在Python中查找一定半径内的纬度和经度对答案

【问题标题】：Finding pairs of latitude and longitude within a certain radius in Python在Python中查找一定半径内的纬度和经度对
【发布时间】：2021-06-02 20:33:08
【问题描述】：

给定一个数据框df，如下：

    id              location        lon       lat
0    1            Onyx Spire  116.35425  39.87760
1    2        Unison Lookout  116.44333  39.93237
2    3       History Lookout  116.14857  39.73727
3    4     Domination Pillar  116.46387  39.96286
4    5           Union Tower  116.36373  39.95064
5    6   Ruby Forest Obelisk  116.35786  39.89463
6    7      Rust Peak Pillar  116.34870  39.98170
7    8      Ash Forest Tower  116.38461  39.94938
8    9  Prestige Mound Tower  116.34052  39.98977
9   10  Sapphire Mound Tower  116.35063  39.92982
10  11       Kinship Lookout  116.43020  39.99997
11  12    Exhibition Obelisk  116.45108  39.94371

对于每个location，如果它们之间的距离小于且等于，我需要找出其他位置名称，例如5 km。

代码基于this link的回答：

from scipy.spatial import distance
from math import sin, cos, sqrt, atan2, radians

def get_distance(point1, point2):
    R = 6370
    lat1 = radians(point1[0])  #insert value
    lon1 = radians(point1[1])
    lat2 = radians(point2[0])
    lon2 = radians(point2[1])

    dlon = lon2 - lon1
    dlat = lat2- lat1

    a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
    c = 2 * atan2(sqrt(a), sqrt(1-a))
    distance = R * c
    return distance

all_points = df[['lat', 'lon']].values
dm = distance.cdist(all_points, all_points, get_distance)
pd.DataFrame(dm, index=df.index, columns=df.index)

输出：

           0          1          2   ...         9          10         11
0    0.000000   9.736316  23.494395  ...   5.813891  15.066709  11.054762
1    9.736316   0.000000  33.222475  ...   7.908015   7.598415   1.423357
2   23.494395  33.222475   0.000000  ...  27.492814  37.822285  34.549129
3   13.312235   3.815179  36.787014  ...  10.327235   5.024900   2.391864
4    8.160542   7.082601  30.000842  ...   2.569988   7.883467   7.484839
5    1.918235   8.409888  25.009951  ...   3.960618  13.235325   9.641336
6   11.583243   9.752599  32.096627  ...   5.770232   7.233093   9.692770
7    8.389761   5.350670  31.017383  ...   3.622002   6.835323   5.700434
8   12.525586  10.838805  32.501864  ...   6.720541   7.722060  10.722467
9    5.813891   7.908015  27.492814  ...   0.000000  10.334273   8.701063
10  15.066709   7.598415  37.822285  ...  10.334273   0.000000   6.502921
11  11.054762   1.423357  34.549129  ...   8.701063   6.502921   0.000000

但我想获得类似于以下数据框的输出。请注意location1、location2、location3 是距离location 的位置名称（配对的位置名称可能不准确，仅用作示例帮助理解），如果是NaN，则不存在这样的location：

    id              location  ...            location2          location3
0    1            Onyx Spire  ...                  NaN                NaN
1    2        Unison Lookout  ...                  NaN                NaN
2    3       History Lookout  ...                  NaN                NaN
3    4     Domination Pillar  ...                  NaN                NaN
4    5           Union Tower  ...                  NaN                NaN
5    6   Ruby Forest Obelisk  ...                  NaN                NaN
6    7      Rust Peak Pillar  ...                  NaN                NaN
7    8      Ash Forest Tower  ...      Kinship Lookout                NaN
8    9  Prestige Mound Tower  ...                  NaN                NaN
9   10  Sapphire Mound Tower  ...                  NaN                NaN
10  11       Kinship Lookout  ...  Ruby Forest Obelisk  Domination Pillar
11  12    Exhibition Obelisk  ...                  NaN                NaN

