Replace values in list using Python [duplicate]

Build a new list with a list comprehension:

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

You can modify the original list in-place if you want, but it doesn't actually save time:

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

Here are (Python 3.6.3) timings demonstrating the non-timesave:

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)

And Python 2.7.6 timings:

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 loopIn [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

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

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]