Python: one-line extend within for-loop

Question

Code:

print([some_data[name]['indices'] for name in some_data.keys()])

[[[0, 0], [1, 0], [2, 0], [2, 1]], [[3, 0], [3, 1], [1, 1], [0, 1]], ...]

[[0, 0], [1, 0], [2, 0], [2, 1], [3, 0], [3, 1], [1, 1], [0, 1], ...]

Trying this method told me that the 'list' object has no attribute 'result': One liner for extend loop python

Is there an alteration that I can apply to my code to obtain a one-line solution?

Thanks in advance

What would `some_data[name]['indices']` contain? list of atomic values? — Nobilis, Mar 07 '19 at 14:38
What is `result` supposed to be? You mention it in the error message but not in your code. — khelwood, Mar 07 '19 at 14:39
@Nobilis `some_data[name]['indices']` contains list of lists (length 2): `[[0, 0], [1, 0], [2, 0], [2, 1]]`, etc — Sterling Butters, Mar 07 '19 at 14:40
@mad_ you're right, I had just copied and pasted from where I am actually using it i.e. `if [b, a] not in [some_data[name]['indices'] for name in some_data.keys()]:` — Sterling Butters, Mar 07 '19 at 14:42
Try this: `[e for name in some_data.keys() for e in some_data[name]['indices']]` — Rocky Li, Mar 07 '19 at 14:45
@khelwood I don't have a `result`, since I'm not working with futures objects or whatever is being done in the other post — Sterling Butters, Mar 07 '19 at 14:47
This should already work as you have it, say you have `d = {'a': {'indices': [0, 0]}, 'b': {'indices': [0, 1]}}`, then calling `[d[i]['indices'] for i in d]` gives you `[[0, 0], [0, 1]]` which is what you want, right? — Nobilis, Mar 07 '19 at 14:49
@RockyLi That works, if you would like to post as an answer, I'd accept it — Sterling Butters, Mar 07 '19 at 15:01
Glad it worked for you. Thought it's more of a comment so it will stay that way. — Rocky Li, Mar 07 '19 at 15:03

score 1 · Answer 1 · answered Mar 07 '19 at 14:45

1

  topList = [[1,2],[3,4]]
  flatList = [item for subList in topList for item in subList]

answered Mar 07 '19 at 14:45

AndrewStone

score 0 · Answer 2 · answered Mar 07 '19 at 14:47

0

You are getting tge list correctly as single list. The print() adds the extra []. Print adds to objects an object type.

answered Mar 07 '19 at 14:47

user3559221

No, it doesn't. – Nobilis Mar 07 '19 at 14:47
You are right. Post it by mistake. Cant delete it. – user3559221 Mar 07 '19 at 14:49

Sterling Butters · Answer 3 · 2019-03-07T15:05:26.810

0

To whoever just deleted their answer using itertools.chain - that worked and I was going to accept your answer. The final code:

list(itertools.chain(*[some_data[name]['indices'] for name in some_data.keys()]))

EDIT: Decided to go without the itertools in @Rocky Li's answer:

[e for name in some_data.keys() for e in some_data[name]['indices']]

edited Mar 07 '19 at 15:05

answered Mar 07 '19 at 14:53

Sterling Butters

Nobilis · Answer 4 · 2019-03-07T15:01:45.773

0

Is this what you need? E.g. given:

some_data = {'a': {'indices': [[0, 0], [1, 1]]}, 'b': {'indices': [[2, 2], [3, 3]]}}

You want the sublists, right? So we can do:

[j for i in some_data for j in some_data[i]['indices']]

which returns:

[[0, 0], [1, 1], [2, 2], [3, 3]]

edited Mar 07 '19 at 15:01

answered Mar 07 '19 at 14:54

Nobilis

Actually `some_data` is in the form `{'Name1': {'indices': [[0, 0], [1, 1]]}, 'Name2': {'indices': [[2, 2], [3, 3]]}}` – Sterling Butters Mar 07 '19 at 14:59
should still be okay, does this not work as it is? – Nobilis Mar 07 '19 at 15:01
If it is the same in implementation as @Rocky Li's comment then yes. I just used his because it was already adapted for my use - and worked – Sterling Butters Mar 07 '19 at 15:03
ah, yes, I didn't see that, got beaten to it – Nobilis Mar 07 '19 at 15:04
1

Thank for your help anyway :) – Sterling Butters Mar 07 '19 at 15:05