我如何在 Python 中做到这一点？谢谢。

【问题讨论】：

标签： python-3.x pandas dataframe levenshtein-distance euclidean-distance

【解决方案1】：

想法是为不是0 的值创建掩码，而不是像5km，然后使用DataFrame.dot 进行矩阵乘法，最后使用Series.str.split 用于连接到原始列的新列：

df1 = pd.DataFrame(dm, index=df.index, columns=df.index)

df = (df.join((df1.ne(0) & df1.lt(5)).dot(df['location']+ ',')
                                     .str[:-1]
                                     .str.split(',', expand=True)
                                     .add_prefix('loc')))

print (df)
    id              location        lon       lat                 loc0  \
0    1            Onyx Spire  116.35425  39.87760  Ruby Forest Obelisk   
1    2        Unison Lookout  116.44333  39.93237    Domination Pillar   
2    3       History Lookout  116.14857  39.73727                        
3    4     Domination Pillar  116.46387  39.96286       Unison Lookout   
4    5           Union Tower  116.36373  39.95064     Rust Peak Pillar   
5    6   Ruby Forest Obelisk  116.35786  39.89463           Onyx Spire   
6    7      Rust Peak Pillar  116.34870  39.98170          Union Tower   
7    8      Ash Forest Tower  116.38461  39.94938          Union Tower   
8    9  Prestige Mound Tower  116.34052  39.98977          Union Tower   
9   10  Sapphire Mound Tower  116.35063  39.92982          Union Tower   
10  11       Kinship Lookout  116.43020  39.99997                        
11  12    Exhibition Obelisk  116.45108  39.94371       Unison Lookout   

                    loc1                  loc2                  loc3  
0                   None                  None                  None  
1     Exhibition Obelisk                  None                  None  
2                   None                  None                  None  
3     Exhibition Obelisk                  None                  None  
4       Ash Forest Tower  Prestige Mound Tower  Sapphire Mound Tower  
5   Sapphire Mound Tower                  None                  None  
6       Ash Forest Tower  Prestige Mound Tower                  None  
7       Rust Peak Pillar  Sapphire Mound Tower                  None  
8       Rust Peak Pillar                  None                  None  
9    Ruby Forest Obelisk      Ash Forest Tower                  None  
10                  None                  None                  None  
11     Domination Pillar                  None                  None

对于排序值，请使用：

df1 = pd.DataFrame(dm, index=df.index, columns=df['location'])

df1 = df.join(df1.apply(lambda x: pd.Series(x[(x!=0)&(x < 5)].sort_values().index), axis=1)
                .add_prefix('loc'))
print (df1)
    id              location        lon       lat                  loc0  \
0    1            Onyx Spire  116.35425  39.87760   Ruby Forest Obelisk   
1    2        Unison Lookout  116.44333  39.93237    Exhibition Obelisk   
2    3       History Lookout  116.14857  39.73727                   NaN   
3    4     Domination Pillar  116.46387  39.96286    Exhibition Obelisk   
4    5           Union Tower  116.36373  39.95064      Ash Forest Tower   
5    6   Ruby Forest Obelisk  116.35786  39.89463            Onyx Spire   
6    7      Rust Peak Pillar  116.34870  39.98170  Prestige Mound Tower   
7    8      Ash Forest Tower  116.38461  39.94938           Union Tower   
8    9  Prestige Mound Tower  116.34052  39.98977      Rust Peak Pillar   
9   10  Sapphire Mound Tower  116.35063  39.92982           Union Tower   
10  11       Kinship Lookout  116.43020  39.99997                   NaN   
11  12    Exhibition Obelisk  116.45108  39.94371        Unison Lookout   

                    loc1                 loc2                  loc3  
0                    NaN                  NaN                   NaN  
1      Domination Pillar                  NaN                   NaN  
2                    NaN                  NaN                   NaN  
3         Unison Lookout                  NaN                   NaN  
4   Sapphire Mound Tower     Rust Peak Pillar  Prestige Mound Tower  
5   Sapphire Mound Tower                  NaN                   NaN  
6            Union Tower     Ash Forest Tower                   NaN  
7   Sapphire Mound Tower     Rust Peak Pillar                   NaN  
8            Union Tower                  NaN                   NaN  
9       Ash Forest Tower  Ruby Forest Obelisk                   NaN  
10                   NaN                  NaN                   NaN  
11     Domination Pillar                  NaN                   NaN

