern):
    return [list(row) for row in pattern.splitlines()]

# Copy a grid, properly!
def copy_grid(grid):
    return [row[:] for row in grid]

# Print a grid
def show_grid(grid):
    for row in grid:
        print(*row)
    print()

def run(grid):
    show_grid(grid)

    # Copy the grid to newgrid.
    newgrid = copy_grid(grid)
    while True:
        for r in range(numrows):
            for c in range(numcols):
                # Count the neighbours, making sure that they are in bounds
                count = 0
                # Above this row
                if(r-1 > -1 and c-1 > -1): 
                    if(grid[r-1][c-1] == 'O'):
                        count += 1
                if(r-1 > -1):
                    if(grid[r-1][c] == 'O'):
                        count += 1
                if(r-1 > -1 and c+1 < numcols):
                    if(grid[r-1][c+1] == 'O'):
                        count += 1

                # On this row
                if(c-1 > -1):
                    if(grid[r][c-1] == 'O'):
                        count += 1
                if(c+1 < numcols): 
                    if(grid[r][c+1] == 'O'):
                        count += 1

                # Below this row
                if(r+1 < numrows and c-1 > -1):
                    if(grid[r+1][c-1] == 'O'):
                        count += 1
                if(r+1 < numrows):
                    if(grid[r+1][c] == 'O'):
                        count += 1
                if(r+1 < numrows and c+1 < numcols):
                    if(grid[r+1][c+1] == 'O'):
                        count += 1

                # Update the cell in the new grid
                if grid[r][c] == '-':
                    if count == 3:
                        newgrid[r][c] ='O'
                else:
                    if count < 2 or count> 3:
                        newgrid[r][c] = '-'

        # Copy the newgrid to grid
        grid = copy_grid(newgrid)
        show_grid(grid)

        answer = input("Continue? [Y/n]: ")
        if not answer in 'yY':
            print(" Hope you had a great life! Goodbye!")
            break

#grid = random_grid()
grid = pattern_grid(glider)
run(grid)

该代码可以正常工作，但是仍有很大的改进空间。 例如，这是run()的改进版本，它通过使用几个循环来压缩邻居计数部分。

def run(grid):
    show_grid(grid)

    # Copy the grid to newgrid.
    newgrid = copy_grid(grid)
    while True:
        for r in range(numrows):
            for c in range(numcols):
                # Count the neighbours, making sure that they are in bounds
                # This includes the cell itself in the count
                count = 0
                for y in range(max(0, r - 1), min(r + 2, numrows)):
                    for x in range(max(0, c - 1), min(c + 2, numcols)):
                        count += grid[y][x] == 'O'

                # Update the cell in the new grid
                if grid[r][c] == '-':
                    if count == 3:
                        newgrid[r][c] ='O'
                else:
                    # Remember, this count includes the cell itself
                    if count < 3 or count > 4:
                        newgrid[r][c] = '-'

        # Copy the newgrid to grid
        grid = copy_grid(newgrid)
        show_grid(grid)

        answer = input("Continue? [Y/n]: ")
        if not answer in 'yY':
            print(" Hope you had a great life! Goodbye!")
            break

Answer 2

要计算邻居数，只需遵循以下简单规则即可。 您将拥有当前单元格的行和列。 使用该表格可以形成两个数组，一个数组用于行，另一个数组用于列。 以下是您的逻辑。

您可以在此处找到带有演示视频的详细实现。

this.neibhours = function() {
  var rows = [row-1, row, row+1];
  var cols = [col-1, col, col+1];
  neibhourCells = [];

  for(var i=0; i < rows.length; i++) {
    for(var j=0; j < cols.length; j++) {
      if(!(col === cols[j] && row === rows[i])) {
        var cell = new Cell(rows[i], cols[j])
        if(cell.isOnGrid()) {
          neibhourCells.push(cell);
        }
      }
    }
  }

  return neibhourCells;
}