第一个

另一种可能的解决方案：

# this is to push up all not NaN values to the top of each column
df.loc[:, 'col_1':'col_3'] = df.groupby(
    ['id_1', 'id_2'], sort=False).transform(lambda x: sorted(x, key=pd.isnull))

# this is to remove all useless rows of NaN
df.loc[~(df.duplicated(['id_1', 'id_2']) &
         df.loc[:, 'col_1':'col_3'].isna().all(axis=1))]

输出量：

id_1  id_2     col_1  col_2   col_3
0    33    64       dog   bike    star
1    33    64  kangaroo    car  meteor
4    22    12       NaN  train    rock
6    88    77       NaN  horse     NaN
7   100   100       NaN    NaN     NaN

为了避免难以辨认的Pandas voodoo，在导入和df示例化之后，可以执行以下操作

def get_max_vals_from_row_sets(row, cols):
  mn = 1
  for col in cols:
    mn = max(mn, len(row[col]))
  return mn

def add_id_row(d, row, ids, cols):
  max_vals = get_max_vals_from_row_sets(row, cols)

  for _ in range(max_vals):
    for id_ in ids:
      d[id_].append(row[id_])

    for col in cols:
      if len(row[col]) != 0:
        d[col].append(row[col].pop())
      else:
        d[col].append(np.nan)

def drop_set_nans(row, cols):
  for col in cols:
    if np.nan in row[col]:
      row[col].remove(np.nan)
  return row

def squash_out_redundant_nans(df, ids, cols):
  d = {k: [] for k in df.columns}
  for _, row in df1.iterrows():
    drop_set_nans(row, cols)
    add_id_row(d, row, ids, cols)

  df = pd.DataFrame(d)
  return df

ids = ['id_1', 'id_2']
cols = ['col_1', 'col_2', 'col_3']
df1 = df.groupby(ids).agg(set).reset_index()
df = squash_out_redundant_nans(df1, ids, cols)
print(df)