如何计算整列的最近纬度经度？

Question

我有一个地点事故的纬度列表，想看看离这些事故最近的警察局是什么。目前，事故的经纬度列表使用 pandas 在两个单独的列中（对不起，我对 python 很陌生，所以我可能会错误地使用单词）。派出所的经纬度目前在一个单独的 json 文件中。我目前的目标是创建一个新列（或文件），其中出现最近警察局的 latlongs。理想情况下，它应该是相应的名称，但那是我遇到它时会跨过的一座桥。

我查看了其他人是如何做到的，但除了询问位置，我只能询问一对 latlongs 的位置，而不是同时询问所有这些。

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']))

#these are the latlons of the police stations
tempDataList = [{"lat": 52.003181, "lon": 4.353068}, 
    {"lat": 52.089416, "lon": 4.377340},
    {"lat": 52.019911, "lon": 4.426602},
    {"lat": 52.054457, "lon": 4.388764},
    {"lat": 52.044536, "lon": 4.332631},
    {"lat": 52.072910, "lon": 4.274784},
    {"lat": 52.066099, "lon": 4.298664},
    {"lat": 52.070030, "lon": 4.317355},
    {"lat": 52.052636, "lon": 4.289576},
    {"lat": 52.060829, "lon": 4.318683},
    {"lat": 52.075680, "lon": 4.306810},
    {"lat": 52.040353, "lon": 4.256946},
    {"lat": 52.089381, "lon": 4.345599},
    {"lat": 52.111719, "lon": 4.283909},
    {"lat": 52.055222, "lon": 4.233827},
    {"lat": 52.046393, "lon": 4.253105},
    {"lat": 52.144177, "lon": 4.405549},
    {"lat": 51.987035, "lon": 4.199314},
    {"lat": 52.061650, "lon": 4.486572}]

v = {'lat': 52.103167, 'lon': 4.317532}
print(closest(tempDataList, v))

这是我失败的地方，因为我根本不知道如何让 v 成为两列并将其放入新列。

我希望有一列显示最近警察局的经纬度。 有时我会遇到问题TypeError: string indices must be integers。

Answer 1

您可以在下面看到如何将索引列表转换为 numpy 数组，之后您可以对其进行操作，而无需遍历每个元素。

import numpy as np
# numpy allows you to work with arrays; math only works with scalars
from numpy import cos, arcsin as asin, sqrt

def distance (lat1, lon1, lat2, lon2):
    """Return the distances between all accidents (lat1,lon1)
    and all police stations (lat2,lon2) as a 2-dimensional array
    """
    # add dummy dimension to p
    lat2 = lat2[:,None]
    lon2 = lon2[:,None]
    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(*args):
    """Returns the index (counting from zero) of the closest
    police station for every accident.
    """
    return np.argmin(distance(*args), axis=0)


tempDataList = [{"lat": 52.003181, "lon": 4.353068},
    {"lat": 52.089416, "lon": 4.377340},
    {"lat": 52.019911, "lon": 4.426602},
    {"lat": 52.054457, "lon": 4.388764},
    {"lat": 52.044536, "lon": 4.332631},
    {"lat": 52.072910, "lon": 4.274784},
    {"lat": 52.066099, "lon": 4.298664},
    {"lat": 52.070030, "lon": 4.317355},
    {"lat": 52.052636, "lon": 4.289576},
    {"lat": 52.060829, "lon": 4.318683},
    {"lat": 52.075680, "lon": 4.306810},
    {"lat": 52.040353, "lon": 4.256946},
    {"lat": 52.089381, "lon": 4.345599},
    {"lat": 52.111719, "lon": 4.283909},
    {"lat": 52.055222, "lon": 4.233827},
    {"lat": 52.046393, "lon": 4.253105},
    {"lat": 52.144177, "lon": 4.405549},
    {"lat": 51.987035, "lon": 4.199314},
    {"lat": 52.061650, "lon": 4.486572}]

# first get lat and lon into arrays
lat, lon = np.transpose([[i['lat'],i['lon']] for i in tempDataList])

# I'm making up the police stations here. Say 5 police stations.
# you should load the actual data in your problem. My station locations
# are randomly located near the first 5 accidents
npolice = 5
# for reproducibility
np.random.seed(3)
plat = lat[:npolice] + np.random.normal(0, 0.02, npolice)
plon = lon[:npolice] + np.random.normal(0, 0.02, npolice)

index = closest(lat, lon, plat, plon)
# index = array([3, 1, 2, 0, 4, 4, 4, 4, 4, 4, 4, 4, 1, 1, 4, 4, 1, 4, 2])

所以最近的警察局位置是

nearest = {'lat': plat[index], 'lon': plon[index]}

如果您还存储了这些信息，则可以使用 index 访问例如每个警察局的名称或地址。

希望对你有帮助。