我有一个包含纬度和经度的地方的数据框.想象一下,例如城市.

df = pd.DataFrame([{'city':"Berlin", 'lat':52.5243700, 'lng':13.4105300},
                   {'city':"Potsdam", 'lat':52.3988600, 'lng':13.0656600},
                   {'city':"Hamburg", 'lat':53.5753200, 'lng':10.0153400}]);

现在我试图让所有城市都在另一个城市的半径范围内.假设距离柏林500公里,距汉堡500公里等所有城市.我会通过复制原始数据帧并使用距离函数连接来完成此操作.

中间结果有点像这样：

Berlin --> Potsdam
Berlin --> Hamburg
Potsdam --> Berlin
Potsdam --> Hamburg
Hamburg --> Potsdam
Hamburg --> Berlin

分组(减少)后的最终结果应该是这样的.备注：如果值列表包含城市的所有列,那将会很酷.

Berlin --> [Potsdam, Hamburg]
Potsdam --> [Berlin, Hamburg]
Hamburg --> [Berlin, Potsdam]

或者只是一个城市周围500公里的城市数量.

Berlin --> 2
Potsdam --> 2
Hamburg --> 2

由于我对Python很陌生,所以我会很感激任何起点.我很熟悉长距离.但不确定Scipy或Pandas中是否有有用的距离/空间方法.

很高兴,如果你能给我一个起点.到目前为止,我尝试过跟随this post.

更新：这个问题背后的原始想法来自Two Sigma Connect Rental Listing Kaggle Competition.这个想法是让那些列在另一个列表周围的100米.其中a)表示密度,因此表示热门区域; b)如果比较地址,您可以查看是否存在交叉,因此是否存在噪声区域.因此,您不需要完整的项目与项目关系,因为您不仅需要比较距离,还需要比较地址和其他元数据. PS：我没有向Kaggle上传解决方案.我只是想学习.

解决方法:

您可以使用：

from math import radians, cos, sin, asin, sqrt

def haversine(lon1, lat1, lon2, lat2):

    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    r = 6371 # Radius of earth in kilometers. Use 3956 for miles
    return c * r

首先需要与merge交叉连接,在boolean indexing中删除city_x和city_y中具有相同值的行：

df['tmp'] = 1
df = pd.merge(df,df,on='tmp')
df = df[df.city_x != df.city_y]
print (df)
    city_x     lat_x     lng_x  tmp   city_y     lat_y     lng_y
1   Berlin  52.52437  13.41053    1  Potsdam  52.39886  13.06566
2   Berlin  52.52437  13.41053    1  Hamburg  53.57532  10.01534
3  Potsdam  52.39886  13.06566    1   Berlin  52.52437  13.41053
5  Potsdam  52.39886  13.06566    1  Hamburg  53.57532  10.01534
6  Hamburg  53.57532  10.01534    1   Berlin  52.52437  13.41053
7  Hamburg  53.57532  10.01534    1  Potsdam  52.39886  13.06566

然后应用haversine功能：

df['dist'] = df.apply(lambda row: haversine(row['lng_x'], 
                                            row['lat_x'], 
                                            row['lng_y'], 
                                            row['lat_y']), axis=1)

过滤距离：

df = df[df.dist < 500]
print (df)
    city_x     lat_x     lng_x  tmp   city_y     lat_y     lng_y        dist
1   Berlin  52.52437  13.41053    1  Potsdam  52.39886  13.06566   27.215704
2   Berlin  52.52437  13.41053    1  Hamburg  53.57532  10.01534  255.223782
3  Potsdam  52.39886  13.06566    1   Berlin  52.52437  13.41053   27.215704
5  Potsdam  52.39886  13.06566    1  Hamburg  53.57532  10.01534  242.464120
6  Hamburg  53.57532  10.01534    1   Berlin  52.52437  13.41053  255.223782
7  Hamburg  53.57532  10.01534    1  Potsdam  52.39886  13.06566  242.464120

最后创建列表或使用groupby获取大小：

df1 = df.groupby('city_x')['city_y'].apply(list)
print (df1)
city_x
Berlin     [Potsdam, Hamburg]
Hamburg     [Berlin, Potsdam]
Potsdam     [Berlin, Hamburg]
Name: city_y, dtype: object

df2 = df.groupby('city_x')['city_y'].size()
print (df2)
city_x
Berlin     2
Hamburg    2
Potsdam    2
dtype: int64

也可以使用numpy haversine solution：

def haversine_np(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees)

    All args must be of equal length.    

    """
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2

    c = 2 * np.arcsin(np.sqrt(a))
    km = 6367 * c
    return km

df['tmp'] = 1
df = pd.merge(df,df,on='tmp')
df = df[df.city_x != df.city_y]
#print (df)

df['dist'] = haversine_np(df['lng_x'],df['lat_x'],df['lng_y'],df['lat_y'])
    city_x     lat_x     lng_x  tmp   city_y     lat_y     lng_y        dist
1   Berlin  52.52437  13.41053    1  Potsdam  52.39886  13.06566   27.198616
2   Berlin  52.52437  13.41053    1  Hamburg  53.57532  10.01534  255.063541
3  Potsdam  52.39886  13.06566    1   Berlin  52.52437  13.41053   27.198616
5  Potsdam  52.39886  13.06566    1  Hamburg  53.57532  10.01534  242.311890
6  Hamburg  53.57532  10.01534    1   Berlin  52.52437  13.41053  255.063541
7  Hamburg  53.57532  10.01534    1  Potsdam  52.39886  13.06566  242.311890