Replace values in list using Python
list, performance, python, replace
Solution
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 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
Problem
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?