问题
I have a dataframe with >2.7MM coordinates, and a separate list of ~2,000 coordinates. I'm trying to return the minimum distance between the coordinates in each individual row compared to every coordinate in the list. The following code works on a small scale (dataframe with 200 rows), but when calculating over 2.7MM rows, it seemingly runs forever.
from haversine import haversine
df
Latitude Longitude
39.989 -89.980
39.923 -89.901
39.990 -89.987
39.884 -89.943
39.030 -89.931
end_coords_list = [(41.342,-90.423),(40.349,-91.394),(38.928,-89.323)]
for row in df.itertuples():
def min_distance(row):
beg_coord = (row.Latitude, row.Longitude)
return min(haversine(beg_coord, end_coord) for end_coord in end_coords_list)
df['Min_Distance'] = df.apply(min_distance, axis=1)
I know the issue lies in the sheer number of calculations that are happening (5.7MM * 2,000 = ~11.4BN), and the fact that running this many loops is incredibly inefficient.
Based on my research, it seems like a vectorized NumPy function might be a better approach, but I'm new to Python and NumPy so I'm not quite sure how to implement this in this particular situation.
Ideal Output:
df
Latitude Longitude Min_Distance
39.989 -89.980 3.7
39.923 -89.901 4.1
39.990 -89.987 4.2
39.884 -89.943 5.9
39.030 -89.931 3.1
Thanks in advance!
回答1:
The haversine func in essence is :
# convert all latitudes/longitudes from decimal degrees to radians
lat1, lng1, lat2, lng2 = map(radians, (lat1, lng1, lat2, lng2))
# calculate haversine
lat = lat2 - lat1
lng = lng2 - lng1
d = sin(lat * 0.5) ** 2 + cos(lat1) * cos(lat2) * sin(lng * 0.5) ** 2
h = 2 * AVG_EARTH_RADIUS * asin(sqrt(d))
Here's a vectorized method leveraging the powerful NumPy broadcasting and NumPy ufuncs
to replace those math-module funcs so that we would operate on entire arrays in one go -
# Get array data; convert to radians to simulate 'map(radians,...)' part
coords_arr = np.deg2rad(coords_list)
a = np.deg2rad(df.values)
# Get the differentiations
lat = coords_arr[:,0] - a[:,0,None]
lng = coords_arr[:,1] - a[:,1,None]
# Compute the "cos(lat1) * cos(lat2) * sin(lng * 0.5) ** 2" part.
# Add into "sin(lat * 0.5) ** 2" part.
add0 = np.cos(a[:,0,None])*np.cos(coords_arr[:,0])* np.sin(lng * 0.5) ** 2
d = np.sin(lat * 0.5) ** 2 + add0
# Get h and assign into dataframe
h = 2 * AVG_EARTH_RADIUS * np.arcsin(np.sqrt(d))
df['Min_Distance'] = h.min(1)
For further performance boost, we can make use of numexpr module to replace the transcendental funcs.
Runtime test and verification
Approaches -
def loopy_app(df, coords_list):
for row in df.itertuples():
df['Min_Distance1'] = df.apply(min_distance, axis=1)
def vectorized_app(df, coords_list):
coords_arr = np.deg2rad(coords_list)
a = np.deg2rad(df.values)
lat = coords_arr[:,0] - a[:,0,None]
lng = coords_arr[:,1] - a[:,1,None]
add0 = np.cos(a[:,0,None])*np.cos(coords_arr[:,0])* np.sin(lng * 0.5) ** 2
d = np.sin(lat * 0.5) ** 2 + add0
h = 2 * AVG_EARTH_RADIUS * np.arcsin(np.sqrt(d))
df['Min_Distance2'] = h.min(1)
Verification -
In [158]: df
Out[158]:
Latitude Longitude
0 39.989 -89.980
1 39.923 -89.901
2 39.990 -89.987
3 39.884 -89.943
4 39.030 -89.931
In [159]: loopy_app(df, coords_list)
In [160]: vectorized_app(df, coords_list)
In [161]: df
Out[161]:
Latitude Longitude Min_Distance1 Min_Distance2
0 39.989 -89.980 126.637607 126.637607
1 39.923 -89.901 121.266241 121.266241
2 39.990 -89.987 126.037388 126.037388
3 39.884 -89.943 118.901195 118.901195
4 39.030 -89.931 53.765506 53.765506
Timings -
In [163]: df
Out[163]:
Latitude Longitude
0 39.989 -89.980
1 39.923 -89.901
2 39.990 -89.987
3 39.884 -89.943
4 39.030 -89.931
In [164]: %timeit loopy_app(df, coords_list)
100 loops, best of 3: 2.41 ms per loop
In [165]: %timeit vectorized_app(df, coords_list)
10000 loops, best of 3: 96.8 µs per loop
来源:https://stackoverflow.com/questions/44681828/efficient-computation-of-minimum-of-haversine-distances