标签归档:random-forest

如何从scikit-learn决策树中提取决策规则?

问题:如何从scikit-learn决策树中提取决策规则?

我可以从决策树中经过训练的树中提取出基本的决策规则(或“决策路径”)作为文本列表吗?

就像是:

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

谢谢你的帮助。

Can I extract the underlying decision-rules (or ‘decision paths’) from a trained tree in a decision tree as a textual list?

Something like:

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

Thanks for your help.


回答 0

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)

这会打印出有效的Python函数。这是尝试返回其输入的树的示例输出,该数字介于0和10之间。

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.]]

这是我在其他答案中看到的一些绊脚石:

  1. 使用tree_.threshold == -2来决定一个节点是否为叶是不是一个好主意。如果它是阈值为-2的真实决策节点怎么办?相反,您应该查看tree.featuretree.children_*
  2. 该行在features = [feature_names[i] for i in tree_.feature]我的sklearn版本中崩溃,因为某些值tree.tree_.feature是-2(特别是对于叶节点)。
  3. 递归函数中不需要有多个if语句,只需一个就可以了。

I believe that this answer is more correct than the other answers here:

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)

This prints out a valid Python function. Here’s an example output for a tree that is trying to return its input, a number between 0 and 10.

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.]]

Here are some stumbling blocks that I see in other answers:

  1. Using tree_.threshold == -2 to decide whether a node is a leaf isn’t a good idea. What if it’s a real decision node with a threshold of -2? Instead, you should look at tree.feature or tree.children_*.
  2. The line features = [feature_names[i] for i in tree_.feature] crashes with my version of sklearn, because some values of tree.tree_.feature are -2 (specifically for leaf nodes).
  3. There is no need to have multiple if statements in the recursive function, just one is fine.

回答 1

我创建了自己的函数,以从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)

此函数首先从节点开始(在子数组中由-1标识),然后递归地找到父节点。我称其为节点的“血统”。一路上,我掌握了创建if / then / else SAS逻辑所需的值:

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

下面的元组集包含创建SAS if / then / else语句所需的所有内容。我不喜欢do在SAS中使用块,这就是为什么我创建描述节点整个路径的逻辑的原因。元组之后的单个整数是路径中终端节点的ID。所有前面的元组组合在一起创建该节点。

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

I created my own function to extract the rules from the decision trees created by sklearn:

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)

This function first starts with the nodes (identified by -1 in the child arrays) and then recursively finds the parents. I call this a node’s ‘lineage’. Along the way, I grab the values I need to create if/then/else SAS logic:

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

The sets of tuples below contain everything I need to create SAS if/then/else statements. I do not like using do blocks in SAS which is why I create logic describing a node’s entire path. The single integer after the tuples is the ID of the terminal node in a path. All of the preceding tuples combine to create that node.

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


回答 2

我修改了Zelazny7提交的代码以打印一些伪代码:

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)

如果调用get_code(dt, df.columns)同一示例,则将获得:

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.]]
}
}
}

I modified the code submitted by Zelazny7 to print some pseudocode:

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)

if you call get_code(dt, df.columns) on the same example you will obtain:

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.]]
}
}
}

回答 3

Scikit Learn引入了一种美味的新方法,称为export_text0.21版(2019年5月),用于从树中提取规则。文档在这里。不再需要创建自定义函数。

拟合模型后,只需两行代码。首先,导入export_text

from sklearn.tree.export import export_text

其次,创建一个包含规则的对象。为了使规则更具可读性,请使用feature_names参数并传递功能名称列表。例如,如果您的模型被调用,model并且您的要素在名为的数据框中命名X_train,则可以创建一个名为的对象tree_rules

tree_rules = export_text(model, feature_names=list(X_train))

然后只需打印或保存tree_rules。您的输出将如下所示:

|--- 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

Scikit learn introduced a delicious new method called export_text in version 0.21 (May 2019) to extract the rules from a tree. Documentation here. It’s no longer necessary to create a custom function.

