由于您是按年龄分组的，因此，请返回每个组的所有排列，然后取乘积(使用itertools的乘积和排列函数) :

Since you're grouping by age, let's do that and return all the permutations for each group and then take the product (using itertools' product and permutation functions):

In [11]: age = df.groupby("age")

如果我们查看单个组的排列:

If we look at the permutations of a single group:

In [12]: age.get_group(21)
Out[12]:
   age   name
2   21  Chris
4   21   Evan

In [13]: list(permutations(age.get_group(21).index))
Out[13]: [(2, 4), (4, 2)]

In [14]: [df.loc[list(p)] for p in permutations(age.get_group(21).index)]
Out[14]:
[   age   name
 2   21  Chris
 4   21   Evan,    age   name
 4   21   Evan
 2   21  Chris]

我们可以通过仅返回每个组的索引来在整个DataFrame上执行此操作(这假设索引是唯一的，如果在执行此操作之前不是reset_index……您可以能够在较低级别上做一些事情):

We can do this on the entire DataFrame by returning just the index for each group (this assumes that the index is unique, if it's not reset_index prior to doing this... you may be able to do something slightly more lower level):

In [21]: [list(permutations(grp.index)) for (name, grp) in age]
Out[21]: [[(1,)], [(2, 4), (4, 2)], [(3,)], [(0,)]]

In [22]: list(product(*[(permutations(grp.index)) for (name, grp) in age]))
Out[22]: [((1,), (2, 4), (3,), (0,)), ((1,), (4, 2), (3,), (0,))]

我们可以将它们加起来:

We can glue these up with sum:

In [23]: [sum(tups, ()) for tups in product(*[(permutations(grp.index)) for (name, grp) in age])]
Out[23]: [(1, 2, 4, 3, 0), (1, 4, 2, 3, 0)]

如果将这些列为列表，则可以应用loc(这将为您提供所需的结果):

If you make these a list you can apply loc (which gets you the desired result):

In [24]: [df.loc[list(sum(tups, ()))] for tups in product(*[list(permutations(grp.index)) for (name, grp) in age])]
Out[24]:
[   age   name
 1   20    Bob
 2   21  Chris
 4   21   Evan
 3   22  David
 0   28    Abe,    age   name
 1   20    Bob
 4   21   Evan
 2   21  Chris
 3   22  David
 0   28    Abe]

以及名称"列(的列表):

And the (list of) the name column:

In [25]: [list(df.loc[list(sum(tups, ())), "name"]) for tups in product(*[(permutations(grp.index)) for (name, grp) in age])]
Out[25]:
[['Bob', 'Chris', 'Evan', 'David', 'Abe'],
 ['Bob', 'Evan', 'Chris', 'David', 'Abe']]

注意:使用 numpy置换矩阵和pd.tools.util.cartesian_product .我怀疑这太多了，除非它变得非常慢(除非它可能会变慢，因为可能会有很多排列)，否则它不会进行探索.

Note: It may be faster to use a numpy permutation matrix and pd.tools.util.cartesian_product. I suspect it's much of a muchness and wouldn't explore this unless this was unusably slow (it's potentially going to be slow anyway because there could be many many permutations)...