下面是一种使用SKompiler库将整个树转换为单个(不一定太容易读懂)python表达式的方法:

from skompiler import skompile
skompile(dtree.predict).to('python/code')

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)

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

下面是一个通过转换export_text的输出从决策树生成Python代码的函数:

import string
from sklearn.tree import export_text

def export_py_code(tree, feature_names, max_depth=100, spacing=4):
    if spacing < 2:
        raise ValueError('spacing must be > 1')

    # Clean up feature names (for correctness)
    nums = string.digits
    alnums = string.ascii_letters + nums
    clean = lambda s: ''.join(c if c in alnums else '_' for c in s)
    features = [clean(x) for x in feature_names]
    features = ['_'+x if x[0] in nums else x for x in features if x]
    if len(set(features)) != len(feature_names):
        raise ValueError('invalid feature names')

    # First: export tree to text
    res = export_text(tree, feature_names=features, 
                        max_depth=max_depth,
                        decimals=6,
                        spacing=spacing-1)

    # Second: generate Python code from the text
    skip, dash = ' '*spacing, '-'*(spacing-1)
    code = 'def decision_tree({}):\n'.format(', '.join(features))
    for line in repr(tree).split('\n'):
        code += skip + "# " + line + '\n'
    for line in res.split('\n'):
        line = line.rstrip().replace('|',' ')
        if '<' in line or '>' in line:
            line, val = line.rsplit(maxsplit=1)
            line = line.replace(' ' + dash, 'if')
            line = '{} {:g}:'.format(line, float(val))
        else:
            line = line.replace(' {} class:'.format(dash), 'return')
        code += skip + line + '\n'

    return code

示例用法:

res = export_py_code(tree, feature_names=names, spacing=4)
print (res)

样例输出:

def decision_tree(f1, f2, f3):
    # DecisionTreeClassifier(class_weight=None, criterion='gini', max_depth=3,
    #                        max_features=None, max_leaf_nodes=None,
    #                        min_impurity_decrease=0.0, min_impurity_split=None,
    #                        min_samples_leaf=1, min_samples_split=2,
    #                        min_weight_fraction_leaf=0.0, presort=False,
    #                        random_state=42, splitter='best')
    if f1 <= 12.5:
        if f2 <= 17.5:
            if f1 <= 10.5:
                return 2
            if f1 > 10.5:
                return 3
        if f2 > 17.5:
            if f2 <= 22.5:
                return 1
            if f2 > 22.5:
                return 1
    if f1 > 12.5:
        if f1 <= 17.5:
            if f3 <= 23.5:
                return 2
            if f3 > 23.5:
                return 3
        if f1 > 17.5:
            if f1 <= 25:
                return 1
            if f1 > 25:
                return 2

上面的示例生成了names = ['f'+str(j+1) for j in range(NUM_FEATURES)]。

一个方便的功能是，它可以生成更小的文件大小与减少间距。只需要设置spacing=2。

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