Question

我想过滤掉列表中仅包含None-elements的所有元组，所以这个列表

listobj = [(None, None, None, None), (None, None, None, None), (None, None, None, None),(None,None,'01/02/2015','25'),(None,None,'01/02/2015',None),(0,None,None,None)]

我想要那样的输出：

  listobj = [(None,None,'01/02/2015','25'),(None,None,'01/02/2015',None),(0,None,None,None)]

Answer 1

>>> [x for x in listobj if any(y is not None for y in x)]
[(None, None, '01/02/2015', '25'), (None, None, '01/02/2015', None), (0, None, None, None)]

Answer 2

listobj= [i for i in listobj if i.count(None)<len(i)]

Answer 3

不确定它是否必须是一个很酷的单线，但这是一个非常基本的方法：

def noneSeq(seq):
  """Helper function that determines if seq is all None."""
  for x in seq:
    if x is not None: return False
  return True

a = [x for x in listobj if not noneSeq(x)]

这给出了：

[(None, None, '01/02/2015', '25'), (None, None, '01/02/2015', None),
 (0, None, None, None)]

如何跳过没有列表中的元组？

3 个答案: