使用 pandas 根据 python 元组的第一部分进行分组

Question

我在通过元组的前两个元素进行分组时遇到了一些麻烦，我已经搜索了很多并尝试了但我无法弄清楚:(

我有这个数据集：

    idi d2  duplicates
0   a   b   (us2, us1, 1)
0   a   b   (us1, us4, 1)
0   a   b   (us4, us2, 1)
0   a   b   (us2, us5, 1)
0   a   b   (us5, us4, 1)
0   a   b   (us4, us1, 1)
0   a   b   (us1, us2, 1)
0   a   b   (us2, us1, 2)
0   a   b   (us1, us4, 4)
0   a   b   (us4, us2, 1)
0   a   b   (us2, us4, 1)
0   a   b   (us4, us2, 1)
1   c   b   (us1, us2, 1)
1   c   b   (us2, us1, 1)
1   c   b   (us1, us2, 1)
1   c   b   (us2, us4, 1)
1   c   b   (us4, us5, 1)
2   v   b   (us4, us5, 1)

我想根据id、id2和'usx'进行分组，所以输出应该是：

    idi d2   duplicates
0   a   b   (us2, us1, 1), (us2, us1, 2)
0   a   b   (us1, us4, 1), (us1, us4, 4)
0   a   b   (us4, us2, 1), (us4, us2, 1), (us4, us2, 1)
0   a   b   (us2, us5, 1)
0   a   b   (us5, us4, 1)
0   a   b   (us4, us1, 1)
0   a   b   (us1, us2, 1)
0   a   b   (us2, us4, 1)
1   c   b   (us1, us2, 1), (us1, us2, 1)
1   c   b   (us2, us1, 1)
1   c   b   (us2, us4, 1)
1   c   b   (us4, us5, 1)
2   v   b   (us4, us5, 1)

生成有效部分的代码是：

d = {'id': [      "a",  "a",   "a", "a",   "a",   "a",   "a",   "a",   "a",   "c",   "c",   "c",   "c",   "c",   "a",   "a",   "a",   "a",   "v",   "v",   "c",   "c"], 
     'id2': ["b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b",   "b"], 
     'userid':         ["us1", "us2", "us1", "us2", "us4", "us4", "us5", "us1", "us2", "us1", "us2", "us1", "us2", "us4", "us4", "us2", "us4", "us2", "us4", "us5", "us4", "us5"],
     "time":            [11,    2,      3,     5,      4,   7,     6,      8,     9,    10,    11,    12,    13,    14,     15,   16,    17,    18,    19,    20,    21,    22]}

df_test = pd.DataFrame(data=d).sort_values('time').reset_index()
df_test  = df_test.groupby(['id','id2']).apply(lambda x: list(zip(x['userid'][:-1], x['userid'][1:],
                                                                                   x['time'][:-1], x['time'][1:]))).reset_index(name = 'duplicates')

df_test['duplicates'] = df_test.apply(lambda x: [(k, v, j - y) for k,v, y,j in x.duplicates if k != v], 1)
df_test['duplicates'] = df_test.apply(lambda x: [(k,v,y) for k,v,y in x.duplicates], 1)
df_test.explode('duplicates')

Answer 1

我相信您需要通过使用str 索引提取的元组的前 2 个值进行分组，它可以工作，因为元组是可迭代的：

df = (df_test.groupby(['id','id2', df_test['duplicates'].str[:2]], sort=False)['duplicates']
             .apply(list)
             .reset_index(level=2, drop=True)
             .reset_index())
print (df)
   id id2                                     duplicates
0   a   b                 [(us2, us1, 1), (us2, us1, 2)]
1   a   b                 [(us1, us4, 1), (us1, us4, 4)]
2   a   b  [(us4, us2, 1), (us4, us2, 1), (us4, us2, 1)]
3   a   b                                [(us2, us5, 1)]
4   a   b                                [(us5, us4, 1)]
5   a   b                                [(us4, us1, 1)]
6   a   b                                [(us1, us2, 1)]
7   a   b                                [(us2, us4, 1)]
8   c   b                 [(us1, us2, 1), (us1, us2, 1)]
9   c   b                                [(us2, us1, 1)]
10  c   b                                [(us2, us4, 1)]
11  c   b                                [(us4, us5, 1)]
12  v   b                                [(us4, us5, 1)]

编辑：

df_test['duplicates'] = df_test.apply(lambda x: [(x['id'], k,v,y) for k,v,y in x.duplicates], 1) 

df_test = df_test.explode('duplicates')
print (df_test)
  id id2        duplicates
0  a   b  (a, us2, us1, 1)
0  a   b  (a, us1, us4, 1)
0  a   b  (a, us4, us2, 1)
0  a   b  (a, us2, us5, 1)
0  a   b  (a, us5, us4, 1)
0  a   b  (a, us4, us1, 1)
0  a   b  (a, us1, us2, 1)
0  a   b  (a, us2, us1, 2)
0  a   b  (a, us1, us4, 4)
0  a   b  (a, us4, us2, 1)
0  a   b  (a, us2, us4, 1)
0  a   b  (a, us4, us2, 1)
1  c   b  (c, us1, us2, 1)
1  c   b  (c, us2, us1, 1)
1  c   b  (c, us1, us2, 1)
1  c   b  (c, us2, us4, 1)
1  c   b  (c, us4, us5, 1)
2  v   b  (v, us4, us5, 1)

---

df = (df_test.groupby(['id','id2', df_test['duplicates'].str[1:3]], sort=False)['duplicates']
              .apply(list)
              .reset_index(level=2, drop=True)
              .reset_index())
print (df)
   id id2                                         duplicates
0   a   b               [(a, us2, us1, 1), (a, us2, us1, 2)]
1   a   b               [(a, us1, us4, 1), (a, us1, us4, 4)]
2   a   b  [(a, us4, us2, 1), (a, us4, us2, 1), (a, us4, ...
3   a   b                                 [(a, us2, us5, 1)]
4   a   b                                 [(a, us5, us4, 1)]
5   a   b                                 [(a, us4, us1, 1)]
6   a   b                                 [(a, us1, us2, 1)]
7   a   b                                 [(a, us2, us4, 1)]
8   c   b               [(c, us1, us2, 1), (c, us1, us2, 1)]
9   c   b                                 [(c, us2, us1, 1)]
10  c   b                                 [(c, us2, us4, 1)]
11  c   b                                 [(c, us4, us5, 1)]
12  v   b                                 [(v, us4, us5, 1)]