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python递归实例变量共享数据

1994-MMMs 发布于 2023-02-08 15:26

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我正在尝试创建一个创建树的递归函数.每个节点保持一个井字游戏的状态,每个节点的子节点是下一个可能的移动.

我将板的状态传递给递归函数.对于每一个可能的动作,我创建一个状态的副本,然后进行移动.这个新状态被传递给递归函数.

#XXX
#O O = state [1,1,1,-1,0,-1,1,1,1]
#XXX

playerX = 1
playerO = -1

class node:
    children = [] #holds all the children
    state = [] #holds the state of the board as a list of ints
    def __init__(self,st):
        self.state = st

    def addChild(self,child):
        self.children.append(child) #if only giving birth was this easy

#returns a node with all it's children filled in
#cState is the state for this node
#currentPlayer flips sign every function call
#stateSize is the size of the board
def makeTreeXO(cState,currentPlayer,stateSize):
    newNode = node(cState)
    for i in range(stateSize):
        print "looking at", i, "in", cState
        if(cState[i] == 0): #found an open space
            print "found an empty space"
            newState = cState #create copy of state
            newState[i] = currentPlayer #make the available move
            print "made new state"
            newNode.addChild(makeTreeXO(newState,currentPlayer*-1,stateSize))
    print "Done with this instance"
    return newNode

root = makeTreeXO([1,0,0,1,1,1,1,1,1],playerX,9)

输出:

looking at 0 in [1, 0, 0, 1, 1, 1, 1, 1, 1]
looking at 1 in [1, 0, 0, 1, 1, 1, 1, 1, 1]
found an empty space
made new state
looking at 0 in [1, 1, 0, 1, 1, 1, 1, 1, 1]
looking at 1 in [1, 1, 0, 1, 1, 1, 1, 1, 1]
looking at 2 in [1, 1, 0, 1, 1, 1, 1, 1, 1]
found an empty space
made new state
looking at 0 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 1 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 2 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 3 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 4 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 5 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 6 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 7 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 8 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
Done with this instance
looking at 3 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 4 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 5 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 6 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 7 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 8 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
Done with this instance
looking at 2 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 3 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 4 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 5 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 6 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 7 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
looking at 8 in [1, 1, -1, 1, 1, 1, 1, 1, 1]
Done with this instance

从print语句可以清楚地看到,对状态所做的更改将被传回到函数的父实例.有谁知道为什么？

1 个回答

问题是,您正在修改类变量,它们将由特定类的所有对象共享.要解决这个问题,make state和children实例变量就像这样
```
class node:
    def __init__(self,st):
        self.state = st
        self.children = []

    def addChild(self,child):
        self.children.append(child) #if only giving birth was this easy
```
按照这条线,
```
newState = cState #create copy of state
```
您正在尝试创建cState并存储它的副本newState.请记住,在Python中,赋值运算符永远不会将一个值复制到另一个.它只是在左侧创建变量以指向赋值语句的右侧表达式的求值结果.

所以,你实际做的是,制作两者newState并cState指向同一个列表.因此,如果您修改newState,它也会cState受到影响.要实际创建列表的副本,可以使用切片运算符,如下所示
```
newState = cState[:] #create copy of state
```
2023-02-08 15:29 回答

华女-1_572

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