Setting a value to one place in a list(b) which is in another list(a) sets that value in the same place in all the lists that are in list(a)

Question

Im trying to make an 9x9 array using nested lists and when I set am trying to set a single value for some reason it is setting that value in every row of that array. It starts with a list like this

[[0,0,0,0,0],
 [0,0,0,0,0],
 [0,0,0,0,0],
 [0,0,0,0,0],
 [0,0,0,0,0]]

I did something like list[2][3]=1 and the list changed to

[[0,0,0,1,0],
 [0,0,0,1,0],
 [0,0,0,1,0],
 [0,0,0,1,0],
 [0,0,0,1,0]]

I can't figure out what I did wrong. My code is

def resetGrids():
    gridHidden=[]
    gridShown=[]
    gridpart=[]
    for i in range(9):
        gridpart.append(0)
    for i in range(9):
        gridHidden.append(gridpart)
        gridShown.append(gridpart)
    for i in range(10):
        searching=True
        while searching:
            checkX=random.randint(0,8)
            checkY=random.randint(0,8)
            print(checkX, checkY)
            if gridHidden[checkY][checkX]!=10:
                print(gridHidden[checkY][checkX])
                gridHidden[checkY][checkX]=10
                print("a")
                #for a in range(-1,2):
                #    if checkY+a>=0 and checkY+a<=8:
                #        for b in range(-1,2):
                #            if checkX+b>=0 and checkX+b<=8:
                #                if gridHidden[checkY+a][checkX+b]!=10:
                #                    gridHidden[checkY+a][checkX+b]+=1
                searching=False
                print("B")
            print(gridHidden)
        print("c")
    print("d")

Does this answer your question? [List of lists changes reflected across sublists unexpectedly](https://stackoverflow.com/questions/240178/list-of-lists-changes-reflected-across-sublists-unexpectedly) You can't append the same list `b` to `a` 9 times and then be surprised that changing one `b` changes all `b`. — Pranav Hosangadi, Oct 07 '20 at 16:22

score 1 · Accepted Answer · answered Oct 07 '20 at 16:25

try to set as the inner lists copy() of the actual list, I guess that it happend because you actually pass a reference to the list. replace to this lines:

gridHidden.append(gridpart.copy())
gridShown.append(gridpart.copy())

when I tried this, and added this part:

print(gridHidden)
gridHidden[1][2] = 3
print(gridHidden)

I got this:

[[0, 10, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 10, 0, 0, 10, 10, 0, 0, 10], [0, 0, 0, 0, 0, 0, 0, 10, 0], [0, 0, 0, 10, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 10, 0, 0, 0], [0, 0, 0, 0, 10, 0, 0, 0, 0], [0, 10, 0, 0, 0, 0, 0, 0, 0]]
[[0, 10, 0, 0, 0, 0, 0, 0, 0], [0, 0, 3, 0, 0, 0, 0, 0, 0], [0, 10, 0, 0, 10, 10, 0, 0, 10], [0, 0, 0, 0, 0, 0, 0, 10, 0], [0, 0, 0, 10, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 10, 0, 0, 0], [0, 0, 0, 0, 10, 0, 0, 0, 0], [0, 10, 0, 0, 0, 0, 0, 0, 0]]

Setting a value to one place in a list(b) which is in another list(a) sets that value in the same place in all the lists that are in list(a)

1 Answers1