python递归实例变量共享数据

Question

我正在尝试创建一个创建树的递归函数。每个节点都持有井字游戏的状态，每个节点的子节点是下一个可能的移动。

我将棋盘的状态传递给递归函数。对于每一个可能的移动，我都会创建一个状态副本，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

从打印语句中可以清楚地看出，对状态所做的更改正在被带回函数的父实例。有谁知道为什么？

Answer 1

问题是，您正在修改类变量，它们将被特定类的所有对象共享。要解决此问题，请像这样制作 state 和 children 实例变量

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

根据这一行，

newState = cState #create copy of state

您正在尝试创建cState 的副本并将其存储在newState 中。请记住，在 Python 中，赋值运算符永远不会将一个的值复制到另一个。它只是将左侧的变量指向赋值语句右侧表达式的评估结果。

所以，你实际上在做的是，让newState 和cState 指向同一个列表。所以，如果你修改newState，它也会影响cState。要实际创建列表的副本，您可以使用切片运算符，如下所示

newState = cState[:] #create copy of state

Answer 2

一个问题是这一行：

newState = cState

从您的评论看来，您正在尝试复制 cState，但是，您只是将 reference 复制到列表，而不是列表的内容。因此，当您修改 newState 时，您也在修改 cState。你应该拥有的是：

 newState = cState.copy()

编辑 由于这是公认的答案，完整的解决方案还包括在节点的构造函数中设置self.children = []，如@thefourtheye 所述