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
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)
其他回答
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。
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)
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
下面是一种使用SKompiler库将整个树转换为单个(不一定太容易读懂)python表达式的方法:
from skompiler import skompile
skompile(dtree.predict).to('python/code')
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)