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
ls = [x if (condition) else None for x in ls]
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]
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]
>>> 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]
In case you want to replace values in place, you can update your original list with values from a list comprehension by assigning to the whole slice of the original.
data = [*range(11)] # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
id_before = id(data)
data[:] = [x if x % 2 else None for x in data]
data
# Out: [None, 1, None, 3, None, 5, None, 7, None, 9, None]
id_before == id(data) # check if list is still the same
# Out: True
If you have multiple names pointing to the original list,
for example you wrote data2=data
before changing the list
and you skip the slice notation for assigning to data
,
data
will rebind to point to the newly created list while data2
still points to the original unchanged list.
data = [*range(11)] # [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
data2 = data
id_before = id(data)
data = [x if x % 2 else None for x in data] # no [:] here
data
# Out: [None, 1, None, 3, None, 5, None, 7, None, 9, None]
id_before == id(data) # check if list is still the same
# Out: False
data2
# Out: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Note: This is no recommendation for generally preferring one over the other (changing list in place or not), but behavior you should be aware of.
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