Once you’ve fit your model, you just need two lines of code. First, import export_text:

from sklearn.tree import export_text

Second, create an object that will contain your rules. To make the rules look more readable, use the feature_names argument and pass a list of your feature names. For example, if your model is called model and your features are named in a dataframe called X_train, you could create an object called tree_rules:

tree_rules = export_text(model, feature_names=list(X_train.columns))

Then just print or save tree_rules. Your output will look like this:

|--- 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

回答 4

0.18.0版本中提供了一种新DecisionTreeClassifier方法。开发人员提供了广泛的(有据可查的)演练decision_path

演练中打印树结构的代码的第一部分似乎还可以。但是,我修改了第二部分中的代码以询问一个样本。我的更改用表示# <--

编辑# <--在拉取请求#8653#10951中指出错误之后,以下代码中标记的更改已在演练链接中更新。现在跟随起来要容易得多。

sample_id = 0
node_index = node_indicator.indices[node_indicator.indptr[sample_id]:
                                    node_indicator.indptr[sample_id + 1]]

print('Rules used to predict sample %s: ' % sample_id)
for node_id in node_index:

    if leave_id[sample_id] == node_id:  # <-- changed != to ==
        #continue # <-- comment out
        print("leaf node {} reached, no decision here".format(leave_id[sample_id])) # <--

    else: # < -- added else to iterate through decision nodes
        if (X_test[sample_id, feature[node_id]] <= threshold[node_id]):
            threshold_sign = "<="
        else:
            threshold_sign = ">"

        print("decision id node %s : (X[%s, %s] (= %s) %s %s)"
              % (node_id,
                 sample_id,
                 feature[node_id],
                 X_test[sample_id, feature[node_id]], # <-- changed i to sample_id
                 threshold_sign,
                 threshold[node_id]))

Rules used to predict sample 0: 
decision id node 0 : (X[0, 3] (= 2.4) > 0.800000011921)
decision id node 2 : (X[0, 2] (= 5.1) > 4.94999980927)
leaf node 4 reached, no decision here

更改sample_id以查看其他样本的决策路径。我没有问过开发人员这些更改,只是在研究示例时看起来更加直观。

There is a new DecisionTreeClassifier method, decision_path, in the 0.18.0 release. The developers provide an extensive (well-documented) walkthrough.

The first section of code in the walkthrough that prints the tree structure seems to be OK. However, I modified the code in the second section to interrogate one sample. My changes denoted with # <--

Edit The changes marked by # <-- in the code below have since been updated in walkthrough link after the errors were pointed out in pull requests #8653 and #10951. It’s much easier to follow along now.

sample_id = 0
node_index = node_indicator.indices[node_indicator.indptr[sample_id]:
                                    node_indicator.indptr[sample_id + 1]]

print('Rules used to predict sample %s: ' % sample_id)
for node_id in node_index:

    if leave_id[sample_id] == node_id:  # <-- changed != to ==
        #continue # <-- comment out
        print("leaf node {} reached, no decision here".format(leave_id[sample_id])) # <--

    else: # < -- added else to iterate through decision nodes
        if (X_test[sample_id, feature[node_id]] <= threshold[node_id]):
            threshold_sign = "<="
        else:
            threshold_sign = ">"

        print("decision id node %s : (X[%s, %s] (= %s) %s %s)"
              % (node_id,
                 sample_id,
                 feature[node_id],
                 X_test[sample_id, feature[node_id]], # <-- changed i to sample_id
                 threshold_sign,
                 threshold[node_id]))

Rules used to predict sample 0: 
decision id node 0 : (X[0, 3] (= 2.4) > 0.800000011921)
decision id node 2 : (X[0, 2] (= 5.1) > 4.94999980927)
leaf node 4 reached, no decision here

Change the sample_id to see the decision paths for other samples. I haven’t asked the developers about these changes, just seemed more intuitive when working through the example.


回答 5

from StringIO import StringIO
out = StringIO()
out = tree.export_graphviz(clf, out_file=out)
print out.getvalue()

您可以看到有向图树。然后,clf.tree_.featureclf.tree_.value分别是节点分割特征数组和节点值数组。您可以从此github源中引用更多详细信息。

from StringIO import StringIO
out = StringIO()
out = tree.export_graphviz(clf, out_file=out)
print out.getvalue()

You can see a digraph Tree. Then, clf.tree_.feature and clf.tree_.value are array of nodes splitting feature and array of nodes values respectively. You can refer to more details from this github source.


回答 6

仅仅因为每个人都非常乐于助人,所以我将对Zelazny7和Daniele的精美解决方案进行修改。这个是针对python 2.7的,带有标签使其更具可读性:

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)

Just because everyone was so helpful I’ll just add a modification to Zelazny7 and Daniele’s beautiful solutions. This one is for python 2.7, with tabs to make it more readable:

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)

回答 7

下面的代码是我在anaconda python 2.7下加上包名称“ pydot-ng”制作带有决策规则的PDF文件的方法。希望对您有所帮助。

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)

一个树形图在这里显示

Codes below is my approach under anaconda python 2.7 plus a package name “pydot-ng” to making a PDF file with decision rules. I hope it is helpful.

