使用列表推导替换列表列表中的无值

Question

我想用list comprehension替换列表列表中的所有None值。我的代码似乎只是返回列表的原始列表。

Names = [['Jon',None,'Bill'],['Andrew','Robert',None]]
Removed_None = [0 if x is None else x for x in Names]
print Removed_None

期望的输出：

[['Jon',0,'Bill'],['Andrew','Robert',0]]

Answer 1

你需要嵌套你的列表理解：

[[v if v is not None else 0 for v in nested] for nested in Names]

演示：

>>> Names = [['Jon',None,'Bill'],['Andrew','Robert',None]]
>>> [[v if v is not None else 0 for v in nested] for nested in Names]
[['Jon', 0, 'Bill'], ['Andrew', 'Robert', 0]]

Answer 2

Removed_None = [[0 if item is None else item for item in each_list] for each_list in Names]

Answer 3

你需要使用嵌套的理解（我为了清晰起见，我写了一个作为函数）

[['Jon', None, 'Bill'], ['Andrew', 'Robert', None]]
>>> def replace(names):
...     return [0 if name is None else name for name in names]
... 
>>> print [replace(names) for names in Names]
[['Jon', 0, 'Bill'], ['Andrew', 'Robert', 0]]