Memory efficient padding a list

Question

I have a list

a = ['a', 'b', 'c']

of given length and I want to insert a certain element 'x' after every item to get

ax = ['a', 'x', 'b', 'x', 'c', 'x']

Since the elements are of large size, I don't want to do a lot of pops or sublists.

Any ideas?

Answer 1

Since the list is large, the best way is to go with a generator, like this

def interleave(my_list, filler):
    for item in my_list:
        yield item
        yield filler

print list(interleave(['a', 'b', 'c'], 'x'))
# ['a', 'x', 'b', 'x', 'c', 'x']

Or you can return a chained iterator like this

from itertools import chain, izip, repeat
def interleave(my_list, filler):
    return chain.from_iterable(izip(my_list, repeat(filler)))

1. repeat(filler) returns an iterator which gives filler infinite number of times.

2. izip(my_list, repeat(filler)) returns an iterator, which picks one value at a time from both my_list and repeat(filler). So, the output of list(izip(my_list, repeat(filler))) would look like this

   [('a', 'x'), ('b', 'x'), ('c', 'x')]
   

3. Now, all we have to do is flatten the data. So, we chain the result of izip, with chain.from_iterable, which gives one value at a time from the iterables.

Answer 2

Have you considered itertools izip?

izip('ABCD', 'xy') --> Ax By

izip_longest can be used with a zero length list, a fillvalue, and combined via chain.from_iterable as follows:

import itertools
list(itertools.chain.from_iterable(itertools.izip_longest('ABCD', '', fillvalue='x'))
>>> ['A', 'x', 'B', 'x', 'C', 'x', 'D', 'x']