Question

我有一个嵌套的python列表，称为：

from test1 import a,b
print a
print b

在该列表中，我将每个元素加倍。

示例：

[[[[['p'], ['q'], ['t'], ['u']], [['r'], ['s'], ['v'], ['w']]]], 
[[[['p'], ['q'], ['t'], ['w']], [['r'], ['s'], ['v'], ['u']]]], 
[[[['p'], ['q'], ['v'], ['u']], [['r'], ['s'], ['t'], ['w']]]], 
[[[['p'], ['q'], ['v'], ['w']], [['r'], ['s'], ['t'], ['u']]]], 
[[[['p'], ['s'], ['t'], ['u']], [['r'], ['q'], ['v'], ['w']]]], 
[[[['p'], ['s'], ['t'], ['w']], [['r'], ['q'], ['v'], ['u']]]], 
[[[['p'], ['s'], ['v'], ['u']], [['r'], ['q'], ['t'], ['w']]]], 
[[[['p'], ['s'], ['v'], ['w']], [['r'], ['q'], ['t'], ['u']]]], 
[[[['r'], ['q'], ['t'], ['u']], [['p'], ['s'], ['v'], ['w']]]], 
[[[['r'], ['q'], ['t'], ['w']], [['p'], ['s'], ['v'], ['u']]]], 
[[[['r'], ['q'], ['v'], ['u']], [['p'], ['s'], ['t'], ['w']]]], 
[[[['r'], ['q'], ['v'], ['w']], [['p'], ['s'], ['t'], ['u']]]], 
[[[['r'], ['s'], ['t'], ['u']], [['p'], ['q'], ['v'], ['w']]]], 
[[[['r'], ['s'], ['t'], ['w']], [['p'], ['q'], ['v'], ['u']]]], 
[[[['r'], ['s'], ['v'], ['u']], [['p'], ['q'], ['t'], ['w']]]], 
[[[['r'], ['s'], ['v'], ['w']], [['p'], ['q'], ['t'], ['u']]]]]

与

相同

[[[['p'], ['q'], ['t'], ['u']], [['r'], ['s'], ['v'], ['w']]]]

如何使用列表理解功能删除双精度元素？我试图将列表转换为集合，但这不起作用。

所需的输出应为：

[[[['r'], ['s'], ['v'], ['w']], [['p'], ['q'], ['t'], ['u']]]]

谢谢

Answer 1

您可以使用每个子列表中的拼合值创建字典：

d = [[[[['p'], ['q'], ['t'], ['u']], [['r'], ['s'], ['v'], ['w']]]], [[[['p'], ['q'], ['t'], ['w']], [['r'], ['s'], ['v'], ['u']]]], [[[['p'], ['q'], ['v'], ['u']], [['r'], ['s'], ['t'], ['w']]]], [[[['p'], ['q'], ['v'], ['w']], [['r'], ['s'], ['t'], ['u']]]], [[[['p'], ['s'], ['t'], ['u']], [['r'], ['q'], ['v'], ['w']]]], [[[['p'], ['s'], ['t'], ['w']], [['r'], ['q'], ['v'], ['u']]]], [[[['p'], ['s'], ['v'], ['u']], [['r'], ['q'], ['t'], ['w']]]], [[[['p'], ['s'], ['v'], ['w']], [['r'], ['q'], ['t'], ['u']]]], [[[['r'], ['q'], ['t'], ['u']], [['p'], ['s'], ['v'], ['w']]]], [[[['r'], ['q'], ['t'], ['w']], [['p'], ['s'], ['v'], ['u']]]], [[[['r'], ['q'], ['v'], ['u']], [['p'], ['s'], ['t'], ['w']]]], [[[['r'], ['q'], ['v'], ['w']], [['p'], ['s'], ['t'], ['u']]]], [[[['r'], ['s'], ['t'], ['u']], [['p'], ['q'], ['v'], ['w']]]], [[[['r'], ['s'], ['t'], ['w']], [['p'], ['q'], ['v'], ['u']]]], [[[['r'], ['s'], ['v'], ['u']], [['p'], ['q'], ['t'], ['w']]]], [[[['r'], ['s'], ['v'], ['w']], [['p'], ['q'], ['t'], ['u']]]]]
def flatten(_d):
   return [[c for [c] in _d[0][0]], [c for [c] in _d[0][1]]]

def match(a, b):
  return all(c == d for c, d in zip(sorted(a), sorted(b)))

_d = {i:flatten(a) for i, a in enumerate(d)}
result = [a for i, a in enumerate(d) if not any(match(_d[i], _d[c]) for c in range(i))]

输出：

[[[[['p'], ['q'], ['t'], ['u']], [['r'], ['s'], ['v'], ['w']]]], 
 [[[['p'], ['q'], ['t'], ['w']], [['r'], ['s'], ['v'], ['u']]]], 
 [[[['p'], ['q'], ['v'], ['u']], [['r'], ['s'], ['t'], ['w']]]], 
 [[[['p'], ['q'], ['v'], ['w']], [['r'], ['s'], ['t'], ['u']]]], 
 [[[['p'], ['s'], ['t'], ['u']], [['r'], ['q'], ['v'], ['w']]]], 
 [[[['p'], ['s'], ['t'], ['w']], [['r'], ['q'], ['v'], ['u']]]], 
 [[[['p'], ['s'], ['v'], ['u']], [['r'], ['q'], ['t'], ['w']]]], 
 [[[['p'], ['s'], ['v'], ['w']], [['r'], ['q'], ['t'], ['u']]]]]

删除嵌套列表中的双重元素

1 个答案: