blocks|key|57548|text|就像所有语言中的所有树：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|57549|每个节点返回1%2B所有子树大小的总和。|57550|在python中，将此方法应用于根：|57551|def+size(tree):
++++return+1+%2B+sum([size(subtree)+for+subtree+in+tree.subtrees])|code-block|syntax|javascript|57552|特别是对于sklearn，请看这里的源代码[https://github.com/scikit-learn/scikit-learn/tree/master/sklearn]|offset|length|57553|我认为这可以尝试一下：|57554|nodeNumber+=+sum(+len(tree.value)+for+tree+in+clf.estimators_+)|57555|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/scikit-learn/scikit-learn/tree/master/sklearn]^0|0|0|0|0|L|1U|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|12|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@$M|15|N|16|1|17]]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|H|7|19|8|@]|9|@]|A|$I|J]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Like for all tree in all language :

each node return 1 + the sum of all the subtrees size.

In python, apply this method on the root :

<pre><code>def size(tree):
 return 1 + sum([size(subtree) for subtree in tree.subtrees])
</code></pre>

Specifically to sklearn,looking at the source code here [<a href="https://github.com/scikit-learn/scikit-learn/tree/master/sklearn]" rel="nofollow">https://github.com/scikit-learn/scikit-learn/tree/master/sklearn]</a>

I think that this could be tried :

<pre><code>nodeNumber = sum( len(tree.value) for tree in clf.estimators_ )
</code></pre>

blocks|key|57606|text|sklearn.tree._tree.Tree对象有一个'node_count‘属性：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|57607|from+sklearn+import+tree
X+=+[[0,+0],+[1,+1]]
Y+=+[0,+1]
clf+=+tree.DecisionTreeClassifier()
clf+=+clf.fit(X,+Y)
treeObj+=+clf.tree_
print+treeObj.node_count|code-block|syntax|javascript|57608|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A sklearn.tree._tree.Tree object has a 'node_count' property:

<pre><code>from sklearn import tree
X = [[0, 0], [1, 1]]
Y = [0, 1]
clf = tree.DecisionTreeClassifier()
clf = clf.fit(X, Y)
treeObj = clf.tree_
print treeObj.node_count
</code></pre>

blocks|key|57632|text|最大深度是一个非常有用的指标，我在API中找不到它，所以我写了这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|57633|def+dectree_max_depth(tree):
++++n_nodes+=+tree.node_count
++++children_left+=+tree.children_left
++++children_right+=+tree.children_right

++++def+walk(node_id):
++++++++if+(children_left[node_id]+!=+children_right[node_id]):
++++++++++++left_max+=+1+%2B+walk(children_left[node_id])
++++++++++++right_max+=+1+%2B+walk(children_right[node_id])
++++++++++++return+max(left_max,+right_max)
++++++++else:+#+leaf
++++++++++++return+1

++++root_node_id+=+0
++++return+walk(root_node_id)|code-block|syntax|javascript|57634|您可以在森林中的所有树(rf)上使用它，如下所示：|offset|length|style|CODE|57635|[dectree_max_depth(t.tree_)+for+t+in+rf.estimators_]|57636|BSD许可证。|57637|entityMap^0|0|0|C|2|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Max depth is a pretty useful metric, which I didn't find in the API so I wrote this:

<pre><code>def dectree_max_depth(tree):
 n_nodes = tree.node_count
 children_left = tree.children_left
 children_right = tree.children_right

 def walk(node_id):
 if (children_left[node_id] != children_right[node_id]):
 left_max = 1 + walk(children_left[node_id])
 right_max = 1 + walk(children_right[node_id])
 return max(left_max, right_max)
 else: # leaf
 return 1

 root_node_id = 0
 return walk(root_node_id)
</code></pre>

You can use it on all trees in a forest (<code>rf</code>) like this:

<pre><code>[dectree_max_depth(t.tree_) for t in rf.estimators_]
</code></pre>

BSD license.

blocks|key|4564008|text|根据前面的答案，scikit+learn中随机森林的正确代码为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4564009|nodeNumber+=+sum(+clf.estimators_中树的tree.tree_.node_count+)|4564010|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Based on previous answers, the correct code for a Random Forest in scikit learn would be:
nodeNumber = sum( tree.tree_.node_count for tree in clf.estimators_ )

I'm doing some feature induction with decision trees and would like to know the size of the tree in terms of number of nodes. How do I do that in python?

Using the stock example from sklearn's website,

<pre><code>x = [[0,0],[0,1]]
y = [0,1] 

from sklearn.ensemble import RandomForestClassifier
clf = RandomForestClassifer(n_estimators = 10)
clf = clf.fit(x,y)
</code></pre>

I can get to individual trees by something like clf[1], clf[...], but how can I determine the size of each tree in terms of total node number?

How to find out the size of a sklearn decision tree in python?

我正在用决策树做一些特征归纳，我想知道树的节点数量的大小。我如何在python中做到这一点？使用sklearn网站上的股票示例，x = [[0,0],[0,1]]y = [0,1] from sklearn.ensemble import RandomForestClassifierclf = RandomForest...

问如何在python中找到sklearn决策树的大小？
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在python中找到sklearn决策树的大小？EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何在python中找到sklearn决策树的大小？
EN