提取规则以预测决策树中的子节点或概率分数

Posted 2023-03-12

【中文标题】提取规则以预测决策树中的子节点或概率分数

【英文标题】：Extracting rules to predict child nodes or probability scores in a Decision Tree [closed]

【发布时间】：2019-08-07 16:42:08

【问题描述】：

我对决策树的 Python 实现相对较新。我正在尝试提取规则来仅预测子节点，并且我需要它能够预测新数据的概率分数（不仅仅是最终分类），并可能将算法传输给其他用户。有没有简单的方法来做到这一点？我在 (How to extract the decision rules from scikit-learn decision-tree?) 找到了一些解决方案。但是，当我测试它们时，由于某种原因，我没有获得所有的子节点（我的树很大很深）。任何建议表示赞赏。谢谢你。

我已经更新了上面链接中的第一个代码来生成节点，它似乎最适合大树。但是，我很难让它与 pd Dataframes 一起工作。这是示例： 将熊猫导入为 pd 将 numpy 导入为 np 从 sklearn.tree 导入决策树分类器

虚拟数据：

df = pd.DataFrame('col1':[0,1,2,3],'col2':[3,4,5,6],'dv':[0,1,0,1])
df
# create decision tree
dt = DecisionTreeClassifier(random_state=0, max_depth=5, min_samples_leaf=1)
dt.fit(df.loc[:,('col1','col2')], df.dv)

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, node))

    recurse(0, 1)

tree_to_code(dt, df.columns)

以上调用产生以下代码：

def tree(col1, col2, dv):
  if col2 <= 3.5:
    return 1
  else:  # if col2 > 3.5
    if col1 <= 1.5:
      return 3
    else:  # if col1 > 1.5
      if col1 <= 2.5:
        return 5
      else:  # if col1 > 2.5
        return 6

而且，当我如下调用上面的代码时，我得到一个错误，我缺少一个参数。如何修改代码以使其适用于 pandas DataFrame？

tree('col1', 'col2', 'dv_pred')

【问题讨论】：

如果您提供一个预期和实际输出的示例，那就太好了。

【参考方案1】：

这是一个可行的解决方案

import pandas as pd
from sklearn.tree import _tree
from sklearn.tree import DecisionTreeClassifier

df = pd.DataFrame('col1':[0,1,2,3],'col2':[3,4,5,6],'dv':[0,1,0,1])

# create decision tree
dt = DecisionTreeClassifier(random_state=0, max_depth=5, min_samples_leaf=1)
features = ['col1','col2']
dt.fit(df.loc[:,features], df.dv)


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(x):")

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

    recurse(0, 1)

tree_to_code(dt,  df[features].columns)

然后得到预测

df.apply(tree, axis=1)

【讨论】：

我刚刚在一棵非常大的树上进行了测试，效果非常好。这个解决方案非常健壮。谢谢！