我相信这个答案比这里的其他答案更正确:
from sklearn.tree import _treedef tree_to_pre(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)
这会打印出有效的Python函数。这是一个试图返回其输入的树的示例输出,该数字介于0和10之间。
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.]]
这是我在其他答案中看到的一些绊脚石:
- 使用
tree_.threshold == -2
来决定一个节点是否为叶是不是一个好主意。如果它是阈值为-2的真实决策节点怎么办?相反,您应该查看tree.feature
或tree.children_*
。 - 该行在
features = [feature_names[i] for i in tree_.feature]
我的sklearn版本中崩溃,因为某些值tree.tree_.feature
是-2(特别是对于叶节点)。 - 递归函数中不需要有多个if语句,只需一个就可以了。
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