python用SR-Tree实现符号回归的代码

对于符号回归问题，SR-Tree的代码实现如下：

import numpy as np
from scipy.spatial.distance import cdist

class Node:
    def __init__(self, data=None, bounds=None, children=None, isleaf=True):
        self.data = data or []
        self.bounds = bounds
        self.children = children or []
        self.isleaf = isleaf

class SR_Tree:
    def __init__(self, data, labels, thres=1, maxdepth=5):
        self.data = data
        self.labels = labels
        self.thres = thres
        self.maxdepth = maxdepth
        self.root = self.build_tree(self.data, self.labels, depth=0)

    def build_tree(self, data, labels, depth):
        if len(data) <= self.thres or depth >= self.maxdepth:
            return Node(data, bounds=self.get_bounds(data), isleaf=True)
        else:
            n = Node(bounds=self.get_bounds(data), isleaf=False)
            children = self.split_data(data, labels)
            for child_data, child_labels in children:
                child_node = self.build_tree(child_data, child_labels, depth+1)
                n.children.append(child_node)
            return n

    def split_data(self, data, labels):
        n = len(labels)
        dim = data.shape[1]
        k = np.random.randint(dim)
        pivot = np.median(data[:, k])
        left = []
        right = []
        for i in range(n):
            if data[i, k] < pivot:
                left.append((data[i], labels[i]))
            else:
                right.append((data[i], labels[i]))
        return [np.array([d for d, _ in left]), np.array([l for _, l in left])], \
               [np.array([d for d, _ in right]), np.array([l for _, l in right])]

    def query(self, x):
        node = self.root
        while not node.isleaf:
            dists = [self.get_dist(x, child.bounds) for child in node.children]
            idx = np.argmin(dists)
            node = node.children[idx]
        return np.mean(node.data[:, -1])

    def get_dist(self, x, bounds):
        if x.ndim == 1:
            x = x.reshape(1, -1)
        dist = 0.0
        for i in range(x.shape[0]):
            for j in range(x.shape[1]):
                if x[i, j] < bounds[j, 0]:
                    dist += (bounds[j, 0] - x[i, j]) ** 2
                elif x[i, j] > bounds[j, 1]:
                    dist += (x[i, j] - bounds[j, 1]) ** 2
        return dist

    def get_bounds(self, data):
        return np.array([[np.min(data[:, j]), np.max(data[:, j])] for j in range(data.shape[1])])

def generate_data(n=1000):
    x = np.random.uniform(-10, 10, (n, 2))
    y = 0.5 * x[:, 0] ** 2 - 0.3 * x[:, 1] ** 2 + 2 * x[:, 0] - 3 * x[:, 1] + 5 + np.random.normal(0, 0.5, n)
    return np.hstack((x, y.reshape(-1, 1)))

if __name__ == '__main__':
    data = generate_data()
    sr_tree = SR_Tree(data[:, :-1], data[:, -1], thres=50, maxdepth=10)
    x = np.array([-3, 5])
    y = sr_tree.query(x)
    print('Result:', y)

在这个代码实现中，我们将符号回归问题的数据和标签作为SR-Tree的输入，然后构建SR-Tree来进行查询。在SR-Tree的构建过程中，我们按照随机选择轴的策略将数据集分成两个子集，然后递归地构建SR-Tree。在查询时，我们从根节点开始遍历树，根据查询点和每个节点的边界计算距离，并移动到最近的子节点，直到达到叶节点为止，然后返回该叶节点中数据的均值作为预测结果。

在这个实现中，我们使用了numpy和scipy库来进行计算和距离计算。为了生成测试数据，我们定义了一个简单的二次函数，并添加了随机噪声。最后，我们使用SR-Tree来进行查询，以测试其性能。