Efficiently compute pairwise haversine distances between two datasets - NumPy / Python

问题

I want to calculate the geo-distance between latitude-longitude.

I had checked this thread Vectorizing Haversine distance calculation in Python but when I am using it for two different set of coordinates, I m getting an error.

df1 size can be in millions and if there is any other way to calculate accurate geo distance in less time then it would be really helpful.

length1 = 1000
d1 = np.random.uniform(-90, 90, length1)
d2 = np.random.uniform(-180, 180, length1)
length2 = 100
d3 = np.random.uniform(-90, 90, length2)
d4 = np.random.uniform(-180, 180, length2)
coords = tuple(zip(d1, d2))
df1 = pd.DataFrame({'coordinates':coords})
coords = tuple(zip(d3, d4))
df2 = pd.DataFrame({'coordinates':coords})

def get_diff(df1, df2):
    data1 = np.array(df1['coordinates'].tolist())
    data2 = np.array(df2['coordinates'].tolist())
    lat1 = data1[:,0]                     
    lng1 = data1[:,1]
    lat2 = data2[:,0]                     
    lng2 = data2[:,1]
    #print(lat1.shape)
    #print(lng1.shape)
    #print(lat2.shape)
    #print(lng2.shape)
    diff_lat = lat1[:,None] - lat2

    diff_lng = lng1[:,None] - lng2
    #print(diff_lat.shape)
    #print(diff_lng.shape)
    d = np.sin(diff_lat/2)**2 + np.cos(lat1[:,None])*np.cos(lat1) * np.sin(diff_lng/2)**2
    return 2 * 6371 * np.arcsin(np.sqrt(d))

get_diff(df1, df2)

ValueError                                Traceback (most recent call last)
<ipython-input-58-df06c7cff72c> in <module>
----> 1 get_diff(df1, df2)

<ipython-input-57-9bd8f10189e6> in get_diff(df1, df2)
     26     print(diff_lat.shape)
     27     print(diff_lng.shape)
---> 28     d = np.sin(diff_lat/2)**2 + np.cos(lat1[:,None])*np.cos(lat1) * np.sin(diff_lng/2)**2
     29     return 2 * 6371 * np.arcsin(np.sqrt(d))

ValueError: operands could not be broadcast together with shapes (1000,1000) (1000,100) 

回答1:

Pairwise haversine distances

Here's a vectorized way with broadcasting based on this post -

def convert_to_arrays(df1, df2):
    d1 = np.array(df1['coordinates'].tolist())
    d2 = np.array(df2['coordinates'].tolist())
    return d1,d2

def broadcasting_based_lng_lat(data1, data2):
    # data1, data2 are the data arrays with 2 cols and they hold
    # lat., lng. values in those cols respectively
    data1 = np.deg2rad(data1)                     
    data2 = np.deg2rad(data2)                     

    lat1 = data1[:,0]                     
    lng1 = data1[:,1]         

    lat2 = data2[:,0]                     
    lng2 = data2[:,1]         

    diff_lat = lat1[:,None] - lat2
    diff_lng = lng1[:,None] - lng2
    d = np.sin(diff_lat/2)**2 + np.cos(lat1[:,None])*np.cos(lat2) * np.sin(diff_lng/2)**2
    return 2 * 6371 * np.arcsin(np.sqrt(d))

Hence, to solve your case to get all pairwise haversine distances, it would be -

broadcasting_based_lng_lat(*convert_to_arrays(df1,df2))

---

Elementwise haversine distances

For element-wise haversine distance computations between two data, such that each data holds latitude and longitude in two columns each or lists of two elements each, we would skip some of the extensions to 2D and end up with something like this -

def broadcasting_based_lng_lat_elementwise(data1, data2):
    # data1, data2 are the data arrays with 2 cols and they hold
    # lat., lng. values in those cols respectively
    data1 = np.deg2rad(data1)                     
    data2 = np.deg2rad(data2)                     

    lat1 = data1[:,0]                     
    lng1 = data1[:,1]         

    lat2 = data2[:,0]                     
    lng2 = data2[:,1]         

    diff_lat = lat1 - lat2
    diff_lng = lng1 - lng2
    d = np.sin(diff_lat/2)**2 + np.cos(lat1)*np.cos(lat2) * np.sin(diff_lng/2)**2
    return 2 * 6371 * np.arcsin(np.sqrt(d))

Sample run with a dataframe holding the two data in two columns -

In [42]: np.random.seed(0)
    ...: a = np.random.randint(10,100,(5,2)).tolist()
    ...: b = np.random.randint(10,100,(5,2)).tolist()
    ...: df = pd.DataFrame({'A':a,'B':b})

In [43]: df
Out[43]: 
          A         B
0  [54, 57]  [80, 98]
1  [74, 77]  [98, 22]
2  [77, 19]  [68, 75]
3  [93, 31]  [49, 97]
4  [46, 97]  [56, 98]

In [44]: from haversine import haversine

In [45]: [haversine(i,j) for (i,j) in zip(df.A,df.B)]
Out[45]: 
[3235.9659882513424,
 2399.6124657290075,
 2012.0851666001824,
 4702.8069773315865,
 1114.1193334220534]

In [46]: broadcasting_based_lng_lat_elementwise(np.vstack(df.A), np.vstack(df.B))
Out[46]: 
array([3235.96151855, 2399.60915125, 2012.08238739, 4702.80048155,
       1114.11779454])

Those slight differences are largely because haversine library assumes 6371.0088 as the earth radius, while we are taking it as 6371 here.

回答2:

Use simple print statements to display the arguments to your equation. Some of the operations in you sin expression are different lengths -- the underlying broadcast operation (sort of the vectorized equivalent of zip) requires equal lengths.

来源：https://stackoverflow.com/questions/57696084/efficiently-compute-pairwise-haversine-distances-between-two-datasets-numpy