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)

a tree graphy show here


回答 8

我已经经历过了,但是我需要规则以这种格式编写

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

因此,我修改了@paulkernfeld的答案(谢谢),您可以根据自己的需要进行自定义

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)

I’ve been going through this, but i needed the rules to be written in this format

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

So I adapted the answer of @paulkernfeld (thanks) that you can customize to your need

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)

回答 9

这是一种使用SKompiler库将整个树转换为单个(不一定是人类可读的)python表达式的方法:

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

Here is a way to translate the whole tree into a single (not necessarily too human-readable) python expression using the SKompiler library:

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

回答 10

这基于@paulkernfeld的答案。如果您有一个具有特征的数据框X和一个具有共振的目标数据框y,并且想要了解哪个y值终止于哪个节点(并相应地对其进行绘制),则可以执行以下操作:

    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,但可以胜任工作…

This builds on @paulkernfeld ‘s answer. If you have a dataframe X with your features and a target dataframe y with your resonses and you you want to get an idea which y value ended in which node (and also ant to plot it accordingly) you can do the following:

    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 

not the most elegant version but it does the job…


回答 11

这是您需要的代码

我已经修改了最喜欢的代码以正确缩进jupyter笔记本python 3

import numpy as np
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, np.argmax(tree_.value[node])))

    recurse(0, 1)

This is the code you need

I have modified the top liked code to indent in a jupyter notebook python 3 correctly

import numpy as np
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, np.argmax(tree_.value[node])))

    recurse(0, 1)

回答 12

这是一个函数,在python 3下打印scikit-learn决策树的规则,并带有条件块的偏移量以使结构更易读:

def print_decision_tree(tree, feature_names=None, offset_unit='    '):
    '''Plots textual representation of rules of a decision tree
    tree: scikit-learn representation of tree
    feature_names: list of feature names. They are set to f1,f2,f3,... if not specified
    offset_unit: a string of offset of the conditional block'''

    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+"return " + str(value[node]))

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

Here is a function, printing rules of a scikit-learn decision tree under python 3 and with offsets for conditional blocks to make the structure more readable:

def print_decision_tree(tree, feature_names=None, offset_unit='    '):
    '''Plots textual representation of rules of a decision tree
    tree: scikit-learn representation of tree
    feature_names: list of feature names. They are set to f1,f2,f3,... if not specified
    offset_unit: a string of offset of the conditional block'''

    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+"return " + str(value[node]))

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

回答 13

您还可以通过区分它属于哪个类,甚至提及其输出值来使它更具信息性。

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)

You can also make it more informative by distinguishing it to which class it belongs or even by mentioning its output value.

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)


回答 14

这是我提取可直接在sql中使用的形式的决策规则的方法,因此可以按节点对数据进行分组。(基于先前海报的方法。)

结果将是CASE可以复制到sql语句(例如)的后续子句。

SELECT COALESCE(*CASE WHEN <conditions> THEN > <NodeA>*, > *CASE WHEN <conditions> THEN <NodeB>*, > ....)NodeName,* > FROM <table or view>


import numpy as np

import pickle
feature_names=.............
features  = [feature_names[i] for i in range(len(feature_names))]
clf= pickle.loads(trained_model)
impurity=clf.tree_.impurity
importances = clf.feature_importances_
SqlOut=""

#global Conts
global ContsNode
global Path
#Conts=[]#
ContsNode=[]
Path=[]
global Results
Results=[]

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,ParentNode=0,IsElse=0):
        global Conts
        global ContsNode
        global Path
        global Results
        global LeftParents
        LeftParents=[]
        global RightParents
        RightParents=[]
        for i in range(len(left)): # This is just to tell you how to create a list.
            LeftParents.append(-1)
            RightParents.append(-1)
            ContsNode.append("")
            Path.append("")


        for i in range(len(left)): # i is node
            if (left[i]==-1 and right[i]==-1):      
                if LeftParents[i]>=0:
                    if Path[LeftParents[i]]>" ":
                        Path[i]=Path[LeftParents[i]]+" AND " +ContsNode[LeftParents[i]]                                 
                    else:
                        Path[i]=ContsNode[LeftParents[i]]                                   
                if RightParents[i]>=0:
                    if Path[RightParents[i]]>" ":
                        Path[i]=Path[RightParents[i]]+" AND not " +ContsNode[RightParents[i]]                                   
                    else:
                        Path[i]=" not " +ContsNode[RightParents[i]]                     
                Results.append(" case when  " +Path[i]+"  then ''" +"{:4d}".format(i)+ " "+"{:2.2f}".format(impurity[i])+" "+Path[i][0:180]+"''")

            else:       
                if LeftParents[i]>=0:
                    if Path[LeftParents[i]]>" ":
                        Path[i]=Path[LeftParents[i]]+" AND " +ContsNode[LeftParents[i]]                                 
                    else:
                        Path[i]=ContsNode[LeftParents[i]]                                   
                if RightParents[i]>=0:
                    if Path[RightParents[i]]>" ":
                        Path[i]=Path[RightParents[i]]+" AND not " +ContsNode[RightParents[i]]                                   
                    else:
                        Path[i]=" not "+ContsNode[RightParents[i]]                      
                if (left[i]!=-1):
                    LeftParents[left[i]]=i
                if (right[i]!=-1):
                    RightParents[right[i]]=i
                ContsNode[i]=   "( "+ features[i] + " <= " + str(threshold[i])   + " ) "

    recurse(left, right, threshold, features, 0,0,0,0)
print_decision_tree(clf,features)
SqlOut=""
for i in range(len(Results)): 
    SqlOut=SqlOut+Results[i]+ " end,"+chr(13)+chr(10)

Here is my approach to extract the decision rules in a form that can be used in directly in sql, so the data can be grouped by node. (Based on the approaches of previous posters.)

The result will be subsequent CASE clauses that can be copied to an sql statement, ex.

SELECT COALESCE(*CASE WHEN <conditions> THEN > <NodeA>*, > *CASE WHEN <conditions> THEN <NodeB>*, > ....)NodeName,* > FROM <table or view>


import numpy as np

import pickle
feature_names=.............
features  = [feature_names[i] for i in range(len(feature_names))]
clf= pickle.loads(trained_model)
impurity=clf.tree_.impurity
importances = clf.feature_importances_
SqlOut=""

#global Conts
global ContsNode
global Path
#Conts=[]#
ContsNode=[]
Path=[]
global Results
Results=[]

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,ParentNode=0,IsElse=0):
        global Conts
        global ContsNode
        global Path
        global Results
        global LeftParents
        LeftParents=[]
        global RightParents
        RightParents=[]
        for i in range(len(left)): # This is just to tell you how to create a list.
            LeftParents.append(-1)
            RightParents.append(-1)
            ContsNode.append("")
            Path.append("")


        for i in range(len(left)): # i is node
            if (left[i]==-1 and right[i]==-1):      
                if LeftParents[i]>=0:
                    if Path[LeftParents[i]]>" ":
                        Path[i]=Path[LeftParents[i]]+" AND " +ContsNode[LeftParents[i]]                                 
                    else:
                        Path[i]=ContsNode[LeftParents[i]]                                   
                if RightParents[i]>=0:
                    if Path[RightParents[i]]>" ":
                        Path[i]=Path[RightParents[i]]+" AND not " +ContsNode[RightParents[i]]                                   
                    else:
                        Path[i]=" not " +ContsNode[RightParents[i]]                     
                Results.append(" case when  " +Path[i]+"  then ''" +"{:4d}".format(i)+ " "+"{:2.2f}".format(impurity[i])+" "+Path[i][0:180]+"''")

            else:       
                if LeftParents[i]>=0:
                    if Path[LeftParents[i]]>" ":
                        Path[i]=Path[LeftParents[i]]+" AND " +ContsNode[LeftParents[i]]                                 
                    else:
                        Path[i]=ContsNode[LeftParents[i]]                                   
                if RightParents[i]>=0:
                    if Path[RightParents[i]]>" ":
                        Path[i]=Path[RightParents[i]]+" AND not " +ContsNode[RightParents[i]]                                   
                    else:
                        Path[i]=" not "+ContsNode[RightParents[i]]                      
                if (left[i]!=-1):
                    LeftParents[left[i]]=i
                if (right[i]!=-1):
                    RightParents[right[i]]=i
                ContsNode[i]=   "( "+ features[i] + " <= " + str(threshold[i])   + " ) "

    recurse(left, right, threshold, features, 0,0,0,0)
print_decision_tree(clf,features)
SqlOut=""
for i in range(len(Results)): 
    SqlOut=SqlOut+Results[i]+ " end,"+chr(13)+chr(10)

