康威的人生游戏没有正确地计算邻居 [英] Conway's Game of Life not counting neighbors correctly
问题描述
我正在使用Python编写标准的Conway的《人生游戏》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if语句的后继以及每个语句的count值.
I am doing the standard Conway's Game of Life program using Python. I am having an issue when trying to count neighbors as I iterate through the array. I created print statements that print the succession of the if statement, as well as the value of count for each statement.
这是我的代码:(我在整个代码中的#内都有问题)
Here is my code: ( I have the questions inside the # throughout the code)
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:
Here is the current output and error message:
>>> initial()
- O - - O - - O - -
- O - - O - - - O O
- O - - O - O O - O
O - - O - - O O O O
O - O O - - - O O -
O - O - O - O - O -
O - O O O O - - O -
- - - - O O O - - O
O O - O - - O - - -
- - O O O - 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，似乎一直很好，直到在检查机器人右邻居的同时移至[0] [1]为止.
As I iterate through the random array to see what the neighbors are, it seems to be fine up until it moves over to [0][1] while checking the bot right neighbor.
此外，右中邻应该活着，所以应该加1来计数.但是，即使连续使用if语句，count仍为0?
Also, the mid right neighbor should + 1 to count as it is alive. However, even with succession of the if statement, count remains 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?
Question 1: How can I possibly know that my if conditions witll suffice for every instance of [r][c] for all sides of the array?
问题2:我目前的检查范围是否最适合我的情况?在我什至检查值之前，是否有办法进行全部检查"?
Question 2: Is my current method of checking out of bounds the best for my situation? Is there a way to make a "check all for out of bounds" before I even check the value?
在这一点上，我已经不知所措了.在此先感谢您抽出宝贵的时间来回答我的问题
I am at my wit's end at this point. Thanks in advance for the time taken to help answer my questions
推荐答案
由于您的 newgrid 仅包含一个空行，因此出现了索引错误.然后您在 newgrid 中而不是在 grid 中测试邻居(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.看起来它现在正在工作，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，使其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.
You are getting that index error because your newgrid only contains a single empty row. And your testing for neighbours in newgrid instead of in grid (as Blckknght mentions in the comments). I've made a few repairs, but there's a lot more that can be done to improve this code. It looks like it's working now, but it's hard to tell when you're working with random Life forms. :) I suggest giving your program some way of using known Life patterns like blinkers and gliders to see that they behave correctly.
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 copy_grid 函数做到了这一点.
The simplest way to ensure that newgrid is valid is to copy it from grid. If we just do newgrid = grid that simply makes newgrid another name for the grid object. To copy a list of lists properly we need to make copies of each of the internal lists. My new code does that with the copy_grid function.
我修复了计数邻居的小节中的 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.我还浓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了该功能，但该程序目前未使用该功能，我想这是一个有用的示例.:)
I've fixed a couple of minor bugs that you had in the if tests in the section that counts neighbours, and I've simplified the logic that updates a cell from its neighbour count. I've also condensed the code that makes a random grid, and I've added a simple function that can read a Life pattern from a string and build a grid from it. This lets us test the code with a Glider. I've also added a function that makes an empty grid. The program doesn't currently use that function although I used it during my tests, and I guess it's a useful example. :)
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)
此代码可以正常工作，但仍有很多的改进空间.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.
This code does work correctly, but there is still plenty of room for improvement. For example, here's an improved version of run() that condenses the neighbour counting section by using a couple of loops.
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
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