这是基于@paulkernfeld的回答。如果你有一个包含特征的数据框架X和一个包含共振的目标数据框架y，你想知道哪个y值结束于哪个节点(并相应地绘制它)，你可以做以下工作:

    def tree_to_code(tree, feature_names):
        from sklearn.tree import _tree
        codelines = []
        codelines.append('def get_cat(X_tmp):\n')
        codelines.append('   catout = []\n')
        codelines.append('   for codelines in range(0,X_tmp.shape[0]):\n')
        codelines.append('      Xin = X_tmp.iloc[codelines]\n')
        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]
                codelines.append ('{}if Xin["{}"] <= {}:\n'.format(indent, name, threshold))
                recurse(tree_.children_left[node], depth + 1)
                codelines.append( '{}else:  # if Xin["{}"] > {}\n'.format(indent, name, threshold))
                recurse(tree_.children_right[node], depth + 1)
            else:
                codelines.append( '{}mycat = {}\n'.format(indent, node))

        recurse(0, 1)
        codelines.append('      catout.append(mycat)\n')
        codelines.append('   return pd.DataFrame(catout,index=X_tmp.index,columns=["category"])\n')
        codelines.append('node_ids = get_cat(X)\n')
        return codelines
    mycode = tree_to_code(clf,X.columns.values)

    # now execute the function and obtain the dataframe with all nodes
    exec(''.join(mycode))
    node_ids = [int(x[0]) for x in node_ids.values]
    node_ids2 = pd.DataFrame(node_ids)

    print('make plot')
    import matplotlib.cm as cm
    colors = cm.rainbow(np.linspace(0, 1, 1+max( list(set(node_ids)))))
    #plt.figure(figsize=cm2inch(24, 21))
    for i in list(set(node_ids)):
        plt.plot(y[node_ids2.values==i],'o',color=colors[i], label=str(i))  
    mytitle = ['y colored by node']
    plt.title(mytitle ,fontsize=14)
    plt.xlabel('my xlabel')
    plt.ylabel(tagname)
    plt.xticks(rotation=70)       
    plt.legend(loc='upper center', bbox_to_anchor=(0.5, 1.00), shadow=True, ncol=9)
    plt.tight_layout()
    plt.show()
    plt.close 

不是最优雅的版本，但它做到了…

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

我修改了Zelazny7提交的代码来打印一些伪代码:

def get_code(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]
        value = tree.tree_.value

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

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

如果你在同一个例子中调用get_code(dt, df.columns)，你会得到:

if ( col1 <= 0.5 ) {
return [[ 1.  0.]]
} else {
if ( col2 <= 4.5 ) {
return [[ 0.  1.]]
} else {
if ( col1 <= 2.5 ) {
return [[ 1.  0.]]
} else {
return [[ 0.  1.]]
}
}
}

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。

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:)