I have a dict of lists in python:

``````content = {88962: [80, 130], 87484: , 53662: [58,80]}
``````

I want to turn it into a list of the unique values

``````[58,64,80,130]
``````

I wrote a manual solution, but it’s a manual solution. I know there are more concise and more elegant way to do this with list comprehensions, map/reduce , itertools , etc. anyone have a clue ?

``````content = {88962: [80, 130], 87484: , 53662: [58,80]}
result = set({})
for k in content.keys() :
for i in content[k]:
# and list/sort/print just to compare the output
r2 = list( result )
r2.sort()
print r2
``````

Double set comprehension:

Python 3:

``````sorted({x for v in content.values() for x in v})
``````

Python 2:

``````sorted({x for v in content.itervalues() for x in v})
``````

In python3.7 you can use a combination of `.values`, and `chain`.

``````from itertools import chain
sorted(set(chain(*content.values())))
# [58, 64, 80, 130]

# another option is `itertools.groupby`
from itertools import groupby
[k for k, g in groupby(sorted(chain(*content.values())))]
``````

In python2.7

``````from itertools import chain
sorted(set(chain.from_iterable(content.itervalues())))
# [58, 64, 80, 130]

# another option is `itertools.groupby`
[k for k, g in groupby(sorted(chain.from_iterable(content.itervalues())))]
``````

use `set()` and `itertools.chain()`:

``````In : content = {88962: [80, 130], 87484: , 53662: [58,80]}

In : from itertools import chain

In : x=set(chain(*content.values()))

In : x
Out: set([58, 64, 80, 130]) # a set, the items may or may not be sorted

In : sorted(x)         #convert set to a sorted list
Out: [58, 64, 80, 130]
``````

``````sorted(set(val
for row in content.itervalues()
for val in row))
``````

`set` gets us all the distinct values (like a dictionary, but without the overhead of storing values). `sorted` then just takes the created `set` and returns a `list` sorted in ascending order.

``````list(reduce(lambda a, b: a.union(set(b)), content.itervalues(), set()))
``````

The `lambda` turns the two input arguments into sets and unions them.

The `reduce` will do a left fold over the list that is passed to it — in this case, the lists that are the values of your dictionaries.

The `reduce` will turn the result of this, which is a `set` back into a list.

This can also be spelled:

``````list(reduce(lambda a, b: a | set(b), content.itervalues(), set()))
``````

``````sorted(set(sum(content.values(), [])))
``````

Use list comprehension to generate a non-unique list, convert it to a set to get the unique values, and then back into a sorted list. Perhaps not the most efficient, but yet another one line solution (this time with no imports).

Python 3:

``````sorted(list(set([val for vals in content.values() for val in vals])))
``````

Python 2.7:

``````sorted(list(set([val for vals in content.itervalues() for val in vals])))
``````