使用每个城市的所有可能组合计算 5 个城市之间的地理距离答案

【问题标题】：Calculate geographical distance between 5 cities with all the possible combinations of each city使用每个城市的所有可能组合计算 5 个城市之间的地理距离
【发布时间】：2019-07-31 05:50:30
【问题描述】：

所以我有一个包含 3 列（城市、纬度、经度）的 csv 文件我已经使用此代码从这个 csv 文件在 python 中创建了一个数据框

data = pd.read_csv("lat_long.csv",nrows=10)
Lat = data.lat.tolist()
Lon = data.lon.tolist()
suburb = data.suburb.tolist()
dict={'Latitude':Lat,'Longitude':Lon}
df = pd.DataFrame(dict,index=(suburb))

输出是这样的

                                 Latitude   Longitude
AUSTRALIAN NATIONAL UNIVERSITY -35.277272  149.117136
BARTON                         -35.201372  149.095065
DARWIN                         -12.801028  130.955789
DARWIN                         -12.801028  130.955789
PARAP                          -12.432181  130.843310
ALAWA                          -12.378451  130.877014
BRINKIN                        -12.367769  130.869808
CASUARINA                      -12.376597  130.850489
JINGILI                        -12.385761  130.873726
LEE POINT                      -12.360865  130.891349

现在我想要的是从 1 个城市到其他 9 个城市的距离的所有可能组合。它应该看起来像

                                              DISTANCE
AUSTRALIAN NATIONAL UNIVERSITY- BARTON
AUSTRALIAN NATIONAL UNIVERSITY - DARWIN
AUSTRALIAN NATIONAL UNIVERSITY - DARWIN
AUSTRALIAN NATIONAL UNIVERSITY - PARAP

我尝试过使用嵌套的 for 循环来执行此操作，它可以工作，但我想要快一点。

【问题讨论】：

标签： python pandas numpy combinations

【解决方案1】：

我从数据框开始

    city         Latitude   Longitude
0   AUSTRAL.    -35.277272  149.117136
1   BARTON      -35.201372  149.095065
2   DARWIN      -12.801028  130.955789
3   DARWIN      -12.801028  130.955789
4   PARAP       -12.432181  130.843310
5   ALAWA       -12.378451  130.877014
6   BRINKIN     -12.367769  130.869808
7   CASUARINA   -12.376597  130.850489
8   JINGILI     -12.385761  130.873726
9   LEE_POINT   -12.360865  130.891349

并创建一个新列，它只是一个帮助创建我们通过将数据框与自身合并获得的笛卡尔积。

df['join'] = 1
df_joined = pd.merge(df, df,on='join')

df_joined['haversine_dist'] = df_joined.apply(lambda x: haversine((x.Latitude_x, x.Longitude_x),(x.Latitude_y,x.Longitude_y)), 1)

结果（仅前 5 列）

    city_x      Latitude_x  Longitude_x join city_y Latitude_y  Longitude_y haversine_dist
0   AUSTRAL.    -35.277272  149.117136  1   AUSTRAL.    -35.277272  149.117136  0.000000
1   AUSTRAL.    -35.277272  149.117136  1   BARTON  -35.201372  149.095065  8.674473
2   AUSTRAL.    -35.277272  149.117136  1   DARWIN  -12.801028  130.955789  3093.972598
3   AUSTRAL.    -35.277272  149.117136  1   DARWIN  -12.801028  130.955789  3093.972598
4   AUSTRAL.    -35.277272  149.117136  1   PARAP   -12.432181  130.843310  3135.034018
5   AUSTRAL.    -35.277272  149.117136  1   ALAWA   -12.378451  130.877014  3138.077950

【讨论】：

这应该是解决方案。我现在没有机会测试它，但这应该可以。如果没有就给我打电话，我今晚会检查一下。

【解决方案2】：

为了测试，我手工构造了原始的DataFrame

import pandas as pd 
import itertools
from haversine import haversine
x = {'city':['AUSTRALIAN NATIONAL UNIVERSITY', 'BARTON', 'DARWIN', 'DARWIN', 'PARAP', 'ALAWA', 'BRINKIN', 'CASUARINA', 'JINGILI', 'LEE_POINT' ]}
la = {'Latitude':[-35.277272,-35.201372, -12.801028 , -12.801028, -12.432181, -12.378451, -12.367769, -12.376597, -12.385761, -12.360865]}
lo = {'Longitude':[149.117136,149.095065, 130.955789 , 130.955789, 130.843310,  130.877014, 130.869808, 130.850489, 130.873726, 130.891349]}
data = {**x, **la, **lo}
df = pd.DataFrame(data)

删除重复项。

df = df.drop_duplicates()

列出所有城市。

city = list(df["city"])

合并两个

TwoCity = list(itertools.combinations(city, 2))

构造新的DataFrame

df1 = pd.DataFrame({'TwoCity':TwoCity})
df1['Distance(km)'] = df1.apply(lambda row: \
          haversine((df[df['city']==row.TwoCity[0]]['Latitude'], df[df['city']==row.TwoCity[0]]['Longitude']),\
                    (df[df['city']==row.TwoCity[1]]['Latitude'], df[df['city']==row.TwoCity[1]]['Longitude'])),axis=1)
print(df1.to_string(index=False))

df1的最终结果是（手动稍作调整）：

   TwoCity                                     Distance(km)
   (AUSTRALIAN NATIONAL UNIVERSITY, BARTON)      8.674473
   (AUSTRALIAN NATIONAL UNIVERSITY, DARWIN)   3093.972598
    (AUSTRALIAN NATIONAL UNIVERSITY, PARAP)   3135.034018
    (AUSTRALIAN NATIONAL UNIVERSITY, ALAWA)   3138.077950
  (AUSTRALIAN NATIONAL UNIVERSITY, BRINKIN)   3139.500311
(AUSTRALIAN NATIONAL UNIVERSITY, CASUARINA)   3139.808790
  (AUSTRALIAN NATIONAL UNIVERSITY, JINGILI)   3137.587038
(AUSTRALIAN NATIONAL UNIVERSITY, LEE_POINT)   3138.882795
                           (BARTON, DARWIN)   3086.264122
                            (BARTON, PARAP)   3127.309536
                            (BARTON, ALAWA)   3130.345201
                          (BARTON, BRINKIN)   3131.767583
                        (BARTON, CASUARINA)   3132.079061
                          (BARTON, JINGILI)   3129.855257
                        (BARTON, LEE_POINT)   3131.146957
                            (DARWIN, PARAP)     42.791471
                            (DARWIN, ALAWA)     47.759804
                          (DARWIN, BRINKIN)     49.071577
                        (DARWIN, CASUARINA)     48.558395
                          (DARWIN, JINGILI)     47.026561
                        (DARWIN, LEE_POINT)     49.441057
                             (PARAP, ALAWA)      7.006568
                           (PARAP, BRINKIN)      7.718791
                         (PARAP, CASUARINA)      6.229645
                           (PARAP, JINGILI)      6.128079
                         (PARAP, LEE_POINT)      9.492285
                           (ALAWA, BRINKIN)      1.422460
                         (ALAWA, CASUARINA)      2.888261
                           (ALAWA, JINGILI)      0.887821
                         (ALAWA, LEE_POINT)      2.499614
                       (BRINKIN, CASUARINA)      2.316553
                         (BRINKIN, JINGILI)      2.045378
                       (BRINKIN, LEE_POINT)      2.462424
                       (CASUARINA, JINGILI)      2.721699
                     (CASUARINA, LEE_POINT)      4.770298
                       (JINGILI, LEE_POINT)      3.365596

【讨论】：