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。
其他回答
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。
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.]]
}
}
}
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
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。
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)