使用Python替换列表中的值

Question

我有一个列表,我想用None替换值,其中condition()返回True.

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

例如,如果条件检查bool(项目%2)应该返回:

[None, 1, None, 3, None, 5, None, 7, None, 9, None]

最有效的方法是什么？

Answer 1

使用列表解析构建新列表:

new_items = [x if x % 2 else None for x in items]

如果需要,您可以就地修改原始列表,但实际上并不节省时间:

items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
for index, item in enumerate(items):
    if not (item % 2):
        items[index] = None

以下是(Python 3.6.3)演示非时间的时间:

In [1]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: for index, item in enumerate(items):
   ...:     if not (item % 2):
   ...:         items[index] = None
   ...:
1.06 µs ± 33.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [2]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: new_items = [x if x % 2 else None for x in items]
   ...:
891 ns ± 13.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

和Python 2.7.6时序:

In [1]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: for index, item in enumerate(items):
   ...:     if not (item % 2):
   ...:         items[index] = None
   ...: 
1000000 loops, best of 3: 1.27 µs per loop
In [2]: %%timeit
   ...: items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
   ...: new_items = [x if x % 2 else None for x in items]
   ...: 
1000000 loops, best of 3: 1.14 µs per loop

Answer 2

ls = [x if (condition) else None for x in ls]

Answer 3

这是另一种方式:

>>> L = range (11)
>>> map(lambda x: x if x%2 else None, L)
[None, 1, None, 3, None, 5, None, 7, None, 9, None]

Answer 4

关于OP在评论中提出的问题,即:

如果我有一个生成器从范围(11)而不是列表中生成值,该怎么办？是否可以替换发电机中的值？

当然,这很简单......:

def replaceiniter(it, predicate, replacement=None):
  for item in it:
    if predicate(item): yield replacement
    else: yield item

只需传递任何可迭代(包括调用生成器的结果)作为第一个arg,谓词决定是否必须将值替换为第二个arg,然后让我们翻录.

例如:

>>> list(replaceiniter(xrange(11), lambda x: x%2))
[0, None, 2, None, 4, None, 6, None, 8, None, 10]