Python: Replace values in list

Question

Hello Stack Overflow Community,

I have a list where I want to replace values with None where condition() returns True.

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

For example, if condition checks bool(item%2) should return:

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

What is the most efficient way to do this?

Answer 1

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

Answer 2

The most efficient:

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

The easiest to read:

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

Answer 3

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

Answer 4

Riffing on a side question asked by the OP in a comment, i.e.:

what if I had a generator that yields the values from range(11) instead of a list. Would it be possible to replace values in the generator?

Sure, it's trivially easy...:

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

Just pass any iterable (including the result of calling a generator) as the first arg, the predicate to decide if a value must be replaced as the second arg, and let 'er rip.

For example:

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

Answer 5

Here's another way:

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