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？
答案 0 :(得分：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