我正在努力解决这个问题，而无需编写一些可怕的循环来检查 groupby 之后的组并将它们合并在一起。我觉得他们必须有一种我不知道的方法来做到这一点。
我想要做的是将数据帧按其两列进行分组，但有时会出现键翻转的这些组的组合(即 [key1, key2] 将有一个组和 [key2, key1] ] 将有一个组。我实际上想将这些组合中的组合并为一个组。
可以在事后循环执行。我也尝试过使用这样的一些方法:
unique combinations of values in selected columns in pandas data frame and count
但无法让它工作。
这是我的 df 示例:
Ves-1 type Ves-2 type Duration
0 cargo tug 898.559993
1 fishing_trawling tug 898.559992
2 fishing_transiting tug 898.559993
3 fishing_transiting tug 898.559993
4 tug tug 898.559992
5 cargo tug 898.560002
6 cargo tug 898.560002
7 passenger tug 907.200008
8 pleasure tug 898.560003
9 cargo tug 898.559993
10 cargo tug 898.559993
11 cargo fishing_transiting 898.560002
12 cargo fishing_transiting 898.559993
13 cargo fishing_transiting 898.560002
14 tug fishing_transiting 898.560003
15 cargo fishing_transiting 907.200008
16 cargo fishing_transiting 907.200008
17 tug fishing_transiting 898.560002
18 cargo fishing_transiting 898.560002
19 fishing_transiting fishing_transiting 898.559993
如果我只是使用两个 Ves 列进行简单的分组:
>>> test.groupby(['Ves-1 type','Ves-2 type'])['Duration'].agg(list)
Ves-1 type Ves-2 type
cargo fishing_transiting [898.560002, 898.5599930000001, 898.560002, 90...
tug [898.5599930000001, 898.560002, 898.560002, 89...
fishing_transiting fishing_transiting [898.5599930000001]
tug [898.5599930000001, 898.5599930000001]
fishing_trawling tug [898.5599920000001]
passenger tug [907.200008]
pleasure tug [898.560003]
tug fishing_transiting [898.560003, 898.560002]
tug [898.5599920000001]
现在的问题是我有一个 Fishing_transiting/tug 组合和一个 tug/fishing_transiting 组合...有什么方法可以将这些组合并在一起吗？
编辑 - 我尝试过的另一种解决方法有效，但想知道是否有办法在 groupby 中处理这个问题:
>>> test['key'] = list(zip(test['Ves-1 type'].values, test['Ves-2 type'].values))
>>> test['key'] = test['key'].apply(sorted).astype(str)
>>> test.groupby('key')['Duration'].agg(list)
key
['cargo', 'fishing_transiting'] [898.560002, 898.5599930000001, 898.560002, 90...
['cargo', 'tug'] [898.5599930000001, 898.560002, 898.560002, 89...
['fishing_transiting', 'fishing_transiting'] [898.5599930000001]
['fishing_transiting', 'tug'] [898.5599930000001, 898.5599930000001, 898.560...
['fishing_trawling', 'tug'] [898.5599920000001]
['passenger', 'tug'] [907.200008]
['pleasure', 'tug'] [898.560003]
['tug', 'tug'] [898.5599920000001]
最佳答案
让我们沿着 axis=1 对 Ves-1 type 和 Ves-2 type 列中的值进行排序，然后按 groupby 这些排序列上的数据框和使用 list 的 agg Duration:
c = ['Ves-1 type', 'Ves-2 type']
df.groupby(np.sort(df[c], axis=1).T.tolist())['Duration'].agg(list)
cargo fishing_transiting [898.5600019999999, 898.559993, 898.5600019999...
tug [898.559993, 898.5600019999999, 898.5600019999...
fishing_transiting fishing_transiting [898.559993]
tug [898.559993, 898.559993, 898.5600029999999, 89...
fishing_trawling tug [898.559992]
passenger tug [907.200008]
pleasure tug [898.5600029999999]
tug tug [898.559992]
Name: Duration, dtype: object
关于python - 如何使用两列进行 pandas groupby 但合并组以获得这两列中键的唯一组合，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/64244640/