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我可以申请我的 invincenty并确定两台连续机器之间的距离。但是，我想在不重复的情况下找到组中所有机器之间的距离。geopydataframepandas

例如，如果我按公司名称分组，并且有 3 台机器与该公司关联，我想找到机器 1 和 2、1 和 3 以及（2 和 3）之间的距离，但不计算（2和 1) 和 (3 和 1) 因为它们是对称的（结果相同）。

import pandas as pd
from geopy.distance import vincenty

df = pd.DataFrame({'ser_no': [1, 2, 3, 4, 5, 6, 7, 8, 9, 0],
                'co_nm': ['aa', 'aa', 'aa', 'bb', 'bb', 'bb', 'bb', 'cc', 'cc', 'cc'],
                'lat': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
                'lon': [21, 22, 23, 24, 25, 26, 27, 28, 29, 30]})

coord_col = ['lat', 'lon']
matching_cust = df['co_nm'] == df['co_nm'].shift(1)
shift_coords = df.shift(1).loc[matching_cust, coord_col]
# join in shifted coords and compute distance
df_shift = df.join(shift_coords, how = 'inner', rsuffix = '_2')
# return distance in miles
df['dist'] = df_shift.apply(lambda x: vincenty((x[1], x[2]), 
    (x[4], x[5])).mi, axis = 1)

这只能找到组中连续机器的距离我该如何扩展以找到组中所有机器的距离？

此代码返回：

  co_nm  lat  lon  ser_no      dist
0    aa    1   21       1       NaN
1    aa    2   22       2  97.47832
2    aa    3   23       3  97.44923
3    bb    4   24       4       NaN
4    bb    5   25       5  97.34752
5    bb    6   26       6  97.27497
6    bb    7   27       7  97.18804
7    cc    8   28       8       NaN
8    cc    9   29       9  96.97129
9    cc   10   30       0  96.84163

编辑：

期望的输出将找到公司相关机器的唯一距离组合；也就是说，因为co_nm aa我们会得到 ser_no (1,2), (1,3), (2,3), (1,3) 之间的距离以及机器的距离 in co_nm bband cc，但我们不会确定不同co_nm组中机器的距离。

这有意义吗？

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1 回答 1

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UPDATE2：使用功能：

def calc_dist(df):
    return pd.DataFrame(
               [ [grp,
                  df.loc[c[0]].ser_no,
                  df.loc[c[1]].ser_no,
                  vincenty(df.loc[c[0], ['lat','lon']], df.loc[c[1], ['lat','lon']])
                 ]
                 for grp,lst in df.groupby('co_nm').groups.items()
                 for c in combinations(lst, 2)
               ],
               columns=['co_nm','machineA','machineB','distance'])

In [27]: calc_dist(df)
Out[27]:
   co_nm  machineA  machineB               distance
0     aa         1         2  156.87614939082016 km
1     aa         1         3   313.7054454472326 km
2     aa         2         3    156.829329105069 km
3     cc         8         9  156.06016539095216 km
4     cc         8         0   311.9109981692541 km
5     cc         9         0  155.85149813446617 km
6     bb         4         5  156.66564183673603 km
7     bb         4         6   313.2143330250297 km
8     bb         4         7   469.6225353388079 km
9     bb         5         6  156.54889741438788 km
10    bb         5         7  312.95759746593706 km
11    bb         6         7   156.4089967703544 km

更新：

In [9]: dist = pd.DataFrame(
   ...:   [ [grp,
   ...:      df.loc[c[0]].ser_no,
   ...:      df.loc[c[1]].ser_no,
   ...:      vincenty(df.loc[c[0], ['lat','lon']], df.loc[c[1], ['lat','lon']])
   ...:     ]
   ...:     for grp,lst in df.groupby('co_nm').groups.items()
   ...:     for c in combinations(lst, 2)
   ...:   ],
   ...:   columns=['co_nm','machineA','machineB','distance'])

In [10]: dist
Out[10]:
   co_nm  machineA  machineB               distance
0     aa         1         2  156.87614939082016 km
1     aa         1         3   313.7054454472326 km
2     aa         2         3    156.829329105069 km
3     cc         8         9  156.06016539095216 km
4     cc         8         0   311.9109981692541 km
5     cc         9         0  155.85149813446617 km
6     bb         4         5  156.66564183673603 km
7     bb         4         6   313.2143330250297 km
8     bb         4         7   469.6225353388079 km
9     bb         5         6  156.54889741438788 km
10    bb         5         7  312.95759746593706 km
11    bb         6         7   156.4089967703544 km

解释：组合部分

In [11]: [c
   ....:  for grp,lst in df.groupby('co_nm').groups.items()
   ....:  for c in combinations(lst, 2)]
Out[11]:
[(0, 1),
 (0, 2),
 (1, 2),
 (7, 8),
 (7, 9),
 (8, 9),
 (3, 4),
 (3, 5),
 (3, 6),
 (4, 5),
 (4, 6),
 (5, 6)]

旧答案：

In [3]: from itertools import combinations

In [4]: import pandas as pd

In [5]: from geopy.distance import vincenty

In [6]: df = pd.DataFrame({'machine': [1,2,3], 'lat': [11, 12, 13], 'lon': [21,22,23]})

In [7]: df
Out[7]:
   lat  lon  machine
0   11   21        1
1   12   22        2
2   13   23        3

In [8]: dist = pd.DataFrame(
   ...:   [ [df.loc[c[0]].machine,
   ...:      df.loc[c[1]].machine,
   ...:      vincenty(df.loc[c[0], ['lat','lon']], df.loc[c[1], ['lat','lon']])
   ...:     ]
   ...:     for c in combinations(df.index, 2)
   ...:   ],
   ...:   columns=['machineA','machineB','distance'])

In [9]: dist
Out[9]:
   machineA  machineB               distance
0         1         2   155.3664523771998 km
1         1         3   310.4557192973811 km
2         2         3  155.09044419651156 km
于 2016-04-19T20:52:42.313 回答