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)
其他回答
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
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。
Thank for the wonderful solution of @paulkerfeld. On top of his solution, for all those who want to have a serialized version of trees, just use tree.threshold, tree.children_left, tree.children_right, tree.feature and tree.value. Since the leaves don't have splits and hence no feature names and children, their placeholder in tree.feature and tree.children_*** are _tree.TREE_UNDEFINED and _tree.TREE_LEAF. Every split is assigned a unique index by depth first search. Notice that the tree.value is of shape [n, 1, 1]
下面是一个通过转换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
示例用法:
res = export_py_code(tree, feature_names=names, spacing=4)
print (res)
样例输出:
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
上面的示例生成了names = ['f'+str(j+1) for j in range(NUM_FEATURES)]。
一个方便的功能是,它可以生成更小的文件大小与减少间距。只需要设置spacing=2。
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)