Python的不断更新阵列 [英] Python continuously updating array
问题描述
我在python解决数独的写了一个简单的回溯算法，一会儿有困难，因为它似乎更新不应该被更新的阵列，从而导致整个程序备份。
从本质上讲，作为code的一部分，我存储在初始值的数组，然后取得数组的副本，这样，当我回溯和分配新值，我没有改变任何数字的在原始数独给予。但是，在这个过程中我莫名其妙地更新我最初的数字数组。我终于想通了，它正在由线而引起格=初始，它应该有我的工作网格（网格）设置为初始值（初始），但只叫在程序的开始。当我把该线路输出和手动分配电网同样的事情作为初始（通过复制/粘贴），该程序正常工作。
我包括我下面的全code，有人知道为什么该行可能已经得到再次呼吁？我不明白。
初始= [0,0,0,0,7,0,1,4,0]
[2,0,0,0,1,5,3,0,6]
[0,6,1,0,3,0,0,0,8]
[0,0,0,7,0,0,0,5,3]
[0,0,9,0,0,0,8,0,0]
[0,7,0,0,0,1,0,0,0]
[1,0,0,0,4,0,0,9,0]
[9,0,6,1,0,0,0,0,2]
[0,2,3,0,9,0,0,0,0]＃系统下面一行曾经是格=初始格= [[0,0,0,0,7,0,1,4,0]，
[2,0,0,0,1,5,3,0,6]
[0,6,1,0,3,0,0,0,8]
[0,0,0,7,0,0,0,5,3]
[0,0,9,0,0,0,8,0,0]
[0,7,0,0,0,1,0,0,0]
[1,0,0,0,4,0,0,9,0]
[9,0,6,1,0,0,0,0,2]
[0,2,3,0,9,0,0,0,0]
高清printBoard（网格）：
因为我在范围（0,8）：
打印（电网[I]）高清checkValidity（NUM，行，列）：
如果num网格[行]：
返回False
因为我在范围（0,8）：
如果电网[I] [COL] == NUM：
返回False
cageRow =行// 3
cageCol =山坳// 3
因为我在范围（0,3）：
如果num网格[3 * cageRow + I] [3 * cageCol：3 * cageCol + 3]：
返回False
返回True高清nextCell（行，列，原路返回）：
如果原路返回== 0：
如果COL == 8：
行+ = 1
山坳= 0
其他：
山坳+ = 1
其他：
如果COL == 0：
按行= 1
山坳= 8
其他：
协作= 1
如果行℃的：
打印（错误：退步太过分了。）
回报（行，列）高清findNewNumber（行，列，NUM，原路返回）：
在范围I（NUM + 1,10）：
如果checkValidity（I，行，列）：
返回（I，0）
返回（0,1）
行= 0
山坳= 0
原路返回= 0
打印（解决......）而排＆LT; 9：
如果电网[行] [COL] ==初始[行] [COL]和初始[行] [山口] = 0！
[行，列] = nextCell（行，列，原路返回）
其他：
NUM =网格[行] [COL]
格[行] [COL] = 99
[NUM，原路返回] = findNewNumber（行，列，NUM，原路返回）
格[行] [山口] = NUM
[行，列] = nextCell（行，列，原路返回）打印（解决了！）
printBoard（初始）
打印（）
printBoard（网格）
格=初始
ID（初始）== ID（网格）
＆GT;＆GT;＆GT;真正
这不会设置电网相同的值初始。它使电网引用同一个对象初始。人们通常做的是使浅拷贝，这样
格=初始[：]
ID（初始）== ID（网格）
＆GT;＆GT;＆GT;假
这不会工作你的情况虽然。这将创建一个新的名单外，将举行同样的内部列表。
ID（初始[0]）== ID（网格[0]）
＆GT;＆GT;＆GT;真正
您需要的是 deepcopy的来获取内部列表的拷贝。
进口副本格= copy.deepcopy（初始）
然后一切都被复制
ID（初始）== ID（网格）
＆GT;＆GT;＆GT;假
ID（初始[0]）== ID（网格[0]）
＆GT;＆GT;＆GT;假
I wrote a simple backtracking algorithm in python to solve sudoku's, and for a while had difficulty because it seemed to update an array that should not have been updated, which caused the entire program to backup.
Essentially, as part of the code, I stored an array of the initial values and then made a copy of the array so that when I was backtracking and assigning new values, I didn't change any of the numbers that were given in the original sudoku. However, in the process I somehow updated my array of initial numbers. I eventually figured out it was being caused by the line grid=initial, which should have set my working grid (grid) to the initial values (initial) but was only called at the start of the program. When I took that line out and manually assigned grid to the same thing as initial (via copy/paste), the program worked fine.
I've included my full code below, anybody know why that line may have been getting called again? I can't figure it out.
initial=[[0,0,0,0,7,0,1,4,0],
[2,0,0,0,1,5,3,0,6],
[0,6,1,0,3,0,0,0,8],
[0,0,0,7,0,0,0,5,3],
[0,0,9,0,0,0,8,0,0],
[0,7,0,0,0,1,0,0,0],
[1,0,0,0,4,0,0,9,0],
[9,0,6,1,0,0,0,0,2],
[0,2,3,0,9,0,0,0,0]]
#The following line used to be grid=initial
grid=[[0,0,0,0,7,0,1,4,0],
[2,0,0,0,1,5,3,0,6],
[0,6,1,0,3,0,0,0,8],
[0,0,0,7,0,0,0,5,3],
[0,0,9,0,0,0,8,0,0],
[0,7,0,0,0,1,0,0,0],
[1,0,0,0,4,0,0,9,0],
[9,0,6,1,0,0,0,0,2],
[0,2,3,0,9,0,0,0,0]]
def printBoard(grid):
for i in range(0,8):
print(grid[i])
def checkValidity(num,row,col):
if num in grid[row]:
return False
for i in range(0,8):
if grid[i][col]==num:
return False
cageRow=row//3
cageCol=col//3
for i in range(0,3):
if num in grid[3*cageRow+i][3*cageCol:3*cageCol+3]:
return False
return True
def nextCell(row,col,backtrack):
if backtrack==0:
if col==8:
row+=1
col=0
else:
col+=1
else:
if col==0:
row-=1
col=8
else:
col-=1
if row<0:
print("Error: Backtracked too far.")
return (row,col)
def findNewNumber(row,col,num,backtrack):
for i in range(num+1,10):
if checkValidity(i,row,col):
return (i,0)
return (0,1)
row=0
col=0
backtrack=0
print("Solving...")
while row<9:
if grid[row][col]==initial[row][col] and initial[row][col]!=0:
[row,col]=nextCell(row,col,backtrack)
else:
num=grid[row][col]
grid[row][col]=99
[num,backtrack]=findNewNumber(row,col,num,backtrack)
grid[row][col]=num
[row,col]=nextCell(row,col,backtrack)
print("Solved!")
printBoard(initial)
print("")
printBoard(grid)
grid = initial
id(initial) == id(grid)
>>> True
This doesn't set grid to the same values as initial. It makes grid refer to the same object as initial. What people commonly do is make a shallow copy, like this
grid = initial[:]
id(initial) == id(grid)
>>> False
This will not work in your case though. It would create a new outer list that will hold the same interior lists.
id(initial[0]) == id(grid[0])
>>> True
What you need is a deepcopy to get copies of the interior lists as well.
import copy
grid = copy.deepcopy(initial)
then everything is copied
id(initial) == id(grid)
>>> False
id(initial[0]) == id(grid[0])
>>> False
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