从这个答案中，您可以得到一个可读且高效的表示:https://stackoverflow.com/a/65939892/3746632

输出如下所示。X为一维向量，表示单个实例的特征。

from numba import jit,njit
@njit
def predict(X):
    ret = 0
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            ret += 1
    else:  # if w_pizza > 0.5
        pass
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        ret += 1
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                ret += 1
        else:  # if w_mexico > 0.5
            ret += 1
    else:  # if w_pizza > 0.5
        pass
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                ret += 1
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        ret += 1
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        pass
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            ret += 1
    else:  # if w_pizza > 0.5
        ret += 1
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        ret += 1
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        pass
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        pass
    if X[0] <= 0.5: # if w_pizza <= 0.5
        if X[1] <= 0.5: # if w_mexico <= 0.5
            if X[2] <= 0.5: # if w_reusable <= 0.5
                ret += 1
            else:  # if w_reusable > 0.5
                pass
        else:  # if w_mexico > 0.5
            pass
    else:  # if w_pizza > 0.5
        pass
    return ret/10

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。

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)

我相信这个答案比这里的其他答案更正确:

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)

这将打印出一个有效的Python函数。下面是一个树的输出示例，它试图返回它的输入，一个0到10之间的数字。

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_。用阈值== -2来判断节点是否是叶节点不是一个好主意。如果它是一个阈值为-2的真实决策节点呢?相反，你应该看看树。Feature or tree.children_*。 对于tree_中的i，行features = [feature_names[i]。我的sklearn版本崩溃了，因为树。树_。特征为-2(特别是叶节点)。 递归函数中不需要有多个if语句，一个就可以了。

我已经经历过这些了，但我需要把规则写成这种形式

if A>0.4 then if B<0.2 then if C>0.8 then class='X' 

所以我改编了@paulkernfeld的答案(谢谢)，你可以根据自己的需要定制

def tree_to_code(tree, feature_names, Y):
    tree_ = tree.tree_
    feature_name = [
        feature_names[i] if i != _tree.TREE_UNDEFINED else "undefined!"
        for i in tree_.feature
    ]
    pathto=dict()

    global k
    k = 0
    def recurse(node, depth, parent):
        global k
        indent = "  " * depth

        if tree_.feature[node] != _tree.TREE_UNDEFINED:
            name = feature_name[node]
            threshold = tree_.threshold[node]
            s= "{} <= {} ".format( name, threshold, node )
            if node == 0:
                pathto[node]=s
            else:
                pathto[node]=pathto[parent]+' & ' +s

            recurse(tree_.children_left[node], depth + 1, node)
            s="{} > {}".format( name, threshold)
            if node == 0:
                pathto[node]=s
            else:
                pathto[node]=pathto[parent]+' & ' +s
            recurse(tree_.children_right[node], depth + 1, node)
        else:
            k=k+1
            print(k,')',pathto[parent], tree_.value[node])
    recurse(0, 1, 0)

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