好吧,所以我终于把它打印出来了,并实际做了一些事情,但规则不适用,对吗?我试过打乱规则,但似乎无法正确打印出来,以下是我的规则:
nCount = 0
for i in range(x-1,x+2):
for j in range(y-1,y+2):
if not(i == x and j == y):
nCount += int(mat[i][j])
if(nCount < 2 or nCount > 3):
return 0
elif(nCount == 2 or nCount == 3):
return 1
else:
return mat[i][j]
我测试了一个中间有3个1的常规5x5作为振荡器,但它不是振荡,而是像方形框一样用零填充5x5边缘,然后停止
以下是我的其余编码:
import time
import os
def buildMatrix(fname):
fp = open(fname, "r")
row = fp.readlines()
matrix = []
for i in range(0, len(row), 1):
token = row[i].split(" ")
token[-1] = token[-1].replace('n', ' ')
matrix.append(token)
fp.close()
return matrix
def getRows(mat):
return len(mat)
def getCols(mat):
return len(mat[0])
def printGen(mat): #this is what i use for printing
os.system('clear')
for i in range(0, len(mat), 1):
for j in range(0, len(mat[0]), 1):
if(mat[i][j] == 1):
print("#", sep=" ", end=" ")
else:
print(" ", sep=" ", end=" ")
print()
def nextGen(cur, nxt): #use this to update to the next matrix
for i in range(0, len(cur)-1, 1):
for j in range(0, len(cur[0])-1, 1):
nxt[i][j] = neighbors(i, j, cur)
return nxt
def neighbors(x, y, mat):
nCount = 0
for i in range(x-1,x+2):
for j in range(y-1,y+2):
if not(i == x and j == y):
nCount += int(mat[i][j])
if(nCount < 2 or nCount > 3):
return 0
elif(nCount == 2 or nCount ==3):
return 1
else:
return mat[i][j]
这并不是我的全部代码,因为我仍在将所有这些导入另一个repo并从该repo运行它,但另一部分我已经完成了
elif(nCount == 2,3):
这不是你想要的。这构建了一个2元素元组,其第一个元素是nCount == 2
的值,第二个元素是3,然后将该元组转换为布尔值以决定该走哪条路。如果你想检查nCount
是否等于2或3,你可以使用
elif nCount in (2, 3):
但它是冗余的,因为nCount
必须是控制流到达elif
的控制流之一。当然,这对于执行生活规则来说是不正确的;你想要的是
elif nCount == 3:
如果一个细胞被另外两个活细胞包围,则保持其当前的活/死状态。无论它目前的生活状态如何,它周围必须有3个活细胞才能在下一代存活。
您的代码似乎没有首先进行复制,因此出生或死亡的结果会自行影响。尝试复制:
import numpy
# this function does all the work
def play_life(a):
xmax, ymax = a.shape
b = a.copy() # copy current life grid
for x in range(xmax):
for y in range(ymax):
n = numpy.sum(a[max(x - 1, 0):min(x + 2, xmax), max(y - 1, 0):min(y + 2, ymax)]) - a[x, y]
if a[x, y]:
if n < 2 or n > 3:
b[x, y] = 0 # living cells with <2 or >3 neighbors die
elif n == 3:
b[x, y] = 1 # dead cells with 3 neighbors ar born
return(b)
life = numpy.zeros((5, 5), dtype=numpy.byte)
# place starting conditions here
life[2, 1:4] = 1 # middle three in middle row to 1
# now let's play
print(life)
for i in range(3):
life = play_life(life)
print(life)
我也用numpy
来表示速度。请注意,当进行复制时,会考虑具有2个或3个邻居的活小区的情况。以下是这个测试用例的结果:
[[0 0 0 0 0]
[0 0 0 0 0]
[0 1 1 1 0]
[0 0 0 0 0]
[0 0 0 0 0]]
[[0 0 0 0 0]
[0 0 1 0 0]
[0 0 1 0 0]
[0 0 1 0 0]
[0 0 0 0 0]]
[[0 0 0 0 0]
[0 0 0 0 0]
[0 1 1 1 0]
[0 0 0 0 0]
[0 0 0 0 0]]
[[0 0 0 0 0]
[0 0 1 0 0]
[0 0 1 0 0]
[0 0 1 0 0]
[0 0 0 0 0]]
现在,如果你没有numpy
,那么你可以使用这样的"2D"阵列:
# this function does all the work
def play_life(a):
xmax = len(a)
ymax = len(a[0])
b = [[cell for cell in row] for row in life]
for x in range(xmax):
for y in range(ymax):
n = 0
for i in range(max(x - 1, 0), min(x + 2, xmax)):
for j in range(max(y - 1, 0), min(y + 2, ymax)):
n += a[i][j]
n -= a[x][y]
if a[x][y]:
if n < 2 or n > 3:
b[x][y] = 0 # living cells with <2 or >3 neighbors die
elif n == 3:
b[x][y] = 1 # dead cells with 3 neighbors ar born
return(b)
# this function just prints the board
def show_life(a):
print('n'.join([' '.join([str(cell) for cell in row]) for row in life]) + 'n')
# create board
x_size, y_size = (5, 5)
life = [[0 for y in range(y_size)] for x in range(x_size)]
# place starting conditions here
for x in range(1, 4): life[x][2] = 1 # middle three in middle row to 1
# now let's play
show_life(life)
for i in range(3):
life = play_life(life)
show_life(life)
它为测试用例输出以下内容:
0 0 0 0 0
0 0 1 0 0
0 0 1 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 1 1 1 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 1 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 1 1 1 0
0 0 0 0 0
0 0 0 0 0