CODEWITHSUNDEEP
pythonlist-comprehension
asm
asm

Reputation: 937

Extending lists using list comprehension in python

So let's say I have list items = ['abc', 'def', 'tre'].

Now I want to insert a flag before each of list items.

E.g the new list items2 should be ['-g', 'abc', '-g', 'def', '-g', 'tre'].

I could read the list and append each one after appended the flag but I want to start doing it the python way.

What I came up with is:

items2.extend(['-g', i] for i in items )

What this gives me a list with smaller lists:

items2 = [['-g', 'abc'], ['-g', 'def'], ['-g', 'tre']]

I understand that this is because it would be equivalent to saying 

items2.extend([ ... ] , [ ... ], [ ... ])

Any ideas about this? Thanks

Upvotes: 2

Views: 269

Answers (4)

user3885927
user3885927

Reputation: 3503

Another simple way is to loop through your list and insert after every element.

for i in range(len(items)):
   items.insert((i*2)+1,'-g')

As you insert your length up to the insertion point doubles, so you need to put your next element at (i*2)+1

Upvotes: 1

Sylvain Leroux
Sylvain Leroux

Reputation: 51990

As a complement to other excellent zip based answers, you might use itertools.repeat() instead of "counting" items:

>>> items = ['abc', 'def', 'tre']
>>> list(itertools.chain(*zip(itertools.repeat('-g'), items)))
#                             ^^^^^^^^^^^^^^^^^^^^^^
#                             repeat as much as needed
['-g', 'abc', '-g', 'def', '-g', 'tre']

And if you don't like zip, an alternative:

>>> list(itertools.chain(*[('-g', i) for i in items]))
['-g', 'abc', '-g', 'def', '-g', 'tre']

Upvotes: 2

miku
miku

Reputation: 188014

One approach would be using zip:

>>> items = ['abc', 'def', 'tre']
>>> zip(['-g', '-g', '-g'], items)
[('-g', 'abc'), ('-g', 'def'), ('-g', 'tre')]

After that, you can use itertools.chain to flatten the list:

>>> list(itertools.chain(*zip(['-g', '-g', '-g'], items)))
['-g', 'abc', '-g', 'def', '-g', 'tre']

Note that due to function call overhead, this is quite slow, compared to a simple list comprehension:

In [17]: %timeit [j for i in items for j in '-g', i]
1000000 loops, best of 3: 923 ns per loop

In [18]: %timeit list(itertools.chain(*zip(['-g', '-g', '-g'], items)))
100000 loops, best of 3: 2.67 µs per loop

So for performance and clarity, I suggest you use Robᵩ answer.

Upvotes: 3

Robᵩ
Robᵩ

Reputation: 168626

[j for i in items for j in '-g', i]

Upvotes: 6

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