生成python列表的二维置换

我有一个python列表,如下所示:

[['row_0','row_0_0','row_0_1'],['row_1','row_1_0','row_1_1']]

其中父列表中的每个列表都称为 row column . ['row_0','row_0_0','row_0_1'] 是一排。 “row_0”、“row_0_0”和“row_0_1”是此行的列。

我想以这样的方式产生排列

总行数与原始父列表中的行数相同,即2。
每行的列的排列仍保留在该行内。例如: ['row_0','row_0_0','row_0_1'] 可以有排列 ['row_0','row_0_1','row_0_0'] , ['row_0','row_0_0','row_0_1']
每行的第一列在排列中从不改变。例如:“row_0”in和“row_1”始终保持其列表中的第一项。

到目前为止,我实现的是这样的:

perm_list =  [['row_0','row_0_0','row_0_1'],['row_1','row_1_0','row_1_1']]
all_permutations = list(itertools.permutations(perm_list, len(perm_list))
print(all_permutations)

编辑

我要的输出是这样的:

[
  [['row_0','row_0_0','row_0_1'],['row_1','row_1_0','row_1_1']],
  [['row_0','row_0_1','row_0_0'],['row_1','row_1_1','row_1_0']],
  [['row_1','row_1_0','row_1_1'],['row_0','row_0_0','row_0_1']],
  [['row_1','row_1_1','row_1_0'],['row_0','row_0_1','row_0_0']],
]

a、每个子列表中的第一项保持不变。 b、每个子列表中的项都保留在该列表中。 c、父列表和子列表的项目总数保持不变。

l1=[[list((x[0],)+y) for y in itertools.permutations(x[1:], len(x)-1)] for x in l ] list(map(list,itertools.product(l1[0],l1[1]))) [[['row_0', 'row_0_0', 'row_0_1'], ['row_1', 'row_1_0', 'row_1_1']], [['row_0', 'row_0_0', 'row_0_1'], ['row_1', 'row_1_1', 'row_1_0']], [['row_0', 'row_0_1', 'row_0_0'], ['row_1', 'row_1_0', 'row_1_1']], [['row_0', 'row_0_1', 'row_0_0'], ['row_1', 'row_1_1', 'row_1_0']]]