我能从决策树中的训练树中提取基本的决策规则(或“决策路径”)作为文本列表吗?
喜欢的东西:
if A>0.4 then if B<0.2 then if C>0.8 then class='X'
当前回答
下面是我以一种可以直接在sql中使用的形式提取决策规则的方法，这样数据就可以按节点分组。(根据之前海报的做法)
结果将是后续的CASE子句，可以复制到sql语句，例如。
SELECT COALESCE(*CASE WHEN <conditions> THEN > <NodeA>*， >* CASE WHEN <条件> THEN <NodeB>*， > ....)* > FROM <表或视图>
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)
其他回答
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.]]
}
}
}
我创建了自己的函数，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)
这个函数首先从节点(在子数组中由-1标识)开始，然后递归地查找父节点。我称之为节点的“沿袭”。在此过程中，我获取了我需要创建if/then/else SAS逻辑的值:
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
下面的元组集包含了创建SAS if/then/else语句所需的所有内容。我不喜欢在SAS中使用do块，这就是为什么我创建逻辑来描述节点的整个路径。元组后的单个整数为路径中终端节点的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
从这个答案中，您可以得到一个可读且高效的表示:https://stackoverflow.com/a/65939892/3746632
输出如下所示。X为一维向量，表示单个实例的特征。
from numba import jit,njit
@njit
def predict(X):
ret = 0
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
ret += 1
else: # if w_pizza > 0.5
pass
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
ret += 1
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
ret += 1
else: # if w_mexico > 0.5
ret += 1
else: # if w_pizza > 0.5
pass
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
ret += 1
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
ret += 1
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
pass
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
ret += 1
else: # if w_pizza > 0.5
ret += 1
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
ret += 1
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
pass
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
pass
if X[0] <= 0.5: # if w_pizza <= 0.5
if X[1] <= 0.5: # if w_mexico <= 0.5
if X[2] <= 0.5: # if w_reusable <= 0.5
ret += 1
else: # if w_reusable > 0.5
pass
else: # if w_mexico > 0.5
pass
else: # if w_pizza > 0.5
pass
return ret/10
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'
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