Question

我在Pandas中有2个数据框，分别包含longitude和latitude。我试图遍历第一行中的每一行，并在第二个数据帧中找到最接近的匹配longitude和latitude。

到目前为止，我在python中拥有此功能，而我在另一篇SO帖子中发现了此信息...

from math import cos, asin, sqrt

def distance(lat1, lon1, lat2, lon2):
    p = 0.017453292519943295
    a = 0.5 - cos((lat2-lat1)*p)/2 + cos(lat1*p)*cos(lat2*p) * (1-cos((lon2-lon1)*p)) / 2
    return 12742 * asin(sqrt(a))

def closest(data, v):
    return min(data, key=lambda p: distance(v['lat'],v['lon'],p['lat'],p['lon']))

tempDataList = [{'lat': 39.7612992, 'lon': -86.1519681}, 
                {'lat': 39.762241,  'lon': -86.158436 }, 
                {'lat': 39.7622292, 'lon': -86.1578917}]

v = {'lat': 39.7622290, 'lon': -86.1519750}
print(closest(tempDataList, v))

我将尝试对其进行修改，以用于我的熊猫数据框，但是例如，有没有更有效的方法使用PyProj来做到这一点？

有人有示例或类似代码吗？

Answer 1

我认为，如果您使用GIS库，则可以更轻松地完成此操作。因此，如果您正在使用Geopandas和Shapely，它将更加舒适。（也使用pyproj。）从下面的代码开始。

import pandas as pd
import geopandas as gpd
from shapely.ops import Point, nearest_points

tempDataList = [{'lat': 39.7612992, 'lon': -86.1519681}, 
                {'lat': 39.762241,  'lon': -86.158436 }, 
                {'lat': 39.7622292, 'lon': -86.1578917}]

df = pd.DataFrame(tempDataList)

#make point geometry for geopandas
geometry = [Point(xy) for xy in zip(df['lon'], df['lat'])]

#use a coordinate system that matches your coordinates. EPSG 4326 is WGS84
gdf = gpd.GeoDataFrame(df, crs = "EPSG:4326", geometry = geometry) 

#change point geometry
v = {'lat': 39.7622290, 'lon': -86.1519750}
tp = Point(v['lon'], v['lat'])

#now you can calculate the distance between v and others.
gdf.distance(tp)

#If you want to get nearest points
multipoints = gdf['geometry'].unary_union
queried_point, nearest_point = nearest_points(tp, multipoints)
print(nearest_point)

熊猫-比较2个数据框并找到最接近的值

1 个答案: