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
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)
其他回答
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:)
这是基于@paulkernfeld的回答。如果你有一个包含特征的数据框架X和一个包含共振的目标数据框架y,你想知道哪个y值结束于哪个节点(并相应地绘制它),你可以做以下工作:
def tree_to_code(tree, feature_names):
from sklearn.tree import _tree
codelines = []
codelines.append('def get_cat(X_tmp):\n')
codelines.append(' catout = []\n')
codelines.append(' for codelines in range(0,X_tmp.shape[0]):\n')
codelines.append(' Xin = X_tmp.iloc[codelines]\n')
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]
codelines.append ('{}if Xin["{}"] <= {}:\n'.format(indent, name, threshold))
recurse(tree_.children_left[node], depth + 1)
codelines.append( '{}else: # if Xin["{}"] > {}\n'.format(indent, name, threshold))
recurse(tree_.children_right[node], depth + 1)
else:
codelines.append( '{}mycat = {}\n'.format(indent, node))
recurse(0, 1)
codelines.append(' catout.append(mycat)\n')
codelines.append(' return pd.DataFrame(catout,index=X_tmp.index,columns=["category"])\n')
codelines.append('node_ids = get_cat(X)\n')
return codelines
mycode = tree_to_code(clf,X.columns.values)
# now execute the function and obtain the dataframe with all nodes
exec(''.join(mycode))
node_ids = [int(x[0]) for x in node_ids.values]
node_ids2 = pd.DataFrame(node_ids)
print('make plot')
import matplotlib.cm as cm
colors = cm.rainbow(np.linspace(0, 1, 1+max( list(set(node_ids)))))
#plt.figure(figsize=cm2inch(24, 21))
for i in list(set(node_ids)):
plt.plot(y[node_ids2.values==i],'o',color=colors[i], label=str(i))
mytitle = ['y colored by node']
plt.title(mytitle ,fontsize=14)
plt.xlabel('my xlabel')
plt.ylabel(tagname)
plt.xticks(rotation=70)
plt.legend(loc='upper center', bbox_to_anchor=(0.5, 1.00), shadow=True, ncol=9)
plt.tight_layout()
plt.show()
plt.close
不是最优雅的版本,但它做到了…
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。
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。
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)