Question

我可以将vincenty中的geopy应用于dataframe中的pandas，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。

例如，如果我按公司名称分组并且有3台与该公司关联的机器，我希望找到机器1和2,1和3之间的距离，以及（2和3）但不计算距离在（2和1）和（3和1）之间，因为它们是对称的（相同的结果）。

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，如何在此处展开以查找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

修改

所需的输出将找到公司相关机器的唯一距离组合;也就是说，对于co_nm aa，我们将得到ser_no（1,2），（1,3），（2,3），（1,3）之间的距离以及{{1}中机器的距离}和co_nm bb也是如此，但我们不会确定不同cc组中机器的距离。

这有意义吗？

Answer 1

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

<强>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)]

OLD回答：

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

python：递归地查找组中各点之间的距离

1 个答案: