用python 来实现经纬度的测算与输出

深碍√TFBOYSˉ_ 2023-06-17 05:52 30阅读 0赞

*本文采用高德地图地理标识规范*

首先导入两个分析包，math和pandas，如果有需要还有numpy，mathplotlib等常用数据分析包

import pandas as pd
    import math

测算一经度和一纬度的距离

# 地球半径
    r  = 12713.504 / 2
    # 1纬度长
    wd = (2 * math.pi * r) / 180
    # 1经度长
    jd = (5 * math.sin(math.atan2(x2-x1,y2-y1))) / (wd * math.cos(x))

![图片来源百度图片][format_png] 该图片来源于百度图片

读写原excel文件并输出

df = pd.read_excel('jwd.xlsx')
    # df = df.head()
    # make a new col
    df['res1'] = df['a_road']
    res1 = [[0,1]]
    
    
    # get index
    for i in range(df.iloc[:,0].size):
        x1 = df.latA[i]
        y1 = df.lngA[i]
        x2 = df.latB[i]
        y2 = df.lngB[i]
        # 通过三角函数进行求值
        # 1纬度 111km 
        # 1经度 111*cos纬度 km
        x  = (5 * math.cos(math.atan2(x2-x1,y2-y1))) / wd
        y  = (5 * math.sin(math.atan2(x2-x1,y2-y1))) / (wd * math.cos(x))
        #print(x,y)
        for j in range(0,int(df.distance[i] / 5) + 1):
            if j == int(df.distance[i] / 5):
                num = df.distance[i] % 5
                x  = (num * math.cos(math.atan2(x2-x1,y2-y1))) / wd
                y  = (num * math.sin(math.atan2(x2-x1,y2-y1))) / (wd * math.cos(x))
                res1.append([x2 - x,y2 - y])
            elif j == 0:
                res1[j][0] = x1 + x 
                res1[j][1] = y1 + y 
            else:
                res1.append([x1 + x * (j + 1),y1 + y * (j + 1)])
        df.loc[i,'res1'] = str(res1)
    
    df.to_excel('经纬度结果表.xlsx',index=False,sheet_name = 'df1')

[format_png]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9zczEuYmRzdGF0aWMuY29tLzcwY0Z2WFNoX1ExWW54R2twb1dLMUhGNmhoeS9pdC91PTUxMDY4ODM2NCwyMjYyNDA2MTc1JmZtPTI2JmdwPTAuanBn?x-oss-process=image/format,png