虚拟数据：

Question

对于决策树的Python实现，我还比较陌生。我试图提取规则以仅预测子节点，并且我需要它能够预测新数据的概率分数（而不仅仅是最终分类），并可能将算法转移给其他用户。有一个简单的方法吗？我在（{{3）}找到了一些解决方案。但是，当我测试它们时，由于某种原因（我的树很大且很深），我无法获得所有的子节点。任何建议表示赞赏。谢谢。

我已经更新了上面链接中的第一个代码以生成节点，它似乎与大树配合使用效果最佳。但是，我很难使它与pd Dataframes一起使用。这是示例：将熊猫作为pd导入将numpy导入为np 从sklearn.tree导入DecisionTreeClassifier

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)

上面的调用产生以下代码：

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，当我如下调用上述代码时，出现错误，提示我缺少一个参数。如何修改代码以使其在pandas DataFrame上运行？

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

Answer 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)

然后获取预测

df.apply(tree, axis=1)

提取规则以预测决策树中的子节点或概率分数

虚拟数据：

1 个答案: