如何从scikit-learn决策树中提取决策规则？
我可以从决策树中的受过训练的树中提取基础决策规则（或“决策路径”）作为文本列表吗？
类似的东西：
if A>0.4 then if B<0.2 then if C>0.8 then class='X'
感谢您的帮助。
23 个答案:
答案 0 :(得分：96)
我相信这个答案比其他答案更正确：
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)
这将打印出有效的Python函数。以下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.]]
以下是我在其他答案中看到的一些绊脚石：
- 使用
tree_.threshold == -2来确定节点是否为叶子不是一个好主意。如果它是一个阈值为-2的真实决策节点怎么办？相反，您应该查看tree.feature或tree.children_*。 - 行
features = [feature_names[i] for i in tree_.feature]与我的sklearn版本崩溃，因为tree.tree_.feature的某些值为-2（特别是对于叶节点）。 - 递归函数中不需要多个if语句，只需要一个。
答案 1 :(得分：45)
我创建了自己的函数来从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（在子数组中由-1标识）开始，然后以递归方式查找父节点。我将此称为节点的“谱系”。一路上，我抓住了我需要创建的值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
下面的元组集包含创建SAS if / then / else语句所需的一切。我不喜欢在SAS中使用do块，这就是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
答案 2 :(得分：33)
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)
如果您在同一示例中致电get_code(dt, df.columns)，您将获得：
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.]]
}
}
}
答案 3 :(得分：12)
DecisionTreeClassifier版本中有一个新的0.18.0方法decision_path。开发人员提供了广泛的（记录完备的）walkthrough。
打印树结构的演练中的第一部分代码似乎没问题。但是，我修改了第二部分中的代码以询问一个样本。我的更改以# <--
修改在拉取请求#8653和{{3}中指出错误后，以下代码中# <--标记的更改已在演练链接中更新}。现在跟进起来要容易得多。
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。我没有向开发人员询问这些变化，在完成示例时似乎更直观。
答案 4 :(得分：11)
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_.value是分别分割特征和节点值数组的节点阵列。您可以参考此github source中的更多详细信息。
答案 5 :(得分：5)
Scikit Learn在0.21版（2019年5月）中引入了一种名为export_text的美味新方法，用于从树中提取规则。 Documentation here。不再需要创建自定义函数。
一旦适合模型，您只需要两行代码。首先，导入export_text：
from sklearn.tree.export import export_text
第二，创建一个包含规则的对象。要使规则看起来更具可读性，请使用feature_names参数并传递功能名称列表。例如，如果您的模型名为model，并且要素在名为X_train的数据框中命名，则可以创建一个名为tree_rules的对象：
tree_rules = export_text(model, feature_names=list(X_train))
然后仅打印或保存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
答案 6 :(得分：3)
仅仅因为每个人都非常乐于助人，我只想对Zelazny7和Daniele的漂亮解决方案进行修改。这个用于python 2.7，带有标签以使其更具可读性：
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)
答案 7 :(得分：3)
现在您可以使用export_text。
delimiter[sklearn] [1]中的完整示例
from sklearn.tree import export_text
r = export_text(loan_tree, feature_names=(list(X_train.columns)))
print(r)
答案 8 :(得分：2)
以下代码是我在anaconda python 2.7下的方法加上包名＆＃34; pydot-ng＆＃34;制作带有决策规则的PDF文件。我希望它有所帮助。
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)
答案 9 :(得分：2)
这建立在@paulkernfeld的答案之上。如果你有一个带有你的特征的数据框X和一个带有你的共鸣的目标数据框y你想知道哪个y值在哪个节点结束（以及相应地绘制它的蚂蚁）你可以做以下事情：
def tree_to_code(tree, feature_names):
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，但它确实起作用......
答案 10 :(得分：2)
这是您需要的代码
我修改了最喜欢的代码以正确地在jupyter笔记本python 3中缩进
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)
答案 11 :(得分：1)
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 用户希望从树中查看确切的规则。
这就是我基于 paulkernfeld 答案实现一个函数的原因。
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
规则按分配给每个规则的训练样本数排序。对于每个规则，都有关于分类任务的预测类名和预测概率的信息。对于回归任务，仅打印有关预测值的信息。
示例
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)
打印规则：
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：Extract Rules from Decision Tree in 3 Ways with Scikit-Learn and Python。
答案 12 :(得分：1)
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。 （基于以前的张贴者的方法。）
结果将是随后的CASE子句，可以将其复制到sql语句，例如。
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)
答案 13 :(得分：1)
这是一种使用SKompiler库将整棵树转换为单个（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')
答案 14 :(得分：1)
我一直在经历这个，但我需要以这种格式编写规则
if A>0.4 then if B<0.2 then if C>0.8 then class='X'
所以我调整了@paulkernfeld的答案（谢谢），你可以根据自己的需要进行定制
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)
答案 15 :(得分：1)
这是一个函数，在python 3下打印scikit-learn决策树的规则，并使用条件块的偏移量来使结构更具可读性：
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)
答案 16 :(得分：0)
只需像这样使用 sklearn.tree 中的功能
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 tree.dot ，复制所有内容并将其粘贴到此处http://www.webgraphviz.com/并生成图形：）
答案 17 :(得分：0)
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)
答案 18 :(得分：0)
显然很久以前有人已经决定尝试将以下功能添加到官方scikit的树导出功能（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.
这是他的完整提交：
不完全确定此评论发生了什么。但您也可以尝试使用该功能。
我认为这需要向scikit的优秀人员提供严格的文档请求 - 学会正确记录sklearn.tree.Tree API，DecisionTreeClassifier API是tree_作为其属性{{1}}公开的基础树结构
答案 19 :(得分：0)
修改了Zelazny7的代码，用于从决策树中获取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'
答案 20 :(得分：0)
感谢@paulkerfeld的精彩解决方案。在他的解决方案之上，对于所有想要树的序列化版本的人，只需使用tree.threshold，tree.children_left，tree.children_right，tree.feature和tree.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.feature和tree.children_***中的占位符为_tree.TREE_UNDEFINED和_tree.TREE_LEAF。 depth first search为每个拆分分配唯一索引。
请注意，tree.value的形状为[n, 1, 1]
答案 21 :(得分：0)
这是一个通过转换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
样品用量：
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names = ['f'+str(j+1) for j in range(NUM_FEATURES)]生成的。
一个方便的功能是它可以生成较小的文件，且间距减小。只需设置spacing=2。
答案 22 :(得分：0)
从这个答案中，您会得到一个可读且有效的表示：https://stackoverflow.com/a/65939892/3746632
输出看起来像这样。 X 是表示单个实例特征的一维向量。
include:
- template: Auto-DevOps.gitlab-ci.yml
test:
variables:
DB_URL: "mongodb://mongo:27017/kubernetes-poc-app"
services:
- name: mongo:4.4.3
alias: mongo
stage: test
image: gliderlabs/herokuish:latest
needs: []
script:
- cp -R . /tmp/app
- /bin/herokuish buildpack test
rules:
- if: '$TEST_DISABLED'
when: never
- if: '$CI_COMMIT_TAG || $CI_COMMIT_BRANCH'
.production: &production_template
extends: .auto-deploy
stage: production
script:
- auto-deploy check_kube_domain
- auto-deploy download_chart
- auto-deploy ensure_namespace
- auto-deploy initialize_tiller
- auto-deploy create_secret
- auto-deploy deploy
- auto-deploy delete canary
- auto-deploy delete rollout
- auto-deploy persist_environment_url
environment:
name: production
url: http://prod.$KUBE_INGRESS_BASE_DOMAIN
artifacts:
paths: [environment_url.txt, tiller.log]
when: always
production:
<<: *production_template
rules:
- if: '$CI_KUBERNETES_ACTIVE == null || $CI_KUBERNETES_ACTIVE == ""'
when: never
- if: '$STAGING_ENABLED'
when: never
- if: '$CANARY_ENABLED'
when: never
- if: '$INCREMENTAL_ROLLOUT_ENABLED'
when: never
- if: '$INCREMENTAL_ROLLOUT_MODE'
when: never
- if: '$CI_COMMIT_BRANCH == "master"'
staging:
extends: .auto-deploy
stage: staging
variables:
DATABASE_URL: "here should be url"
DATABASE_NAME: "here should be name"
script:
- auto-deploy check_kube_domain
- auto-deploy download_chart
- auto-deploy ensure_namespace
- auto-deploy initialize_tiller
- auto-deploy create_secret
- auto-deploy deploy
- auto-deploy persist_environment_url
artifacts:
paths: [ environment_url.txt, tiller.log ]
when: always
environment:
name: staging
url: http://staging.$KUBE_INGRESS_BASE_DOMAIN
rules:
- if: '$CI_KUBERNETES_ACTIVE == null || $CI_KUBERNETES_ACTIVE == ""'
when: never
- if: '$CI_COMMIT_BRANCH != "develop"'
when: never
- if: '$STAGING_ENABLED'
review:
extends: .auto-deploy
stage: review
script:
- auto-deploy check_kube_domain
- auto-deploy download_chart
- auto-deploy ensure_namespace
- auto-deploy initialize_tiller
- auto-deploy create_secret
- auto-deploy deploy
- auto-deploy persist_environment_url
environment:
name: review/$CI_COMMIT_REF_NAME
url: http://review.$KUBE_INGRESS_BASE_DOMAIN
on_stop: stop_review
artifacts:
paths: [environment_url.txt, tiller.log]
when: always
rules:
- if: '$CI_KUBERNETES_ACTIVE == null || $CI_KUBERNETES_ACTIVE == ""'
when: never
- if: '$CI_COMMIT_BRANCH == "master" || $CI_COMMIT_BRANCH == "develop"'
when: never
- if: '$REVIEW_DISABLED'
when: never
- if: '$CI_COMMIT_TAG || $CI_COMMIT_BRANCH'
when: manual
allow_failure: true
stop_review:
extends: .auto-deploy
stage: cleanup
variables:
GIT_STRATEGY: none
script:
- auto-deploy initialize_tiller
- auto-deploy delete
environment:
name: review/$CI_COMMIT_REF_NAME
action: stop
allow_failure: true
rules:
- if: '$CI_KUBERNETES_ACTIVE == null || $CI_KUBERNETES_ACTIVE == ""'
when: never
- if: '$CI_COMMIT_BRANCH == "master" || $CI_COMMIT_BRANCH == "develop"'
when: never
- if: '$REVIEW_DISABLED'
when: never
- if: '$CI_COMMIT_TAG || $CI_COMMIT_BRANCH'
when: manual
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