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
只需使用sklearn中的函数。像这样的树
from sklearn.tree import export_graphviz
export_graphviz(tree,
out_file = "tree.dot",
feature_names = tree.columns) //or just ["petal length", "petal width"]
然后在项目文件夹中查找文件树。点，复制所有的内容，并粘贴到这里http://www.webgraphviz.com/，并生成您的图形:)
其他回答
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)
这将打印出一个有效的Python函数。下面是一个树的输出示例，它试图返回它的输入，一个0到10之间的数字。
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_。用阈值== -2来判断节点是否是叶节点不是一个好主意。如果它是一个阈值为-2的真实决策节点呢?相反，你应该看看树。Feature or tree.children_*。 对于tree_中的i，行features = [feature_names[i]。我的sklearn版本崩溃了，因为树。树_。特征为-2(特别是叶节点)。 递归函数中不需要有多个if语句，一个就可以了。
只需使用sklearn中的函数。像这样的树
from sklearn.tree import export_graphviz
export_graphviz(tree,
out_file = "tree.dot",
feature_names = tree.columns) //or just ["petal length", "petal width"]
然后在项目文件夹中查找文件树。点，复制所有的内容，并粘贴到这里http://www.webgraphviz.com/，并生成您的图形:)
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)
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.]]
}
}
}
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'
推荐文章
- 证书验证失败:无法获得本地颁发者证书
- 当使用pip3安装包时，“Python中的ssl模块不可用”
- 无法切换Python与pyenv
- Python if not == vs if !=
- 如何从scikit-learn决策树中提取决策规则?
- 为什么在Mac OS X v10.9 (Mavericks)的终端中apt-get功能不起作用?
- 将旋转的xtick标签与各自的xtick对齐
- 为什么元组可以包含可变项?
- 如何合并字典的字典?
- 如何创建类属性?
- 数据挖掘中分类和聚类的区别?
- 不区分大小写的“in”
- 在Python中获取迭代器中的元素个数
- 解析日期字符串并更改格式
- 使用try和。Python中的if
