Thank for the wonderful solution of @paulkerfeld. On top of his solution, for all those who want to have a serialized version of trees, just use tree.threshold, tree.children_left, tree.children_right, tree.feature and tree.value. Since the leaves don't have splits and hence no feature names and children, their placeholder in tree.feature and tree.children_*** are _tree.TREE_UNDEFINED and _tree.TREE_LEAF. Every split is assigned a unique index by depth first search. Notice that the tree.value is of shape [n, 1, 1]

from StringIO import StringIO
out = StringIO()
out = tree.export_graphviz(clf, out_file=out)
print out.getvalue()

你可以看到一个有向图树。然后,clf.tree_。Feature和clf.tree_。值分别为节点数组拆分特征和节点数组值。你可以参考这个github来源的更多细节。

我创建了自己的函数，从sklearn创建的决策树中提取规则:

import pandas as pd
import numpy as np
from sklearn.tree import DecisionTreeClassifier

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

# create decision tree
dt = DecisionTreeClassifier(max_depth=5, min_samples_leaf=1)
dt.fit(df.ix[:,:2], df.dv)

这个函数首先从节点(在子数组中由-1标识)开始，然后递归地查找父节点。我称之为节点的“沿袭”。在此过程中，我获取了我需要创建if/then/else SAS逻辑的值:

def get_lineage(tree, feature_names):
     left      = tree.tree_.children_left
     right     = tree.tree_.children_right
     threshold = tree.tree_.threshold
     features  = [feature_names[i] for i in tree.tree_.feature]

     # get ids of child nodes
     idx = np.argwhere(left == -1)[:,0]     

     def recurse(left, right, child, lineage=None):          
          if lineage is None:
               lineage = [child]
          if child in left:
               parent = np.where(left == child)[0].item()
               split = 'l'
          else:
               parent = np.where(right == child)[0].item()
               split = 'r'

          lineage.append((parent, split, threshold[parent], features[parent]))

          if parent == 0:
               lineage.reverse()
               return lineage
          else:
               return recurse(left, right, parent, lineage)

     for child in idx:
          for node in recurse(left, right, child):
               print node

下面的元组集包含了创建SAS if/then/else语句所需的所有内容。我不喜欢在SAS中使用do块，这就是为什么我创建逻辑来描述节点的整个路径。元组后的单个整数为路径中终端节点的ID。所有前面的元组组合起来创建该节点。

In [1]: get_lineage(dt, df.columns)
(0, 'l', 0.5, 'col1')
1
(0, 'r', 0.5, 'col1')
(2, 'l', 4.5, 'col2')
3
(0, 'r', 0.5, 'col1')
(2, 'r', 4.5, 'col2')
(4, 'l', 2.5, 'col1')
5
(0, 'r', 0.5, 'col1')
(2, 'r', 4.5, 'col2')
(4, 'r', 2.5, 'col1')
6

现在可以使用export_text了。

from sklearn.tree import export_text

r = export_text(loan_tree, feature_names=(list(X_train.columns)))
print(r)

来自[sklearn][1]的完整示例

from sklearn.datasets import load_iris
from sklearn.tree import DecisionTreeClassifier
from sklearn.tree import export_text
iris = load_iris()
X = iris['data']
y = iris['target']
decision_tree = DecisionTreeClassifier(random_state=0, max_depth=2)
decision_tree = decision_tree.fit(X, y)
r = export_text(decision_tree, feature_names=iris['feature_names'])
print(r)

Scikit learn在0.21版(2019年5月)中引入了一个名为export_text的有趣的新方法，用于从树中提取规则。这里的文档。不再需要创建自定义函数。

一旦你适应了你的模型，你只需要两行代码。首先，导入export_text:

from sklearn.tree import export_text

其次，创建一个包含规则的对象。为了使规则看起来更具可读性，使用feature_names参数并传递一个特性名称列表。例如，如果你的模型是model，你的特征是在一个名为X_train的数据框架中命名的，你可以创建一个名为tree_rules的对象:

tree_rules = export_text(model, feature_names=list(X_train.columns))

然后打印或保存tree_rules。输出如下所示:

|--- Age <= 0.63
|   |--- EstimatedSalary <= 0.61
|   |   |--- Age <= -0.16
|   |   |   |--- class: 0
|   |   |--- Age >  -0.16
|   |   |   |--- EstimatedSalary <= -0.06
|   |   |   |   |--- class: 0
|   |   |   |--- EstimatedSalary >  -0.06
|   |   |   |   |--- EstimatedSalary <= 0.40
|   |   |   |   |   |--- EstimatedSalary <= 0.03
|   |   |   |   |   |   |--- class: 1

只需使用sklearn中的函数。像这样的树

from sklearn.tree import export_graphviz
    export_graphviz(tree,
                out_file = "tree.dot",
                feature_names = tree.columns) //or just ["petal length", "petal width"]

然后在项目文件夹中查找文件树。点，复制所有的内容，并粘贴到这里http://www.webgraphviz.com/，并生成您的图形:)