python - 向量化 Pandas 中的函数

Question

我有一个包含纬度/经度坐标列表的数据框:

d = {'Provider ID': {0: '10001',
  1: '10005',
  2: '10006',
  3: '10007',
  4: '10008',
  5: '10011',
  6: '10012',
  7: '10016',
  8: '10018',
  9: '10019'},
 'latitude': {0: '31.215379379000467',
  1: '34.22133455500045',
  2: '34.795039606000444',
  3: '31.292159523000464',
  4: '31.69311635000048',
  5: '33.595265517000485',
  6: '34.44060759100046',
  7: '33.254429322000476',
  8: '33.50314015000049',
  9: '34.74643089500046'},
 'longitude': {0: ' -85.36146587999968',
  1: ' -86.15937514799964',
  2: ' -87.68507485299966',
  3: ' -86.25539902199966',
  4: ' -86.26549483099967',
  5: ' -86.66531866799966',
  6: ' -85.75726760699968',
  7: ' -86.81407933399964',
  8: ' -86.80242858299965',
  9: ' -87.69893502799965'}}
df = pd.DataFrame(d)

我的目标是使用 haversine 函数计算出 KM 中每个项目之间的距离:

from math import radians, cos, sin, asin, sqrt
def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """

    # convert decimal degrees to radians 
    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)) 

    # 6367 km is the radius of the Earth
    km = 6367 * c
    return km

我的目标是获得一个类似于下面的 result_df 的数据框，其中的值是每个提供者 ID 之间的距离:

 result_df = pd.DataFrame(columns = df['Provider ID'], index=df['Provider ID'])

我可以循环执行此操作，但速度非常慢。我正在寻找一些帮助将其转换为矢量化方法:

for first_hospital_coordinates in result_df.columns:
    for second_hospital_coordinates in result_df['Provider ID']:
        if first_hospital_coordinates == 'Provider ID':
            pass
        else:
            L1 = df[df['Provider ID'] == first_hospital_coordinates]['latitude'].astype('float64').values
            O1 = df[df['Provider ID'] == first_hospital_coordinates]['longitude'].astype('float64').values
            L2 = df[df['Provider ID'] == second_hospital_coordinates]['latitude'].astype('float64').values
            O2 = df[df['Provider ID'] == second_hospital_coordinates]['longitude'].astype('float64').values

            distance = haversine(O1, L1, O2, L2)

            crit = result_df['Provider ID'] == second_hospital_coordinates
            result_df.loc[crit, first_hospital_coordinates] = distance

Answer 1

要矢量化此代码，您需要对完整的数据帧进行操作，而不是对单个的经纬度进行操作。我已经尝试过这个。我需要结果 df 和一个新函数 h2，

import numpy as np
def h2(df, p):
    inrad = df.applymap(radians)
    dlon = inrad.longitude-inrad.longitude[p]
    dlat = inrad.latitude-inrad.latitude[p]
    lat1 = pd.Series(index = df.index, data = [df.latitude[p] for i in range(len(df.index))])
    a = np.sin(dlat/2)*np.sin(dlat/2) + np.cos(df.latitude) * np.cos(lat1) * np.sin(dlon/2)**2
    c = 2 * 1/np.sin(np.sqrt(a))
    km = 6367 * c
    return km

df = df.set_index('Provider ID')
df = df.astype(float)
df2 = pd.DataFrame(index = df.index, columns = df.index)
for c in df2.columns:
    df2[c] = h2(df, c)

print (df2)

这应该会产生，(我不确定我是否有正确的答案......我的目标是向量化代码)

Provider ID         10001         10005         10006         10007  \
Provider ID                                                           
10001                 inf  5.021936e+05  5.270062e+05  1.649088e+06   
10005        5.021936e+05           inf  9.294868e+05  4.985233e+05   
10006        5.270062e+05  9.294868e+05           inf  4.548412e+05   
10007        1.649088e+06  4.985233e+05  4.548412e+05           inf   
10008        1.460299e+06  5.777248e+05  5.246954e+05  3.638231e+06   
10011        6.723581e+05  2.004199e+06  1.027439e+06  6.394402e+05   
10012        4.559090e+05  3.265536e+06  7.573411e+05  4.694125e+05   
10016        7.680036e+05  1.429573e+06  9.105474e+05  7.517467e+05   
10018        7.096548e+05  1.733554e+06  1.020976e+06  6.701920e+05   
10019        5.436342e+05  9.278739e+05  2.891822e+07  4.638858e+05   

Provider ID         10008         10011         10012         10016  \
Provider ID                                                           
10001        1.460299e+06  6.723581e+05  4.559090e+05  7.680036e+05   
10005        5.777248e+05  2.004199e+06  3.265536e+06  1.429573e+06   
10006        5.246954e+05  1.027439e+06  7.573411e+05  9.105474e+05   
10007        3.638231e+06  6.394402e+05  4.694125e+05  7.517467e+05   
10008                 inf  7.766998e+05  5.401081e+05  9.496953e+05   
10011        7.766998e+05           inf  1.341775e+06  4.220911e+06   
10012        5.401081e+05  1.341775e+06           inf  1.119063e+06   
10016        9.496953e+05  4.220911e+06  1.119063e+06           inf   
10018        8.236437e+05  1.242451e+07  1.226941e+06  5.866259e+06   
10019        5.372119e+05  1.051748e+06  7.514774e+05  9.362341e+05   

Provider ID         10018         10019  
Provider ID                              
10001        7.096548e+05  5.436342e+05  
10005        1.733554e+06  9.278739e+05  
10006        1.020976e+06  2.891822e+07  
10007        6.701920e+05  4.638858e+05  
10008        8.236437e+05  5.372119e+05  
10011        1.242451e+07  1.051748e+06  
10012        1.226941e+06  7.514774e+05  
10016        5.866259e+06  9.362341e+05  
10018                 inf  1.048895e+06  
10019        1.048895e+06           inf  

[10 rows x 10 columns]