我在scikit-learn文档中找到了此代码，并对其进行了修改以适合您的问题。
由于
RandomForestClassifier是的集合，
DecisionTreeClassifier我们可以遍历不同的树并检索每个树中的样本的决策路径。希望能帮助到你：
import numpy as npfrom sklearn.model_selection import train_test_splitfrom sklearn.datasets import make_classificationfrom sklearn.ensemble import RandomForestClassifierX, 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_countn_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_]))为了在随机森林中打印不同的树，您可以通过以下方式遍历估计器：
for i,e in enumerate(estimator.estimators_): print("Tree %dn"%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)这是在第一树输出
RandomForestClassifier为
sample_id = 0：
Tree 1The 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 indexdecision id node 113 : (X_test[1, 'Feature_4'] (= 2.2422482391211678) > -2.0)Prediction for sample 1: 1.0The following samples [1, 1] share the node [ 0 86 92 102 106 110 112 113] in the treeIt is 6.956521739130435 % of all nodes.Prediction for sample 1: 1.0Prediction for sample 1: 1.0Tree 2The 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 indexdecision id node 132 : (X_test[1, 'Feature_4'] (= 2.2422482391211678) > -2.0)Prediction for sample 1: 0.0The following samples [1, 1] share the node [ 0 88 102 110 130 131 132] in the treeIt is 5.185185185185185 % of all nodes.Prediction for sample 1: 0.0Prediction for sample 1: 0.0