回答 15

现在您可以使用export_text。

from sklearn.tree import export_text

r = export_text(loan_tree, feature_names=(list(X_train.columns)))
print(r)

[sklearn] [1]中的完整示例

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)

Now you can use export_text.

from sklearn.tree import export_text

r = export_text(loan_tree, feature_names=(list(X_train.columns)))
print(r)

A complete example from [sklearn][1]

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)

回答 16

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'

Modified Zelazny7’s code to fetch SQL from the decision tree.

# 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'

回答 17

显然,很久以前,已经有人决定尝试将以下功能添加到官方scikit的树导出功能中(该功能基本上仅支持export_graphviz)

def export_dict(tree, feature_names=None, max_depth=None) :
    """Export a decision tree in dict format.

这是他的全部承诺:

https://github.com/scikit-learn/scikit-learn/blob/79bdc8f711d0af225ed6be9fdb708cea9f98a910/sklearn/tree/export.py

不确定该评论发生了什么。但是您也可以尝试使用该功能。

我认为这对scikit-learn的优秀人员提出了严肃的文档要求,以正确地记录sklearn.tree.TreeAPI,API是DecisionTreeClassifier作为其属性公开的底层树结构tree_

Apparently a long time ago somebody already decided to try to add the following function to the official scikit’s tree export functions (which basically only supports export_graphviz)

def export_dict(tree, feature_names=None, max_depth=None) :
    """Export a decision tree in dict format.

Here is his full commit:

https://github.com/scikit-learn/scikit-learn/blob/79bdc8f711d0af225ed6be9fdb708cea9f98a910/sklearn/tree/export.py

Not exactly sure what happened to this comment. But you could also try to use that function.

I think this warrants a serious documentation request to the good people of scikit-learn to properly document the sklearn.tree.Tree API which is the underlying tree structure that DecisionTreeClassifier exposes as its attribute tree_.


回答 18

像这样使用sklearn.tree中的函数

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tree.dot文件,复制所有内容并将其粘贴到此处http://www.webgraphviz.com/并生成图形:)

Just use the function from sklearn.tree like this

from sklearn.tree import export_graphviz
    export_graphviz(tree,
                out_file = "tree.dot",
                feature_names = tree.columns) //or just ["petal length", "petal width"]

And then look in your project folder for the file tree.dot, copy the ALL the content and paste it here http://www.webgraphviz.com/ and generate your graph :)


回答 19

感谢@paulkerfeld的出色解决方案。在他的解决方案之上,为所有那些谁希望有树木序列化版本,只要使用tree.thresholdtree.children_lefttree.children_righttree.featuretree.value。由于叶子没有分裂,因此没有要素名称和子元素,因此它们在tree.featuretree.children_***中的占位符为_tree.TREE_UNDEFINEDand _tree.TREE_LEAF。每个分割均由分配唯一索引depth first search
请注意,tree.value形状为[n, 1, 1]

Thank for the wonderful solution of @paulkerfeld. On top of his solution, for all those who want to have a serialized version of trees, just use tree.threshold, tree.children_left, tree.children_right, tree.feature and tree.value. Since the leaves don’t have splits and hence no feature names and children, their placeholder in tree.feature and tree.children_*** are _tree.TREE_UNDEFINED and _tree.TREE_LEAF. Every split is assigned a unique index by depth first search.
Notice that the tree.value is of shape [n, 1, 1]


回答 20

这是一个通过转换以下内容的决策树生成Python代码的函数export_text

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

用法示例:

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

样本输出:

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

上面的示例是使用生成的names = ['f'+str(j+1) for j in range(NUM_FEATURES)]

一个方便的功能是,它可以生成较小的文件,且间距减小。刚设定spacing=2

Here is a function that generates Python code from a decision tree by converting the output of export_text:

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

Sample usage:

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

Sample output:

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

The above example is generated with names = ['f'+str(j+1) for j in range(NUM_FEATURES)].

One handy feature is that it can generate smaller file size with reduced spacing. Just set spacing=2.


