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:)

修改了Zelazny7的代码以从决策树中获取SQL。

# SQL from decision tree

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]
     le='<='               
     g ='>'
     # 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)
     print 'case '
     for j,child in enumerate(idx):
        clause=' when '
        for node in recurse(left, right, child):
            if len(str(node))<3:
                continue
            i=node
            if i[1]=='l':  sign=le 
            else: sign=g
            clause=clause+i[3]+sign+str(i[2])+' and '
        clause=clause[:-4]+' then '+str(j)
        print clause
     print 'else 99 end as clusters'

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:)

显然，很久以前就有人决定尝试将以下函数添加到官方scikit的树导出函数中(基本上只支持export_graphviz)

def export_dict(tree, feature_names=None, max_depth=None) :
    """Export a decision tree in dict format.

以下是他的全部承诺:

https://github.com/scikit-learn/scikit-learn/blob/79bdc8f711d0af225ed6be9fdb708cea9f98a910/sklearn/tree/export.py

不太确定这条评论发生了什么。但是你也可以尝试使用这个函数。

我认为这为scikit-learn的优秀人员提供了一个严肃的文档需求，以正确地记录sklearn.tree.Tree API，这是一个底层的树结构，DecisionTreeClassifier将其作为属性tree_公开。

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