如何从scikit-learn决策树中提取决策规则？

我可以从决策树中的受过训练的树中提取基础决策规则(或“决策路径”)作为文本列表吗？

就像是：

如果A> 0.4,那么如果B <0.2那么如果C> 0.8则那么class ='X' 谢谢你的帮助.

解决方法:

我相信这个答案比其他答案更正确：

from sklearn.tree import _tree

def tree_to_code(tree, feature_names):
    tree_ = tree.tree_
    feature_name = [
        feature_names[i] if i != _tree.TREE_UNDEFINED else "undefined!"
        for i in tree_.feature
    ]
    print "def tree({}):".format(", ".join(feature_names))

    def recurse(node, depth):
        indent = "  " * depth
        if tree_.feature[node] != _tree.TREE_UNDEFINED:
            name = feature_name[node]
            threshold = tree_.threshold[node]
            print "{}if {} <= {}:".format(indent, name, threshold)
            recurse(tree_.children_left[node], depth + 1)
            print "{}else:  # if {} > {}".format(indent, name, threshold)
            recurse(tree_.children_right[node], depth + 1)
        else:
            print "{}return {}".format(indent, tree_.value[node])

    recurse(0, 1)

这将打印出有效的Python函数.以下是尝试返回其输入的树的示例输出,该数字介于0和10之间.

def tree(f0):
  if f0 <= 6.0:
    if f0 <= 1.5:
      return [[ 0.]]
    else:  # if f0 > 1.5
      if f0 <= 4.5:
        if f0 <= 3.5:
          return [[ 3.]]
        else:  # if f0 > 3.5
          return [[ 4.]]
      else:  # if f0 > 4.5
        return [[ 5.]]
  else:  # if f0 > 6.0
    if f0 <= 8.5:
      if f0 <= 7.5:
        return [[ 7.]]
      else:  # if f0 > 7.5
        return [[ 8.]]
    else:  # if f0 > 8.5
      return [[ 9.]]

以下是我在其他答案中看到的一些绊脚石：

>使用tree_.threshold == -2来确定节点是否为叶子不是一个好主意.如果它是一个阈值为-2的真实决策节点怎么办？相反,您应该查看tree.feature或tree.children_ *.
> line feature = [feature_ames [i] for i in tree_.feature]与我的sklearn版本崩溃,因为tree.tree_.feature的某些值为-2(特别是对于叶节点).
>递归函数中不需要多个if语句,只需一个就可以了.

标签：decision-tree,random-forest,python,scikit-learn,machine-learning
来源： https://codeday.me/bug/20190915/1804838.html