How to flatten a dictionary into a tuple of two lists?

Question

Now I have a dictionary,

{2.9: [0.66], 3.3: [0.82, 0.48]}

and I want to flatten it to a tuple consisting of two lists:

([3.3, 3.3, 2.9], [0.82, 0.48, 0.66])

3.3 appears twice because there were two elements in the corresponding dictionary entry.

How do I do it?

There are two elements in the list corresbonding to 3.3, so each of the element has to match with 3.3 in the flattened tuple. — Febday, Mar 10 '16 at 22:27
He is expanding it. So think: `{2.9: 0.66, 3.3: 0.82, 3.3:0.48}` (if that were possible) — Goodies, Mar 10 '16 at 22:29

John La Rooy · Answer 1 · 2016-03-10T23:11:19.287

If tuple of tuples is ok, it's slightly simpler

>>> D = {2.9: [0.66], 3.3: [0.82, 0.48]}
>>> tuple(zip(*((k, v) for k in D for v in D[k])))
((3.3, 3.3, 2.9), (0.82, 0.48, 0.66))

If you really need lists

>>> tuple(map(list, zip(*((k, v) for k in D for v in D[k]))))
([3.3, 3.3, 2.9], [0.82, 0.48, 0.66])

My variation of @Steven's and @Padraic's answers. (I avoid using dict.items because it's behaviour is so different between Python2 and Python3)

keys, vals = result = [], []
for k in D:
    v = D[k]
    vals += v
    keys += [k] * len(v)
print(result)

score 2 · Answer 2 · edited May 23 '17 at 12:09

2

>>> d = {2.9: [0.66], 3.3: [0.82, 0.48]}

First let's distribute the two keys of the dictionary into a list of tuples:

>>> ts = [(k, x) for k, l in d.items() for x in l]
>>> ts
[(3.3, 0.82), (3.3, 0.48), (2.9, 0.66)]

I'm using a two-level list comprehension, which is syntactic sugar for the equivalent nested loop.

>>> result = zip(*ts)
>>> list(result)
[(3.3, 3.3, 2.9), (0.82, 0.48, 0.66)]

Or as a single expression, using a lazy generator expression instead of a list comprehension:

result = zip(*((k, x) for k, l in d.items() for x in l))

edited May 23 '17 at 12:09

Community

answered Mar 10 '16 at 22:30

Benjamin Hodgson

Minor nit: OP wants a tuple of lists, not a list of tuples. – Steven Rumbalski Mar 10 '16 at 22:36
I'm of the opinion that it's likely OP doesn't mind whether it's a tuple or a list :) – Benjamin Hodgson Mar 10 '16 at 22:38

score 2 · Answer 3 · answered Mar 10 '16 at 22:57

A variation of Stevens answer using extend as opposed to append as the dict keys are immutable so it is safe to just multiply.

d = {2.9: [0.66], 3.3: [0.82, 0.48]}

keys, vals = result = [], []
for k, v in d.items():
    vals.extend(v)
    keys.extend([k] * len(v))

score 1 · Answer 4 · answered Mar 10 '16 at 22:44

1

The best answer is John La Rooy's one-liner using map, zip, and a generator expression. Here is a simplistic alternative:

d = {2.9: [0.66], 3.3: [0.82, 0.48]}

keys, vals = result = [], []
for k in d:
    for v in d[k]:
        keys.append(k)
        vals.append(v)

answered Mar 10 '16 at 22:44

Steven Rumbalski

I like this one better than mine - and it's probably faster too – John La Rooy Mar 10 '16 at 22:52

4 Answers4