【讨论】：

谢谢，顺便说一句，我们可以将loc0, loc1, ... loc3 从最短到最长的距离排列吗？
@ahbon - 这很复杂，但可能。我正在回答。
顺便说一句，我发现有些配对的位置不在5 km的距离范围内，可能是地球半径设置不准确，R = 6370?
@ahbon - 不幸的是这个区域对我来说是未知的，所以不知道。
@ahbon - 也许需要一些 geopandas 解决方案，从不使用它，所以不知道。我添加了从低到高排序的解决方案。

【解决方案2】：

这是一个使用 BallTree 的方法，从最短到最长距离排序

from sklearn.neighbors import BallTree
import pandas as pd
import numpy as np


data = { 'lon' : [116.35425, 116.44333, 116.14857, 116.46387, 116.36373, 116.35786, 116.34870, 116.38461, 116.34052, 116.35063, 116.43020, 116.45108],
'lat' : [39.87760, 39.93237, 39.73727, 39.96286, 39.95064, 39.89463, 39.98170, 39.94938, 39.98977, 39.92982, 39.99997, 39.94371],
'location' : ["Onyx Spire", "Unison Lookout", "History Lookout", "Domination Pillar", "Union Tower", "Ruby Forest Obelisk", "Rust Peak Pillar", "Ash Forest Tower", "Prestige Mound Tower", "Sapphire Mound Tower", "Kinship Lookout", "Exhibition Obelisk"]}

locations = pd.DataFrame.from_dict(data)

创建球树

locations_radians =  np.radians(locations[["lat","lon"]].values)
tree = BallTree(locations_radians, leaf_size=12, metric='haversine')

distance_in_meters = 5000
earth_radius = 6371000
    
radius = distance_in_meters / earth_radius

请注意，我首先对is_within 中的is_within_sorted 进行排序

is_within, distances = tree.query_radius(locations_radians, r=radius, count_only=False, return_distance=True) 

is_within_sorted = [ iw[ np.argsort(di) ] for iw,di in zip(is_within, distances) ]
distances_sorted = [np.sort(d) for d in distances]

is_within 包含不同长度的数组，它们将返回半径内位置的索引。您可以将这些与实际距离一起存储。

现在我用 Nan 填充并创建一个 DF，以便以后加入

pad_with_nans = [ np.pad(locations.location[iw], (0,locations.lat.size), 'constant', constant_values=np.nan)[:locations.lat.size] for iw in is_within_sorted]
location_names = [ 'location_{}'.format(i) for i in range(locations.lat.size) ]

within_radius = pd.DataFrame(pad_with_nans, index=locations.index, columns=location_names)

我们有

locations.join(within_radius)

给予

         lon       lat           location         location_0  \
0  116.35425  39.87760         Onyx Spire         Onyx Spire   
1  116.44333  39.93237     Unison Lookout     Unison Lookout   
2  116.14857  39.73727    History Lookout    History Lookout   
3  116.46387  39.96286  Domination Pillar  Domination Pillar   
4  116.36373  39.95064        Union Tower        Union Tower   

            location_1            location_2        location_3  \
0  Ruby Forest Obelisk                   NaN               NaN   
1   Exhibition Obelisk     Domination Pillar               NaN   
2                  NaN                   NaN               NaN   
3   Exhibition Obelisk        Unison Lookout               NaN   
4     Ash Forest Tower  Sapphire Mound Tower  Rust Peak Pillar   

             location_4  location_5  location_6  location_7  location_8  \
0                   NaN         NaN         NaN         NaN         NaN   
1                   NaN         NaN         NaN         NaN         NaN   
2                   NaN         NaN         NaN         NaN         NaN   
3                   NaN         NaN         NaN         NaN         NaN   
4  Prestige Mound Tower         NaN         NaN         NaN         NaN   

   location_9  location_10  location_11  
0         NaN          NaN          NaN  
1         NaN          NaN          NaN  
2         NaN          NaN          NaN  
3         NaN          NaN          NaN  
4         NaN          NaN          NaN

点本身始终在其内部，因此您可以删除第一列。

【讨论】：