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
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)
其他回答
我已经经历过这些了,但我需要把规则写成这种形式
if A>0.4 then if B<0.2 then if C>0.8 then class='X'
所以我改编了@paulkernfeld的答案(谢谢),你可以根据自己的需要定制
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)
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'
现在可以使用export_text了。
from sklearn.tree import export_text
r = export_text(loan_tree, feature_names=(list(X_train.columns)))
print(r)
来自[sklearn][1]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)
在0.18.0版本中,有一个新的DecisionTreeClassifier方法decision_path。开发人员提供了一个广泛的(文档良好的)演练。
演练中打印树结构的第一部分代码似乎没有问题。但是,我修改了第二节中的代码来检查一个示例。我的更改用# <——表示
在拉取请求#8653和#10951中指出错误后,下面代码中由# <——标记的更改已在演练链接中更新。现在就容易多了。
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以查看其他示例的决策路径。我没有向开发人员询问这些更改,只是在示例中看起来更直观。
因为每个人都很乐于助人,所以我将对Zelazny7和Daniele的漂亮解决方案进行修改。这是针对python 2.7的,使用tab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)