create a dictionary from two lists based on correct indexing

Asked Oct 23 '22 at 23:03

Active Oct 23 '22 at 23:13

Viewed 29 times

I have two arrays, lets say L1 and L2 takes the following form.

L1 = array([2, 3, 5, 2, 5, 2, 3])
L2 = array([1.2, 3.3, 5.7, 1.6, 5.5, 1.5, 3.4])

Now I want to create a dictionary in the following form;

dict_ = {2: [1.2, 1.6, 1.5], 3: [3.3, 3.4], 5: [5.7, 5.5]}

This is what I did;

dict_ = {}

for ts, te in zip(L1, L2):
  if ts not in dict_.keys():
    dict_[ts] = ts
    dict_.update({ts: te})  

  if ts in dict_.keys():
    dict_[ts] = [dict_[ts], te]

The format I received is not very elegant.. and not as I expected.

edited Oct 23 '22 at 23:13

Michael M.

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asked Oct 23 '22 at 23:03

Mass17

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1

`if ts not in dict_: dict_[ts] = []; dict_[ts].append(te)` – deceze Oct 23 '22 at 23:06
1

Or: `dict_.setdefault(ts, []).append(te)` – deceze Oct 23 '22 at 23:07
Are you using numpy? – Michael M. Oct 23 '22 at 23:08
may I ask what you mean by "correct indexing"? and what is the relation between L1, L2 and your desired dictionary? – Dariush Mazlumi Oct 23 '22 at 23:09
yes I use numpy – Mass17 Oct 23 '22 at 23:09
1

Alternatively, you could try [`defaultdict`](https://docs.python.org/3/library/collections.html#collections.defaultdict), which uses a default value type. That way, you can append things to the list without needing to initialize the key – TimH Oct 23 '22 at 23:10
2

@learner the article you referenced explains why you **shouldn't** use your proposal – bn_ln Oct 23 '22 at 23:49
@bn_In Thanks for correcting me. Here is the correct snippet that I tested successfully: x={k:[] for k in set(L1)} for ts, te in zip(L1, L2): if not te in x[ts]: x[ts].append(te) – learner Oct 23 '22 at 23:56

create a dictionary from two lists based on correct indexing

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