纬度和经度地址

Question

我有大约8000 +纬度和经度。 我需要提取（反向代码），他们到位置，可能是城市。

最初我使用Google地图进行了检查，但根据他们的条款，我们不应该使用我们的脚本来打他们的服务。

即使是OpenStreetMaps也不允许我们反复打他们的服务器。他们有一些时间限制。

因此，我下载了位置的纬度和经度。我写了一个python脚本，

import tabular as tb 
import csv 



citiesLatLongData = tb.tabarray(SVfile="D:/latitude-longitude/citieslatlong.csv") 

allData = tb.tabarray(SVfile="C:/Users/User/Desktop/alldata.csv") 

latlonglocs = {'a1':"Car Nicobar",'a2':"Port Blair",'a3':"Hyderabad",'a4':"Kadapa",'a5':"Puttaparthi", 
'a6':"Rajahmundry",'a7':"Tirupati",'a8':"Vijayawada",'a9':"Vishakhapatnam",'a10':"Itanagar", 
'a11':"Dibrugarh",'a12':"Dispur",'a13':"Guwahati",'a14':"North Lakhimpur",'a15':"Silchar", 
'a16':"Gaya",'a17':"Patna",'a18':"Chandigarh",'a19':"Raipur",'a20':"Silvassa", 
'a21':"Daman",'a22':"Bawana",'a23':"New Delhi",'a24':"Mormugao",'a25':"Panaji", 
'a26':"Ahmedabad",'a27':"Bhavnagar",'a28':"Bhuj",'a29':"Gandhinagar",'a30':"Jamnagar", 
'a31':"Kandla",'a32':"Rajkot",'a33':"Vadodara",'a34':"Hisar",'a35':"Bilaspur", 
'a36':"Dharamsala",'a37':"Kulu",'a38':"Shimla",'a39':"Jammu",'a40':"Srinagar",'a41':"Jamshedpur", 
'a42':"Ranchi",'a43':"Bangalore",'a44':"Belgaum",'a45':"Bellary",'a46':"Hubli Dharwad", 
'a47':"Mandya",'a48':"Mangalore",'a49':"Mysore",'a50':"Cochin",'a51':"Kozhikode", 
'a52':"Thiruvananthapuram",'a53':"Bingaram Island ",'a54':"Kavaratti",'a55':"Bhopal",'a56':"Gwalior", 
'a57':"Indore",'a58':"Jabalpur",'a59':"Khandwa",'a60':"Satna",'a61':"Ahmadnagar", 
'a62':"Akola",'a63':"Aurangabad",'a64':"Jalna",'a65':"Kolhapur",'a66':"Mumbai", 
'a67':"Nagpur",'a68':"Nasik",'a69':"Pimpri",'a70':"Pune",'a71':"Solapur", 
'a72':"Imphal",'a73':"Shillong",'a74':"Aizawl",'a75':"Kohima",'a76':"Bhubaneswar", 
'a77':"Jharsuguda",'a78':"Karaikal",'a79':"Mahe",'a80':"Pondicherry",'a81':"Yanam", 
'a82':"Amritsar",'a83':"Pathankot",'a84':"Jaipur",'a85':"Jodhpur",'a86':"Kota", 
'a87':"Udaipur",'a88':"Gangtok",'a89':"Chennai",'a90':"Coimbatore",'a91':"Madurai", 
'a92':"Nagercoil",'a93':"Thiruchendur",'a94':"Thiruvannaamalai",'a95':"Thoothukudi", 
'a96':"Tiruchirappalli",'a97':"Tirunelveli",'a98':"Vellore",'a99':"Agartala", 
'a100':"Agra",'a101':"Allahabad",'a102':"Bareilly",'a103':"Gorakhpur",'a104':"Jhansi", 
'a105':"Kanpur",'a106':"Lucknow",'a107':"Varanasi",'a108':"Dehradun",'a109':"Pantnagar", 
'a110':"Kolkata",'a111':"Siliguri"} 



