筛选元组列表

问题描述

  oldList = [[（1，None），（2,45 ），（3,67）]，[（1，None），（2，None），（3，None），（4,56），（5,78）]，[（1，None）， ，98）]] 
  

我想筛选任何None的实例：

  newList = [[（2,45），（3,67）]，[（4,56），（5,78）] ，[（2，98）]] 
  

最接近的就是这个循环，它不会删除整个元组（只有'None'），它也会销毁元组结构列表：

  newList = [] 
用于oldList中的数据：
用于数据中的点：$ b​​ $ b newList.append（filter（None，point））
  pre 
 
解决方案
最简单的方法是使用嵌套的列表解析：
 
 
 >>> newList = [[for t in l if if None not in t] for oldList] 
>>> newList 
 [[（2，45），（3，67）]，[（4,56），（5,78）]，[（2,98）]] 
  
 
 $ b 
您需要嵌套两个列表解析，因为您正在处理列表列表。列表理解的外部部分 [[...] for oldList] 负责遍历包含的每个内部列表的外部列表。然后在内部列表理解中，你有 [t for t in l if None not in t] ，这是一个非常简单的说法，希望列表中的每个元组不包含 None 。
 
 
 
（可以说，您应该选择比 l更好的名称和 t ，但这取决于您的问题域。我选择了单字母名称来更好地突出显示代码结构。  
 
 
如果您对列表解析不熟悉或不舒服，则在逻辑上等同于以下内容： 
 
 
 >>> newList = [] 
>>> for l in oldList：
 ... temp = [] 
 ... for t in l：
 ... if if not in t：
 ... temp。 append（t）
 ... newList.append（temp）
 ... 
>>> newList 
 [[（2，45），（3，67）]，[（4,56），（5,78）]，[（2,98）]] 
  
 
I have a list of lists of tuples:
oldList = [[(1,None),(2,45),(3,67)],[(1,None), (2,None), (3,None),(4,56),(5,78)],[(1, None),(2, 98)]]
I would like to filter any instance of "None":
newList = [[(2,45),(3,67)], [(4,56),(5,78)], [(2, 98)]]
The closest I've come is with this loop, but it does not drop the entire tuple (only the 'None') and it also destroys the list of lists of tuples structure:
newList = []
for data in oldList:
    for point in data:
        newList.append(filter(None,point))
 解决方案 
The shortest way to do this is with a nested list comprehension:
>>> newList = [[t for t in l if None not in t] for l in oldList]
>>> newList
[[(2, 45), (3, 67)], [(4, 56), (5, 78)], [(2, 98)]]
You need to nest two list comprehensions because you are dealing with a list of lists. The outer part of the list comprehension [[...] for l in oldList] takes care of iterating through the outer list for each inner list contained. Then inside the inner list comprehension you have [t for t in l if None not in t], which is a pretty straightforward way of saying you want each tuple in the list which does not contain a None.
(Arguably, you should choose better names than l and t, but that would depend on your problem domain. I've chosen single-letter names to better highlight the structure of the code.)
If you are unfamiliar or uncomfortable with list comprehensions, this is logically equivalent to the following:
>>> newList = []
>>> for l in oldList:
...     temp = []
...     for t in l:
...         if None not in t:
...             temp.append(t)
...     newList.append(temp)
...
>>> newList
[[(2, 45), (3, 67)], [(4, 56), (5, 78)], [(2, 98)]]
                        
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