python - 树中的递归

Question

我正在用 Python 构建一个简单的二元决策树。我正在使用递归来构建树，但是，作为一个没有牢固掌握这个概念的人，我遇到了一些麻烦。我想在树达到一定深度时停止递归，但我不确定在哪里增加该值，因此它一次会构建多个分支。现在，树只是向右分支，直到达到 5，然后停止。我应该在哪里/如何增加该值？现在我正在函数底部的 for 循环中执行此操作。

def buildTree(currentNode, maxDepth, currentDepth, minGain, currentGain, allFeatures):
print(currentNode.data)
if maxDepth <= currentDepth:
    return None
else:
    splitOn, hasFeat, noFeat, allFeatures, maxGain = split(currentNode, allFeatures)
    print(len(hasFeat), len(noFeat))
    currentNode.left = Tree()
    currentNode.left.vectors = hasFeat
    currentNode.left.data = splitOn
    currentNode.left.entropy = getEntropy(getInstances(currentNode.left.vectors))
    currentNode.right = Tree()
    currentNode.right.vectors = noFeat
    currentNode.right.data = "!" + splitOn
    currentNode.right.entropy = getEntropy(getInstances(currentNode.right.vectors))
    nodeList = [currentNode.right, currentNode.left]
    for node in nodeList:
        return buildTree(node, maxDepth, currentDepth + 1, minGain, maxGain, allFeatures)

Answer 1

应该在 currentNode.left 和 currentNode.right 上使用递归调用。

代码应该是这样的:

def buildTree(currentNode, maxDepth, currentDepth, minGain, currentGain, allFeatures):
    print(currentNode.data)
    if maxDepth <= currentDepth:
        return None
    else:
        splitOn, hasFeat, noFeat, allFeatures, maxGain = split(currentNode, allFeatures)
        print(len(hasFeat), len(noFeat))
        currentNode.left = buildTree(Tree(), maxDepth, currentDepth + 1, minGain, maxGain, allFeatures)
        currentNode.left.vectors = hasFeat
        currentNode.left.data = splitOn
        currentNode.left.entropy = getEntropy(getInstances(currentNode.left.vectors))
        currentNode.right = buildTree(Tree(), maxDepth, currentDepth + 1, minGain, maxGain, allFeatures)
        currentNode.right.vectors = noFeat
        currentNode.right.data = "!" + splitOn
        currentNode.right.entropy = getEntropy(getInstances(currentNode.right.vectors))
        nodeList = [currentNode.right, currentNode.left]
        return nodeList