Tpot-使用遗传编程优化机器学习管道的Python自动机器学习工具

TPOT代表T基于REE的PipelineO优化T哦哦。将TPOT视为您的数据科学助理TPOT是一种Python自动机器学习工具,可使用遗传编程优化机器学习管道

TPOT将通过智能地探索数千个可能的管道来找到最适合您数据的管道,从而自动化机器学习中最繁琐的部分

一个机器学习流水线示例

一旦TPOT完成搜索(或者您厌倦了等待),它就会为您提供它找到的最佳管道的Python代码,这样您就可以从那里修补管道了

TPOT构建在SCRICKIT-LEARN之上,因此它生成的所有代码看起来都应该很熟悉。如果你熟悉SCRICKIT-不管怎样,还是要学

TPOT仍在积极发展中我们鼓励您定期检查此存储库是否有更新

有关TPOT的更多信息,请参阅project documentation

许可证

请参阅repository license有关TPOT的许可和使用信息

通常,我们已经授权TPOT使其尽可能广泛使用

安装

我们坚持TPOT installation instructions在文档中。TPOT需要Python的正常安装

用法

可以使用TPOTon the command linewith Python code

单击相应的链接以在文档中查找有关TPOT用法的更多信息

示例

分类

以下是光学识别手写数字数据集的最小工作示例

from tpot import TPOTClassifier
from sklearn.datasets import load_digits
from sklearn.model_selection import train_test_split

digits = load_digits()
X_train, X_test, y_train, y_test = train_test_split(digits.data, digits.target,
                                                    train_size=0.75, test_size=0.25, random_state=42)

tpot = TPOTClassifier(generations=5, population_size=50, verbosity=2, random_state=42)
tpot.fit(X_train, y_train)
print(tpot.score(X_test, y_test))
tpot.export('tpot_digits_pipeline.py')

运行此代码应该会发现达到约98%测试准确率的管道,并且相应的Python代码应该导出到tpot_digits_pipeline.py文件,如下所示:

import numpy as np
import pandas as pd
from sklearn.ensemble import RandomForestClassifier
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline, make_union
from sklearn.preprocessing import PolynomialFeatures
from tpot.builtins import StackingEstimator
from tpot.export_utils import set_param_recursive

# NOTE: Make sure that the outcome column is labeled 'target' in the data file
tpot_data = pd.read_csv('PATH/TO/DATA/FILE', sep='COLUMN_SEPARATOR', dtype=np.float64)
features = tpot_data.drop('target', axis=1)
training_features, testing_features, training_target, testing_target = \
            train_test_split(features, tpot_data['target'], random_state=42)

# Average CV score on the training set was: 0.9799428471757372
exported_pipeline = make_pipeline(
    PolynomialFeatures(degree=2, include_bias=False, interaction_only=False),
    StackingEstimator(estimator=LogisticRegression(C=0.1, dual=False, penalty="l1")),
    RandomForestClassifier(bootstrap=True, criterion="entropy", max_features=0.35000000000000003, min_samples_leaf=20, min_samples_split=19, n_estimators=100)
)
# Fix random state for all the steps in exported pipeline
set_param_recursive(exported_pipeline.steps, 'random_state', 42)

exported_pipeline.fit(training_features, training_target)
results = exported_pipeline.predict(testing_features)

回归

同样,TPOT可以针对回归问题优化管道。下面是使用Practice波士顿房价数据集的最小工作示例

from tpot import TPOTRegressor
from sklearn.datasets import load_boston
from sklearn.model_selection import train_test_split

housing = load_boston()
X_train, X_test, y_train, y_test = train_test_split(housing.data, housing.target,
                                                    train_size=0.75, test_size=0.25, random_state=42)

tpot = TPOTRegressor(generations=5, population_size=50, verbosity=2, random_state=42)
tpot.fit(X_train, y_train)
print(tpot.score(X_test, y_test))
tpot.export('tpot_boston_pipeline.py')

这将导致管道达到约12.77的均方误差(MSE),并且中的Python代码tpot_boston_pipeline.py应与以下内容类似:

import numpy as np
import pandas as pd
from sklearn.ensemble import ExtraTreesRegressor
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline
from sklearn.preprocessing import PolynomialFeatures
from tpot.export_utils import set_param_recursive

# NOTE: Make sure that the outcome column is labeled 'target' in the data file
tpot_data = pd.read_csv('PATH/TO/DATA/FILE', sep='COLUMN_SEPARATOR', dtype=np.float64)
features = tpot_data.drop('target', axis=1)
training_features, testing_features, training_target, testing_target = \
            train_test_split(features, tpot_data['target'], random_state=42)

