如何打印随机森林的决策路径，而不是特定样本的随机森林中单个树木的路径。
import numpy as np
import pandas as pd
from sklearn.datasets import make_classification
from sklearn.ensemble import RandomForestClassifier
X, y = make_classification(n_samples=1000,
n_features=6,
n_informative=3,
n_classes=2,
random_state=0,
shuffle=False)
# Creating a dataFrame
df = pd.DataFrame({'Feature 1':X[:,0],
'Feature 2':X[:,1],
'Feature 3':X[:,2],
'Feature 4':X[:,3],
'Feature 5':X[:,4],
'Feature 6':X[:,5],
'Class':y})
y_train = df['Class']
X_train = df.drop('Class',axis = 1)
rf = RandomForestClassifier(n_estimators=10,
random_state=0)
rf.fit(X_train, y_train)
随机森林的decision_path在v0.18中引入。 （http://scikit-learn.org/stable/modules/generated/sklearn.ensemble.RandomForestClassifier.html）
然而，它输出一个稀疏矩阵，我不确定如何理解。任何人都可以建议如何最好地打印特定样本的决策路径，然后将其可视化吗？
#Extracting the decision path for instance i = 12
i_data = X_train.iloc[12].values.reshape(1,-1)
d_path = rf.decision_path(i_data)
print(d_path)
输出：
（＆lt; 1x1432类型为''的稀疏矩阵 压缩稀疏行格式的96个存储元素＆gt;，数组（[0,133，＆gt; 282,415,588,761,910,1041,1182,1309,1432]，dtype = int32））
答案 0 :(得分：6)
我在scikit-learn文档中找到了这个code并对其进行了修改以适合您的问题。
由于RandomForestClassifier是DecisionTreeClassifier的集合，我们可以迭代不同的树并检索每个树中的样本的决策路径。
希望它有所帮助：
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.datasets import make_classification
from sklearn.ensemble import RandomForestClassifier
X, y = make_classification(n_samples=1000,
n_features=6,
n_informative=3,
n_classes=2,
random_state=0,
shuffle=False)
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)
estimator = RandomForestClassifier(n_estimators=10,
random_state=0)
estimator.fit(X_train, y_train)
# The decision estimator has an attribute called tree_ which stores the entire
# tree structure and allows access to low level attributes. The binary tree
# tree_ is represented as a number of parallel arrays. The i-th element of each
# array holds information about the node `i`. Node 0 is the tree's root. NOTE:
# Some of the arrays only apply to either leaves or split nodes, resp. In this
# case the values of nodes of the other type are arbitrary!
#
# Among those arrays, we have:
# - left_child, id of the left child of the node
# - right_child, id of the right child of the node
# - feature, feature used for splitting the node
# - threshold, threshold value at the node
#
# Using those arrays, we can parse the tree structure:
#n_nodes = estimator.tree_.node_count
n_nodes_ = [t.tree_.node_count for t in estimator.estimators_]
children_left_ = [t.tree_.children_left for t in estimator.estimators_]
children_right_ = [t.tree_.children_right for t in estimator.estimators_]
feature_ = [t.tree_.feature for t in estimator.estimators_]
threshold_ = [t.tree_.threshold for t in estimator.estimators_]
def explore_tree(estimator, n_nodes, children_left,children_right, feature,threshold,
suffix='', print_tree= False, sample_id=0, feature_names=None):
if not feature_names:
feature_names = feature
assert len(feature_names) == X.shape[1], "The feature names do not match the number of features."
# The tree structure can be traversed to compute various properties such
# as the depth of each node and whether or not it is a leaf.
node_depth = np.zeros(shape=n_nodes, dtype=np.int64)
is_leaves = np.zeros(shape=n_nodes, dtype=bool)
stack = [(0, -1)] # seed is the root node id and its parent depth
while len(stack) > 0:
node_id, parent_depth = stack.pop()
node_depth[node_id] = parent_depth + 1
# If we have a test node
if (children_left[node_id] != children_right[node_id]):
stack.append((children_left[node_id], parent_depth + 1))
stack.append((children_right[node_id], parent_depth + 1))
else:
is_leaves[node_id] = True
print("The binary tree structure has %s nodes"
% n_nodes)
if print_tree:
print("Tree structure: \n")
for i in range(n_nodes):
if is_leaves[i]:
print("%snode=%s leaf node." % (node_depth[i] * "\t", i))
else:
print("%snode=%s test node: go to node %s if X[:, %s] <= %s else to "
"node %s."
