Appending an element in nested list comprehension python

Question

My (derailed) mind would like to do the following:

list1 = [1,2,3]
list2 = ['a','b','c']
list3 = [list([a for a in list2]).append(n) for n in list1]

to output this:

[['a', 'b', 'c', '1'], ['a', 'b', 'c', '2'], ['a', 'b', 'c', '3']]

only using single line list comprehensions (yes I'm in blinded by Haskell love).

Instead it outputs a list of 3 None type items which is understandable as I'm getting the output of append 3 times.

I think there's a key python idea I'm missing here on how I could make this work (or me being completely illogical), any help would be appreciated :)

Michael Bianconi · Accepted Answer · 2022-01-09 21:28:17Z

4

No need to over-complicate things.

>>> list1 = [1,2,3]
>>> list2 = ['a','b','c']
>>> [list2 + [x] for x in list1]
[['a', 'b', 'c', 1], ['a', 'b', 'c', 2], ['a', 'b', 'c', 3]]

answered Jan 9, 2022 at 21:28

Michael Bianconi

5,2522 gold badges12 silver badges26 bronze badges

Sign up to request clarification or add additional context in comments.

12 Comments

Also can do: [[*list2, x] for x in list1]

If you did that, the result would be [['a','b','c',1],['a','b','c',1,2],['a','b','c',1,2,3]]

@Aris list3 = [(a := list([a for a in list2])).append(n) or a for n in list1] works (but it's bad)

@Aris Ha, just realized you even make a copy of a copy. Making double sure, eh?

@Aris Only you know why. You first copy with a list comprehension, then with the list call.

|