# Average CV score on the training set was: -10.812040755234403
exported_pipeline = make_pipeline(
    PolynomialFeatures(degree=2, include_bias=False, interaction_only=False),
    ExtraTreesRegressor(bootstrap=False, max_features=0.5, min_samples_leaf=2, min_samples_split=3, n_estimators=100)
)
# Fix random state for all the steps in exported pipeline
set_param_recursive(exported_pipeline.steps, 'random_state', 42)

exported_pipeline.fit(training_features, training_target)
results = exported_pipeline.predict(testing_features)

请查看文档以了解more examples and tutorials

对TPOT的贡献

我们欢迎您的光临check the existing issues以获取要处理的错误或增强功能。如果您有扩展TPOT的想法,请file a new issue这样我们就可以讨论一下了

在提交任何投稿之前,请审阅我们的contribution guidelines

对TPOT有问题或有疑问吗?

check the existing open and closed issues看看您的问题是否已经得到处理。如果没有,file a new issue在此存储库上,以便我们可以检查您的问题

引用TPOT

如果您在科学出版物中使用TPOT,请考虑至少引用以下一篇论文:

陈天乐,傅维轩,杰森·H·摩尔(2020)。Scaling tree-based automated machine learning to biomedical big data with a feature set selector生物信息学36(1):250-256

BibTeX条目:

@article{le2020scaling,
  title={Scaling tree-based automated machine learning to biomedical big data with a feature set selector},
  author={Le, Trang T and Fu, Weixuan and Moore, Jason H},
  journal={Bioinformatics},
  volume={36},
  number={1},
  pages={250--256},
  year={2020},
  publisher={Oxford University Press}
}

兰德尔·S·奥尔森、瑞安·J·厄巴诺维茨、彼得·C·安德鲁斯、妮可·A·拉文德、拉克里斯·基德和杰森·H·摩尔(2016)。Automating biomedical data science through tree-based pipeline optimization进化计算的应用,第123-137页

BibTeX条目:

@inbook{Olson2016EvoBio,
    author={Olson, Randal S. and Urbanowicz, Ryan J. and Andrews, Peter C. and Lavender, Nicole A. and Kidd, La Creis and Moore, Jason H.},
    editor={Squillero, Giovanni and Burelli, Paolo},
    chapter={Automating Biomedical Data Science Through Tree-Based Pipeline Optimization},
    title={Applications of Evolutionary Computation: 19th European Conference, EvoApplications 2016, Porto, Portugal, March 30 -- April 1, 2016, Proceedings, Part I},
    year={2016},
    publisher={Springer International Publishing},
    pages={123--137},
    isbn={978-3-319-31204-0},
    doi={10.1007/978-3-319-31204-0_9},
    url={http://dx.doi.org/10.1007/978-3-319-31204-0_9}
}

兰德尔·S·奥尔森、内森·巴特利、瑞安·J·厄巴诺维奇和杰森·H·摩尔(2016)。Evaluation of a Tree-based Pipeline Optimization Tool for Automating Data ScienceGECCO 2016论文集,第485-492页

BibTeX条目:

@inproceedings{OlsonGECCO2016,
    author = {Olson, Randal S. and Bartley, Nathan and Urbanowicz, Ryan J. and Moore, Jason H.},
    title = {Evaluation of a Tree-based Pipeline Optimization Tool for Automating Data Science},
    booktitle = {Proceedings of the Genetic and Evolutionary Computation Conference 2016},
    series = {GECCO '16},
    year = {2016},
    isbn = {978-1-4503-4206-3},
    location = {Denver, Colorado, USA},
    pages = {485--492},
    numpages = {8},
    url = {http://doi.acm.org/10.1145/2908812.2908918},
    doi = {10.1145/2908812.2908918},
    acmid = {2908918},
    publisher = {ACM},
    address = {New York, NY, USA},
}

或者,您也可以使用以下DOI直接引用存储库:

支持TPOT

TPOT是在Computational Genetics LabUniversity of Pennsylvania有了来自NIH在赠款R01 AI117694项下。我们非常感谢美国国立卫生研究院和宾夕法尼亚大学在这个项目的发展过程中给予的支持

TPOT标志是由托德·纽穆伊斯(Todd Newmuis)设计的,他慷慨地为该项目贡献了时间