Objects between objects python

Question

I want to make from the list:

L=[1,2,3,4,5,6,7,8,9]

This:

L=[1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9]

This is, put the 1 between the objects in list. Can someone help me?

• Martijn Pieters: Isn't there a 1 too many, at the start? That is to say; you prepended a 1 and put 1s between the original values.

Answer 1

For the result in your example (1 before each object):

L = [y for x in L for y in (1, x)]

For the result described by your text (1 between the objects):

L = [y for x in L for y in (x, 1)]
L.pop()

If you hate multiple for clauses in comprehensions:

L = list(itertools.chain.from_iterable((1, x) for x in L))

• Steve Jessop: @DavidUnric: well, I wouldn't promise that it wins a performance shootout, but it might well be THE first version :-)
• mgilson: It's very nice usage of nested list-comps. If I didn't really dislike nested list comprehensions, I would really like this answer. +1 :)
• Steve Jessop: @mgilson: would it help at all to write it across several lines, so each for clause is on its own line? Most comprehensions with multiple for clauses that I use are still way simpler than the average SQL statement with an equal number of joins. They can be easier or harder to read according to how they're presented :-)
• mgilson: @SteveJessop -- maybe. For some reason, these have always just been mind-benders for me. The itertools solution is nice too. Both ways, it's a great answer. And yeah, definitely simpler than an average SQL statement. A lot simpler than the average regex too ... But I hope our bar for readability is a lot higher than the average readability in those two languages ;-)

Answer 2

print ([i for t in zip([1] * len(L), L) for i in t])

Output

[1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9]

• Martijn Pieters: Yes, I discovered this too; see my comment on the question

Answer 3

I've always really loved slice assignment:

>>> L = range(1, 10)
>>> L
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> # Done with setup, let the fun commence!
>>> LL = [None] * (len(L)*2) # Make space for the output
>>> LL[1::2] = L
>>> LL[::2] = [1] * len(L)
>>> LL # Lets check the results, shall we?
[1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9]

Answer 4

A simple for loop would suffice

for item in list:
 newList.append(1)
 newList.append(item)
list = newList

• Martijn Pieters: Just newList.pop() removes the last element. Swap the appends() and lose the .pop() however, to match the OP sample output.
• Dylan Lawrence: @MartijnPieters Ah, I was reading the sample output opposite it seems.

Answer 5

Using itertools:

>>> L=[1,2,3,4,5,6,7,8,9]
>>> from itertools import chain, izip, repeat
>>> list(chain.from_iterable(izip(repeat(1), L)))
[1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9]