latlongs = {'a1':[9.15,92.8167],'a2':[11.6667,92.7167],'a3':[17.45,78.4667],'a4':[14.4833,78.8333], 
      'a5':[14.1333,77.7833],'a6':[16.9667,81.7667],'a7':[13.65,79.4167],'a8':[16.5333,80.8], 
      'a9':[17.7,83.3],'a10':[27.0833,93.5667],'a11':[27.4833,95.0167],'a12':[26.0833,91.8333], 
      'a13':[26.1667,91.5833],'a14':[27.2333,94.1167],'a15':[24.8167,92.8],'a16':[24.75,84.95], 
      'a17':[25.6,85.1],'a18':[30.7333,76.75],'a19':[21.2333,81.6333],'a20':[20.2833,73], 
      'a21':[20.4167,72.85],'a22':[28.7833,77.0333],'a23':[28.5667,77.1167],'a24':[15.3833,73.8167],  
      'a25':[15.3833,73.8167],'a26':[23.0333,72.6167],'a27':[21.75,72.2],'a28':[23.25,69.6667], 
      'a29':[23.3333,72.5833],'a30':[22.4667,70.0667],'a31':[23.0333,70.2167],'a32':[22.3,70.7833], 
      'a33':[22.3,73.2667],'a34':[29.1667,75.7333],'a35':[31.25,76.6667],'a36':[32.2,76.4], 
      'a37':[31.9667,77.1],'a38':[31.1,77.1667],'a39':[32.7,74.8667],'a40':[34.0833,74.8167], 
      'a41':[22.8167,86.1833],'a42':[23.3167,85.3167],'a43':[12.9833,77.5833],'a44':[15.85,74.6167], 
      'a45':[15.15,76.85],'a46':[15.35,75.1667],'a47':[12.55,76.9],'a48':[12.9167,74.8833], 
      'a49':[12.3,76.65],'a50':[9.95,76.2667],'a51':[11.25,75.7667],'a52':[8.46667,76.95], 
      'a53':[10.9167,72.3333],'a54':[10.5833,72.65],'a55':[23.2833,77.35],'a56':[26.2333,78.2333], 
      'a57':[22.7167,75.8],'a58':[23.2,79.95],'a59':[21.8333,76.3667],'a60':[24.5667,80.8333], 
      'a61':[19.0833,74.7333],'a62':[20.7,77.0667],'a63':[19.85,75.4],'a64':[19.8333,75.8833], 
      'a65':[16.7,74.2333],'a66':[19.1167,72.85],'a67':[21.1,79.05],'a68':[19.8933,73.8], 
      'a69':[18.55,73.8167],'a70':[18.5333,73.8667],'a71':[17.6667,75.9],'a72':[24.7667,93.9], 
      'a73':[25.55,91.85],'a74':[23.6667,92.6667],'a75':[25.6667,94.1167],'a76':[20.25,85.8333], 
      'a77':[21.5833,84.08333],'a78':[10.95,79.7833],'a79':[11.7,75.5333],'a80':[11.9333,79.8833], 
      'a81':[16.7333,82.2167],'a82':[31.6333,74.8667],'a83':[32.2833,75.65],'a84':[26.8167,75.8], 
      'a85':[29.1667,75.7333],'a86':[25.15,75.85],'a87':[24.5667,73.6167],'a88':[27.3333,88.6167], 
      'a89':[13,80.1833],'a90':[11.0333,77.05],'a91':[9.83333,78.0833],'a92':[8.16667,77.4333], 
      'a93':[8.48333,78.1167],'a94':[12.2167,79.0667],'a95':[8.78333,78.1333],'a96':[10.7667,78.7167], 
      'a97':[8.73333,77.7],'a98':[12.9167,79.15],'a99':[23.8833,91.25],'a100':[27.15,77.9667], 
      'a101':[25.45,81.7333],'a102':[28.3667,79.4],'a103':[26.75,83.3667],'a104':[29.1667,75.7333], 
      'a105':[26.4,80.4],'a106':[26.75,80.8833],'a107':[25.45,83],'a108':[30.3167,78.0333], 
      'a109':[29.0833,79.5],'a110':[22.65,88.45],'a111':[26.6333,88.3167] 
      } 

for eachOne in allData: 
    for eachTwo in latlongs: 
     eachOne_Coordinates_Latitude = eachOne['COORDINATES-Latitude'] 
     latlongs_eachTwo_Latitude_Plus = int(latlongs[eachTwo][0]) + 0.18 
     latlongs_eachTwo_Latitude_Minus = int(latlongs[eachTwo][0]) - 0.18 

     eachOne_Coordinates_Longitude = eachOne['COORDINATES-Longitude'] 
     latlongs_eachTwo_Longitude_Plus = int(latlongs[eachTwo][1]) + 0.18 
     latlongs_eachTwo_Longitude_Minus = int(latlongs[eachTwo][1]) - 0.18 

     if ((eachOne_Coordinates_Latitude < latlongs_eachTwo_Latitude_Plus) and (latlongs_eachTwo_Latitude_Plus > latlongs_eachTwo_Latitude_Minus)) and ((eachOne_Coordinates_Longitude < latlongs_eachTwo_Longitude_Plus) and (eachOne_Coordinates_Longitude > latlongs_eachTwo_Longitude_Minus)): 
      someDict.setdefault((eachOne_Coordinates_Latitude,eachOne_Coordinates_Longitude),[]).append(latlongs[eachTwo]) 


for each in someDict: 
    print each,':', min(someDict[each]) 

我的问题： 如你所知，我们从外部来源获取不完全与纬度和经度，我们有匹配的经度和纬度。我听说他们不会匹配，会有一些错误百分比或什么。

我需要一些人的指导。我请某人请向正确的方向指出我的意思，或者如果您知道这样做的任何软件包或脚本。

我会非常感谢你。

Answer 1

这听起来很像“最近点问题”。你有N个点（城市）和M个位置（你的8000个坐标）。对于M个地点中的每个地点，您都希望按其最近的城市对地点进行分类。 Nearest Neighbor Search有很多解决方案，但最简单的是线性搜索：

function getClosestCity(Coordinate location){ 
    bestCity = cities[0]; 
    foreach(city in cities){ 
     if (distance(bestCity.location, location) < distance(city.location, location)){ 
      bestCity = city; 
     } 
    } 
    return bestCity; 
}