% (node_depth[i] * "\t",
i,
children_left[i],
feature[i],
threshold[i],
children_right[i],
))
print("\n")
print()
# First let's retrieve the decision path of each sample. The decision_path
# method allows to retrieve the node indicator functions. A non zero element of
# indicator matrix at the position (i, j) indicates that the sample i goes
# through the node j.
node_indicator = estimator.decision_path(X_test)
# Similarly, we can also have the leaves ids reached by each sample.
leave_id = estimator.apply(X_test)
# Now, it's possible to get the tests that were used to predict a sample or
# a group of samples. First, let's make it for the sample.
#sample_id = 0
node_index = node_indicator.indices[node_indicator.indptr[sample_id]:
node_indicator.indptr[sample_id + 1]]
print(X_test[sample_id,:])
print('Rules used to predict sample %s: ' % sample_id)
for node_id in node_index:
# tabulation = " "*node_depth[node_id] #-> makes tabulation of each level of the tree
tabulation = ""
if leave_id[sample_id] == node_id:
print("%s==> Predicted leaf index \n"%(tabulation))
#continue
if (X_test[sample_id, feature[node_id]] <= threshold[node_id]):
threshold_sign = "<="
else:
threshold_sign = ">"
print("%sdecision id node %s : (X_test[%s, '%s'] (= %s) %s %s)"
% (tabulation,
node_id,
sample_id,
feature_names[feature[node_id]],
X_test[sample_id, feature[node_id]],
threshold_sign,
threshold[node_id]))
print("%sPrediction for sample %d: %s"%(tabulation,
sample_id,
estimator.predict(X_test)[sample_id]))
# For a group of samples, we have the following common node.
sample_ids = [sample_id, 1]
common_nodes = (node_indicator.toarray()[sample_ids].sum(axis=0) ==
len(sample_ids))
common_node_id = np.arange(n_nodes)[common_nodes]
print("\nThe following samples %s share the node %s in the tree"
% (sample_ids, common_node_id))
print("It is %s %% of all nodes." % (100 * len(common_node_id) / n_nodes,))
for sample_id_ in sample_ids:
print("Prediction for sample %d: %s"%(sample_id_,
estimator.predict(X_test)[sample_id_]))
要在随机森林中打印不同的树，您可以通过这种方式迭代估算器：
for i,e in enumerate(estimator.estimators_):
print("Tree %d\n"%i)
explore_tree(estimator.estimators_[i],n_nodes_[i],children_left_[i],
children_right_[i], feature_[i],threshold_[i],
suffix=i, sample_id=1, feature_names=["Feature_%d"%i for i in range(X.shape[1])])
print('\n'*2)
这是RandomForestClassifier中sample_id = 0的第一棵树的输出：
Tree 1
The binary tree structure has 115 nodes
[ 2.36609963 1.32658511 -0.08002818 0.88295736 2.24224824 -0.71469736]
Rules used to predict sample 1:
decision id node 0 : (X_test[1, 'Feature_3'] (= 0.8829573603562209) > 0.7038955688476562)
decision id node 86 : (X_test[1, 'Feature_2'] (= -0.08002817952064323) > -1.4465678930282593)
decision id node 92 : (X_test[1, 'Feature_0'] (= 2.366099632530947) > 0.7020512223243713)
decision id node 102 : (X_test[1, 'Feature_5'] (= -0.7146973587899221) > -1.2842652797698975)
decision id node 106 : (X_test[1, 'Feature_2'] (= -0.08002817952064323) > -0.4031955599784851)
decision id node 110 : (X_test[1, 'Feature_0'] (= 2.366099632530947) > 0.717217206954956)
decision id node 112 : (X_test[1, 'Feature_4'] (= 2.2422482391211678) <= 3.0181679725646973)
==> Predicted leaf index
decision id node 113 : (X_test[1, 'Feature_4'] (= 2.2422482391211678) > -2.0)
Prediction for sample 1: 1.0
The following samples [1, 1] share the node [ 0 86 92 102 106 110 112 113] in the tree
It is 6.956521739130435 % of all nodes.
Prediction for sample 1: 1.0
Prediction for sample 1: 1.0
Tree 2
The binary tree structure has 135 nodes
[ 2.36609963 1.32658511 -0.08002818 0.88295736 2.24224824 -0.71469736]
Rules used to predict sample 1:
decision id node 0 : (X_test[1, 'Feature_3'] (= 0.8829573603562209) > 0.5484486818313599)
decision id node 88 : (X_test[1, 'Feature_2'] (= -0.08002817952064323) > -0.7239605188369751)
decision id node 102 : (X_test[1, 'Feature_5'] (= -0.7146973587899221) > -1.6143207550048828)
decision id node 110 : (X_test[1, 'Feature_0'] (= 2.366099632530947) > 2.3399271965026855)
decision id node 130 : (X_test[1, 'Feature_5'] (= -0.7146973587899221) <= -0.5680553913116455)
decision id node 131 : (X_test[1, 'Feature_0'] (= 2.366099632530947) <= 2.4545814990997314)
==> Predicted leaf index
decision id node 132 : (X_test[1, 'Feature_4'] (= 2.2422482391211678) > -2.0)
Prediction for sample 1: 0.0
The following samples [1, 1] share the node [ 0 88 102 110 130 131 132] in the tree
It is 5.185185185185185 % of all nodes.
Prediction for sample 1: 0.0
Prediction for sample 1: 0.0