关于gis:数据处理车辆飞机轨迹点位抽稀处理

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背景需要：

须要实现飞机仿真实时挪动地位，然而提供的数据量较为宏大，而咱们的数据都是通过 kafka 发送，再由 ue4 承接数据来做渲染。然而数据量较为宏大，在增大飞行速度的同时渲染比拟吃力，于是想到点位个数压缩，数据推送频率不变来实现。

构思原理：利用普克拉斯算法，通过该点位压缩算法来压缩点的个数
这里提供的数据是 excel，我打算把他解决成 geosjon 格局数据，而后用 leaflet 加载，判断是否和原数据是否重合。

'''
Author: nico
Date: 2023-06-07 10:21:42
LastEditTime: 2023-06-07 10:21:46
Description: 
'''
import math
import pandas as pd

def douglas_peucker(coords, epsilon):
    # 找到间隔端点最远的点
    dmax = 0
    index = 0
    for i in range(1, len(coords)-1):
        d = distance(coords[i], coords[0], coords[-1])
        if d > dmax:
            index = i
            dmax = d

    # 如果最大间隔大于阈值，则递归地对两个子段进行解决
    if dmax > epsilon:
        results1 = douglas_peucker(coords[:index+1], epsilon)
        results2 = douglas_peucker(coords[index:], epsilon)

        # 将后果合并
        return results1[:-1] + results2
    else:
        # 如果最大间隔小于阈值，则保留该段的终点和起点
        return [coords[0], coords[-1]]
        
def distance(p, p1, p2):
    # 计算点到线段的间隔
    x0, y0 = p
    x1, y1 = p1
    x2, y2 = p2
    dx = x2 - x1
    dy = y2 - y1
    if dx == 0 and dy == 0:
        return math.sqrt((x0 - x1)**2 + (y0 - y1)**2)
    else:
        t = ((x0 - x1) * dx + (y0 - y1) * dy) / (dx*dx + dy*dy)
        if t < 0:
            px, py = x1, y1
        elif t > 1:
            px, py = x2, y2
        else:
            px, py = x1 + t*dx, y1 + t*dy
        return math.sqrt((x0 - px)**2 + (y0 - py)**2)

# 示例用法

def getData():
    # 读取 Excel 数据到 DataFrame
    df = pd.read_excel('data.xlsx')

    # 将经纬度转换为坐标点
    coordinates = []
    for index, row in df.iterrows():
        coordinates.append([row['lat'], row['lon']])
    return coordinates

coords = getData()
epsilon = 0.0000009
print(len(coords))
simplified_coords = douglas_peucker(coords, epsilon)
new_list = [[sublist[1], sublist[0]] for sublist in simplified_coords]
print(len(new_list),new_list)
simplified_geojson = {
    "type": "FeatureCollection",
    "features": [
        {
            "type": "Feature",
            "properties": {},
            "geometry": {
                "type": "LineString",
                "coordinates":new_list
            }
        }
    ]
}

# 将 GeoJSON 写入文件
with open('chouxi10.json', 'w') as f:
    f.write(str(simplified_geojson))

原作者地址：https://segmentfault.com/u/yourena_c
这里我用 leaflet 来测试造成的 geojson 是否和未解决的元数据合乎，所以经纬度对位也同时做了解决。

<!DOCTYPE html>
<html>
<head>
    <title>Leaflet GeoJSON Example</title>
    <meta charset="utf-8" />
    <meta name="viewport" content="width=device-width, initial-scale=1.0">
    <script src="https://cdn.bootcdn.net/ajax/libs/leaflet/1.9.3/leaflet.min.js"></script>
    <link href="https://cdn.bootcdn.net/ajax/libs/leaflet/1.9.3/leaflet.min.css" rel="stylesheet">
    <script src="./chouxi1.js"></script>
    <script src="./chouxi2.js"></script>
    <script src="./chouxi4.js"></script>
    <script src="./chouxi10.js"></script>
</head>
<body>

    <div id="mapid" style="height: 500px;"></div>

    <script>
        var mymap = L.map('mapid').setView([39.519334,116.437151], 19);

        L.tileLayer('https://{s}.tile.openstreetmap.org/{z}/{x}/{y}.png', {
            maxZoom: 19,
            attribution: 'Map data © <a href="https://openstreetmap.org">OpenStreetMap</a> contributors'
        }).addTo(mymap);

        var myStyle = {
            "color": "#ff0000",
            "weight": 15,
            "opacity": 1
        };

        var myStyle2 = {
            "color": "#00FF00",
            "weight": 10,
            "opacity": 1
        };

        var myStyle4 = {
            "color": "#800000",
            "weight": 5,
            "opacity": 1
        };

        var myStyle10 = {
            "color": "#00FF00",
            "weight": 1,
            "opacity": 1
        };
        
        console.log(geoData);
        L.geoJSON(geoData, {style: myStyle}).addTo(mymap);
        
        L.geoJSON(geoData2, {style: myStyle2}).addTo(mymap);

        
        L.geoJSON(geoData4, {style: myStyle4}).addTo(mymap);

        
        L.geoJSON(geoData10, {style: myStyle10}).addTo(mymap);
    </script>

</body>
</html>

最初后果是抽稀 10 倍，局部特色隐没，然而根本吻合。

正文完