千万别用sql查询经纬度数据了，有更快的方式

最近需要查询学校周边的零售户有哪些，几千个学校，几十万零售户，用sql查询费了十几分钟才查询出来，后来问的AI，使用python代码，十几秒就能搞定。以下是完整的代码，可以作为参考。

def aaaaa():
    import numpy as np
    import pandas as pd
    from sklearn.neighbors import BallTree
    from time import time
    # 1. 从 Doris 读取数据
    from sqlalchemy import create_engine
    engine = create_engine(
        "mysql+pymysql://root:mysql.7334453@192.168.1.10:9030/hx_theme_data?charset=utf8mb4"
    )
    start_time = time()
    merchants = pd.read_sql("""
        SELECT retailer_uuid, label_ids, label_names, company_name,
               label_list_fields, lic_no,
               CAST(latitude AS DOUBLE) AS lat_d,
               CAST(longitude AS DOUBLE) AS lng_d
        FROM t_license
        WHERE longitude IS NOT NULL AND longitude != 0
          AND latitude  IS NOT NULL AND latitude  != 0
          AND lic_status IN ('01','02','10')
    """, engine)

    schools = pd.read_sql("""
        SELECT name, type,
               CAST(lat AS DOUBLE) AS lat_d,
               CAST(lng AS DOUBLE) AS lng_d
        FROM t_data_school_info
        WHERE lat IS NOT NULL AND lat != '' AND lat != '0'
          AND lng IS NOT NULL AND lng != '' AND lng != '0'
    """, engine)

    # 2. 构建 BallTree（使用弧度，haversine 距离）
    school_coords = np.radians(schools[['lat_d', 'lng_d']].values)
    tree = BallTree(school_coords, metric='haversine')

    # 3. 批量查询每个商家最近的学校，距离阈值 200m
    merchant_coords = np.radians(merchants[['lat_d', 'lng_d']].values)
    DISTANCE_M = 200
    EARTH_RADIUS = 6371000

    distances, indices = tree.query(merchant_coords, k=1)  # 找最近1所
    distances_m = distances[:, 0] * EARTH_RADIUS           # 转换为米

    # 4. 过滤距离 < 200m 的结果
    mask = distances_m < DISTANCE_M
    result = merchants[mask].copy()
    result['distance']   = distances_m[mask]
    result['school_name'] = schools.iloc[indices[mask, 0]]['name'].values
    result['school_type'] = schools.iloc[indices[mask, 0]]['type'].values
    result['school_lat']  = schools.iloc[indices[mask, 0]]['lat_d'].values
    result['school_lng']  = schools.iloc[indices[mask, 0]]['lng_d'].values

    # 5. 结果写回 Doris
    # result.to_sql('t_merchant_nearest_school', engine, if_exists='replace')
    print(f"耗时 {time() - start_time:.2f} 秒")
    print(f"找到 {len(result)} 个200m内有学校的商家")