Replacing every other position in a list

user_18004567 0
Viewed 124

I'm trying to make something that allows me to replace every other position in a list with a single item:

l = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
l[::2] = "A"
print(l)

I'm expecting something like:

["A", 1, "A," 3, "A", 5, "A", 7, "A", 9, "A"]

I've tried different positions of indexing [::] but either get an error or a result that doesn't include the rest of the list.

Instead I get this:

ValueError: attempt to assign sequence of size 1 to extended slice of size 2
indexing list python
3 Answers

The slice is correct, but you need to provide a sequence with enough elements to fill all the elements.

l[::2] = ["A"] * math.ceil(len(l)/2)
user_14918950

Edit: Barmar's answer is much better, so long as you're okay with importing the math library.

Rough speed comparison at 10,000 loops (assuming you don't turn the list into a generator, otherwise the list comprehension becomes slightly faster overall):

import math, time

def slicing(n):
    for _ in range(n):
        l = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
        l[::2] = ["A"] * math.ceil(len(l)/2)

def comprehension(n):
    for _ in range(n):
        l = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
        l = ["A" if i % 2 == 0 else elem for i, elem in enumerate(l)]

start = time.time()
slicing(10000)
print(f'Using slicing and math.ceil: {time.time() - start} seconds')

start = time.time()
comprehension(10000)
print(f'Using list comprehension:    {time.time() - start} seconds')

>>> Using slicing and math.ceil: 0.004003286361694336 seconds
>>> Using list comprehension:    0.012019157409667969 seconds

Edit #2: For anyone wondering, Alain's answer comes in at around 0.0075 seconds per 10,000 loops on average, so basically right in the middle.

Old answer: Try using a list comprehension paired with enumerate:

l = ["A" if i % 2 == 0 else elem for i, elem in enumerate(l)]

This replaces every element with an odd index (odd numbers % 2 will equal 0) with the letter "A", and leaves the remaining elements as they are. Enumerate loops over the original iterable, but includes an index element as well.

user_130109560

You must provide as many elements as there are items to assign:

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

l[::2] = ("A" for _ in l[::2])

print(l)
['A', 1, 'A', 3, 'A', 5, 'A', 7, 'A', 9, 'A']

Or this way:

l[::2] = ["A"]*len(range(0,len(l),2))
user_52375600
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