我正在尝试将函数转换为更优化的方式，以减少计算时间。更具体地说，我想相对于plot2的所有点（其中plot1，plot2数据帧带有经度，纬度列）保持对plot1的每个点的最小正弦距离。这是我的代码：
def calculate_min_haversine_distance(plot1,plot2):
for index,row in plot1.iterrows():
minimum = 100000000
for index2,row2 in plot2.iterrows():
dis = haversine_distance(row.latitude,row.longitude,row2.latitude,row2.longitude)
if (dis<minimum):
minimum=dis
plot1.loc(index,'Min Haversine Distance') = minimum
return plot1
我不确定如何摆脱第一个循环，但这应该可以帮助您摆脱第二个循环：
def calculate_min_haversine_distance(plot1,plot2):
for index,row in plot1.iterrows():
plot2['dist'] = plot2.apply(lambda x: haversine_distance(row.latitude,row.longitude,x.latitude,x.longitude),axis=1)
plot1.loc[index,'Min Haversine Distance'] = min(plot2['dist'])
plot2.drop('dist',axis=1,inplace=True) # Delete the temporary column created
return plot1
我会尝试执行以下操作： 希望对您有所帮助。
import pandas as pd
import numpy as np
df1 = pd.DataFrame(data={'lat': [1,2,3,4],'lon': [5,6,7,8]})
df2 = pd.DataFrame(data={'lat': [9,10,11,12],'lon': [13,14,15,16]})
df1['key'],df2['key'] = 1,1
df_c = pd.merge(df1,df2,on='key').drop('key',axis=1)
# below function is copied from: https://stackoverflow.com/a/43577275/4450090
def haversine(lat1,lon1,lat2,lon2,to_radians=True,earth_radius=6371):
if to_radians:
lat1,lon2 = np.radians([lat1,lon2])
a = np.sin((lat2-lat1)/2.0)**2 + \
np.cos(lat1) * np.cos(lat2) * np.sin((lon2-lon1)/2.0)**2
return earth_radius * 2 * np.arcsin(np.sqrt(a))
df_c['dist'] = df_c.apply(lambda x: haversine(x['lat_x'],x['lon_x'],x['lat_y'],x['lon_y']),axis=1)
min_val = 1000000
df_c['dist'] = df_c['dist'].apply(lambda x: x if x < min_val else min_val)