您还可以通过区分它属于哪个类，甚至通过提到它的输出值，使它具有更丰富的信息。

def print_decision_tree(tree, feature_names, offset_unit='    '):    
left      = tree.tree_.children_left
right     = tree.tree_.children_right
threshold = tree.tree_.threshold
value = tree.tree_.value
if feature_names is None:
    features  = ['f%d'%i for i in tree.tree_.feature]
else:
    features  = [feature_names[i] for i in tree.tree_.feature]        

def recurse(left, right, threshold, features, node, depth=0):
        offset = offset_unit*depth
        if (threshold[node] != -2):
                print(offset+"if ( " + features[node] + " <= " + str(threshold[node]) + " ) {")
                if left[node] != -1:
                        recurse (left, right, threshold, features,left[node],depth+1)
                print(offset+"} else {")
                if right[node] != -1:
                        recurse (left, right, threshold, features,right[node],depth+1)
                print(offset+"}")
        else:
                #print(offset,value[node]) 

                #To remove values from node
                temp=str(value[node])
                mid=len(temp)//2
                tempx=[]
                tempy=[]
                cnt=0
                for i in temp:
                    if cnt<=mid:
                        tempx.append(i)
                        cnt+=1
                    else:
                        tempy.append(i)
                        cnt+=1
                val_yes=[]
                val_no=[]
                res=[]
                for j in tempx:
                    if j=="[" or j=="]" or j=="." or j==" ":
                        res.append(j)
                    else:
                        val_no.append(j)
                for j in tempy:
                    if j=="[" or j=="]" or j=="." or j==" ":
                        res.append(j)
                    else:
                        val_yes.append(j)
                val_yes = int("".join(map(str, val_yes)))
                val_no = int("".join(map(str, val_no)))

                if val_yes>val_no:
                    print(offset,'\033[1m',"YES")
                    print('\033[0m')
                elif val_no>val_yes:
                    print(offset,'\033[1m',"NO")
                    print('\033[0m')
                else:
                    print(offset,'\033[1m',"Tie")
                    print('\033[0m')

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

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。

下面的代码是我在anaconda python 2.7下的方法，加上一个包名“pydot-ng”，以制作具有决策规则的PDF文件。希望对大家有所帮助。

from sklearn import tree

clf = tree.DecisionTreeClassifier(max_leaf_nodes=n)
clf_ = clf.fit(X, data_y)

feature_names = X.columns
class_name = clf_.classes_.astype(int).astype(str)

def output_pdf(clf_, name):
    from sklearn import tree
    from sklearn.externals.six import StringIO
    import pydot_ng as pydot
    dot_data = StringIO()
    tree.export_graphviz(clf_, out_file=dot_data,
                         feature_names=feature_names,
                         class_names=class_name,
                         filled=True, rounded=True,
                         special_characters=True,
                          node_ids=1,)
    graph = pydot.graph_from_dot_data(dot_data.getvalue())
    graph.write_pdf("%s.pdf"%name)

output_pdf(clf_, name='filename%s'%n)

这是一个树形图

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)

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

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)