How to dynamically add elements at index to Python list?

Question

Given some Python list,

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g']

for some of its characters contained in another list, e.g.

list2 = ['a', 'd', 'e']

I need to add a character {char}1 to the index right before the original character in list1, such that the updated list1 looks like this:

list1 = ['a1', 'a', 'b', 'c', 'd1', 'd', 'e1', 'e', 'f', 'g']

My initial idea was to store the indices of original elements in a dictionary, and then insert new ones to [i-1] in list1,
in range (1, len(list1)+1). However, I then realised that this would only work for the first element, because the list would then grow and the indices would shift, so I would get wrong results.

Is there an efficient way to do it using insert?

What happens if the item isn't in list1? What about if {char}1 is already in list1? — Axe319
– Axe319, Commented Dec 28, 2022 at 18:19
@JJKam because time complexity will get the better of you if you don't have something efficient. You can probably live with a simple solution for small lists, but it'll blow up on larger ones — roganjosh
– roganjosh, Commented Dec 28, 2022 at 18:20
I don't think using insert is going to be good, inserting is going to make the solution O(n*m) in contrast creating the list from scratch is O(n + m) using a set — Dani Mesejo
– Dani Mesejo, Commented Dec 28, 2022 at 18:27

Dani Mesejo · Accepted Answer · 2022-12-28 18:58:26Z

One approach:

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
list2 = ['a', 'd', 'e']

result = []
for e in list1:
    if e in list2:
        result.extend([f"{e}1", e])
    else:
        result.append(e)
print(result)

Output

['a1', 'a', 'b', 'c', 'd1', 'd', 'e1', 'e', 'f', 'g']

If list2 is large consider transforming it to a set. Like below:

result = []
set2 = set(list2)
for char in list1:
    if char in set2:
        result.extend([f"{char}1", char])
    else:
        result.append(char)

The second solution is O(n + m) expected, where n and m are the size of list1 and list2.

append is O(1) (this means the cost of the operation is constant), extend is a repeated append (and in the context of the question is also constant). From the documentation:

Extend the list by appending all the items from the iterable. Equivalent to a[len(a):] = iterable.

Note: Using insert is going to make the approach O(n * m), since inserting in a list has a linear complexity (see here). The linear complexity comes from the fact that the list has to shift the elements.

Plagon · Accepted Answer · 2022-12-28 18:23:01Z

0

An alternative approach with insert:

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
list2 = ['a', 'd', 'e'] 


for e in list2:
    list1.insert(list1.index(e), f'{e}1')

list1
>>> ['a1', 'a', 'b', 'c', 'd1', 'd', 'e1', 'e', 'f', 'g']

answered Dec 28, 2022 at 18:23

Plagon

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Comments

Arifa Chan · Accepted Answer · 2022-12-28 20:05:40Z

0

Alternatives with itertools:

from itertools import chain

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
list2 = ['a', 'd', 'e']

list1 = [*chain.from_iterable((i,[i+'1',i])[i in list2] for i in list1)]

# ['a1', 'a', 'b', 'c', 'd1', 'd', 'e1', 'e', 'f', 'g']

Or join and split:

list1 = ['a', 'b', 'c', 'd', 'e', 'f', 'g']
list2 = ['a', 'd', 'e']

list1 = ' '.join([(i,' '.join([i+'1',i]))[i in list2] for i in list1]).split()

# ['a1', 'a', 'b', 'c', 'd1', 'd', 'e1', 'e', 'f', 'g']

answered Dec 28, 2022 at 20:05

Arifa Chan

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