Vectorised Haversine公式与pandas数据框
我知道要find两个纬度，经度点之间的距离，我需要使用haversine函数：
def haversine(lon1, lat1, lon2, lat2): lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2]) dlon = lon2 - lon1 dlat = lat2 - lat1 a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2 c = 2 * asin(sqrt(a)) km = 6367 * c return km 我有一个DataFrame其中一列是纬度和另一列是经度。 我想找出这些点离设定点有多远，-56.7213600，37.2175900。 如何从DataFrame中取值并将其放入函数中？
示例DataFrame：
SEAZ LAT LON 1 296.40, 58.7312210, 28.3774110 2 274.72, 56.8148320, 31.2923240 3 192.25, 52.0649880, 35.8018640 4 34.34, 68.8188750, 67.1933670 5 271.05, 56.6699880, 31.6880620 6 131.88, 48.5546220, 49.7827730 7 350.71, 64.7742720, 31.3953780 8 214.44, 53.5192920, 33.8458560 9 1.46, 67.9433740, 38.4842520 10 273.55, 53.3437310, 4.4716664
我无法确认计算是否正确，但以下工作：
In [11]: def haversine(row): lon1 = -56.7213600 lat1 = 37.2175900 lon2 = row['LON'] lat2 = row['LAT'] lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2]) dlon = lon2 - lon1 dlat = lat2 - lat1 a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2 c = 2 * arcsin(sqrt(a)) km = 6367 * c return km df['distance'] = df.apply(lambda row: haversine(row), axis=1) df Out[11]: SEAZ LAT LON distance index 1 296.40 58.731221 28.377411 6275.791920 2 274.72 56.814832 31.292324 6509.727368 3 192.25 52.064988 35.801864 6990.144378 4 34.34 68.818875 67.193367 7357.221846 5 271.05 56.669988 31.688062 6538.047542 6 131.88 48.554622 49.782773 8036.968198 7 350.71 64.774272 31.395378 6229.733699 8 214.44 53.519292 33.845856 6801.670843 9 1.46 67.943374 38.484252 6418.754323 10 273.55 53.343731 4.471666 4935.394528
下面的代码在这么小的dataframe上实际上比较慢，但是我将它应用到了100,000行df：
In [35]: %%timeit df['LAT_rad'], df['LON_rad'] = np.radians(df['LAT']), np.radians(df['LON']) df['dLON'] = df['LON_rad'] - math.radians(-56.7213600) df['dLAT'] = df['LAT_rad'] - math.radians(37.2175900) df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin(df['dLAT']/2)**2 + math.cos(math.radians(37.2175900)) * np.cos(df['LAT_rad']) * np.sin(df['dLON']/2)**2)) 1 loops, best of 3: 17.2 ms per loop
相比于使用4.3s的应用函数快了将近250倍，将来还有一些值得注意的地方
如果我们把所有的上述内容压缩成一行：
In [39]: %timeit df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin((np.radians(df['LAT']) - math.radians(37.21759))/2)**2 + math.cos(math.radians(37.21759)) * np.cos(np.radians(df['LAT']) * np.sin((np.radians(df['LON']) - math.radians(-56.72136))/2)**2))) 100 loops, best of 3: 12.6 ms per loop
我们现在进一步加速了约341倍的速度。
