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。

我需要一种更人性化的决策树规则格式。我正在构建开源AutoML Python包，很多时候MLJAR用户希望从树中看到确切的规则。

这就是为什么我实现了一个基于paulkernfeld答案的函数。

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

    paths = []
    path = []
    
    def recurse(node, path, paths):
        
        if tree_.feature[node] != _tree.TREE_UNDEFINED:
            name = feature_name[node]
            threshold = tree_.threshold[node]
            p1, p2 = list(path), list(path)
            p1 += [f"({name} <= {np.round(threshold, 3)})"]
            recurse(tree_.children_left[node], p1, paths)
            p2 += [f"({name} > {np.round(threshold, 3)})"]
            recurse(tree_.children_right[node], p2, paths)
        else:
            path += [(tree_.value[node], tree_.n_node_samples[node])]
            paths += [path]
            
    recurse(0, path, paths)

    # sort by samples count
    samples_count = [p[-1][1] for p in paths]
    ii = list(np.argsort(samples_count))
    paths = [paths[i] for i in reversed(ii)]
    
    rules = []
    for path in paths:
        rule = "if "
        
        for p in path[:-1]:
            if rule != "if ":
                rule += " and "
            rule += str(p)
        rule += " then "
        if class_names is None:
            rule += "response: "+str(np.round(path[-1][0][0][0],3))
        else:
            classes = path[-1][0][0]
            l = np.argmax(classes)
            rule += f"class: {class_names[l]} (proba: {np.round(100.0*classes[l]/np.sum(classes),2)}%)"
        rule += f" | based on {path[-1][1]:,} samples"
        rules += [rule]
        
    return rules

规则按照分配给每个规则的训练样本的数量进行排序。对于每条规则，都有关于预测的类名和分类任务预测概率的信息。对于回归任务，只打印关于预测值的信息。

例子

from sklearn import datasets
from sklearn.tree import DecisionTreeRegressor
from sklearn import tree

# Prepare the data data
boston = datasets.load_boston()
X = boston.data
y = boston.target

# Fit the regressor, set max_depth = 3
regr = DecisionTreeRegressor(max_depth=3, random_state=1234)
model = regr.fit(X, y)

# Print rules
rules = get_rules(regr, boston.feature_names, None)
for r in rules:
    print(r)

印刷规则:

if (RM <= 6.941) and (LSTAT <= 14.4) and (DIS > 1.385) then response: 22.905 | based on 250 samples
if (RM <= 6.941) and (LSTAT > 14.4) and (CRIM <= 6.992) then response: 17.138 | based on 101 samples
if (RM <= 6.941) and (LSTAT > 14.4) and (CRIM > 6.992) then response: 11.978 | based on 74 samples
if (RM > 6.941) and (RM <= 7.437) and (NOX <= 0.659) then response: 33.349 | based on 43 samples
if (RM > 6.941) and (RM > 7.437) and (PTRATIO <= 19.65) then response: 45.897 | based on 29 samples
if (RM <= 6.941) and (LSTAT <= 14.4) and (DIS <= 1.385) then response: 45.58 | based on 5 samples
if (RM > 6.941) and (RM <= 7.437) and (NOX > 0.659) then response: 14.4 | based on 3 samples
if (RM > 6.941) and (RM > 7.437) and (PTRATIO > 19.65) then response: 21.9 | based on 1 samples

我在我的文章中总结了从决策树中提取规则的方法:用Scikit-Learn和Python以3种方式从决策树中提取规则。

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。

因为每个人都很乐于助人，所以我将对Zelazny7和Daniele的漂亮解决方案进行修改。这是针对python 2.7的，使用tab使其更具可读性:

def get_code(tree, feature_names, tabdepth=0):
    left      = tree.tree_.children_left
    right     = tree.tree_.children_right
    threshold = tree.tree_.threshold
    features  = [feature_names[i] for i in tree.tree_.feature]
    value = tree.tree_.value

    def recurse(left, right, threshold, features, node, tabdepth=0):
            if (threshold[node] != -2):
                    print '\t' * tabdepth,
                    print "if ( " + features[node] + " <= " + str(threshold[node]) + " ) {"
                    if left[node] != -1:
                            recurse (left, right, threshold, features,left[node], tabdepth+1)
                    print '\t' * tabdepth,
                    print "} else {"
                    if right[node] != -1:
                            recurse (left, right, threshold, features,right[node], tabdepth+1)
                    print '\t' * tabdepth,
                    print "}"
            else:
                    print '\t' * tabdepth,
                    print "return " + str(value[node])

    recurse(left, right, threshold, features, 0)

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')

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'