Pandas Dataframe: join items in range based on their geo coordinates (longitude and latitude)

Question

I got a dataframe that contains places with their latitude and longitude. Imagine for example cities.

df = pd.DataFrame([{\'city\':\"Berlin\", \'lat\':52.524         


        

Answer 1

You can use:

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

First need cross join with merge, remove row with same values in city_x and city_y by boolean indexing:

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

Then apply haversine function:

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

Filter distance:

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

And last create list or get size with groupby:

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

Also is possible use 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

Answer 2

UPDATE: I would suggest first to buiuld a distance DataFrame:

from scipy.spatial.distance import squareform, pdist
from itertools import combinations

# see definition of "haversine_np()" below     
x = pd.DataFrame({'dist':pdist(df[['lat','lng']], haversine_np)},
                 index=pd.MultiIndex.from_tuples(tuple(combinations(df['city'], 2))))

efficiently produces pairwise distance DF (without duplicates):

In [106]: x
Out[106]:
                       dist
Berlin  Potsdam   27.198616
        Hamburg  255.063541
Potsdam Hamburg  242.311890

---

Old answer:

Here is a bit optimized version, which uses scipy.spatial.distance.pdist method:

from scipy.spatial.distance import squareform, pdist

# slightly modified version: of http://stackoverflow.com/a/29546836/2901002
def haversine_np(p1, p2):
    """
    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees)

    All args must be of equal length.    

    """
    lat1, lon1, lat2, lon2 = np.radians([p1[0], p1[1],
                                         p2[0], p2[1]])
    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

x = pd.DataFrame(squareform(pdist(df[['lat','lng']], haversine_np)),
                 columns=df.city.unique(),
                 index=df.city.unique())

this gives us:

In [78]: x
Out[78]:
             Berlin     Potsdam     Hamburg
Berlin     0.000000   27.198616  255.063541
Potsdam   27.198616    0.000000  242.311890
Hamburg  255.063541  242.311890    0.000000

let's count number of cities where the distance is greater than 30:

In [81]: x.groupby(level=0, as_index=False) \
    ...:  .apply(lambda c: c[c>30].notnull().sum(1)) \
    ...:  .reset_index(level=0, drop=True)
Out[81]:
Berlin     1
Hamburg    2
Potsdam    1
dtype: int64