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'
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
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
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.]]
}
}
}
这是您需要的代码
我已经修改了顶部喜欢的代码缩进在一个jupyter笔记本python 3正确
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)
下面是一个函数,在python3下打印scikit-learn决策树的规则,并对条件块进行偏移,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)
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'
您还可以通过区分它属于哪个类,甚至通过提到它的输出值,使它具有更丰富的信息。
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)