Python - Flatten a dict of lists into unique values?

Question

I have a dict of lists in python:

content = {88962: [80, 130], 87484: [64], 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: [64], 53662: [58,80]}
result = set({})
for k in content.keys() :
    for i in content[k]:
        result.add(i)
# and list/sort/print just to compare the output
r2 = list( result )
r2.sort()
print r2

Within 3 minutes there are 4 great answers, which all prove i need to read more of the standard library docs. — Jonathan Vanasco, Oct 22 '12 at 17:03
*Chuckles* That feeling doesn't really ever go away - there are a *lot* of batteries included in the standard lib. — Sean Vieira, Oct 22 '12 at 17:06

nneonneo · Accepted Answer · 2022-08-25T18:15:10.127

59

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})

edited Aug 25 '22 at 18:15

answered Oct 22 '12 at 17:03

nneonneo

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A link to the set comprehension documentation would make this perfect. – Sam Mussmann Oct 22 '12 at 17:12
this is really concise and nice to look at. You get the best answer out of sea of really excellent answers. – Jonathan Vanasco Nov 05 '12 at 17:12
1

Note: This won't work in Python3, itervalues has been removed. – Tom Myddeltyn Dec 01 '16 at 13:55
2

@TomMyddeltyn substitute `itervalues` for `values` in python 3 (just like you'd substitute `iteritems` for `items` with dictionaries) since Python 3 favors lazy iteration – Marc Nov 14 '17 at 19:56
The answer by @Jon Clements is better at leveraging ptyhon standard libs. – Elmer Jul 03 '19 at 10:35

score 17 · Answer 2 · edited Jul 03 '19 at 11:37

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())))]

Ashwini Chaudhary · Answer 3 · 2012-10-22T17:06:24.987

7

use set() and itertools.chain():

In [83]: content = {88962: [80, 130], 87484: [64], 53662: [58,80]}

In [84]: from itertools import chain

In [94]: x=set(chain(*content.values()))

In [95]: x
Out[95]: set([58, 64, 80, 130]) # a set, the items may or may not be sorted

In [96]: sorted(x)         #convert set to a sorted list
Out[96]: [58, 64, 80, 130]

edited Oct 22 '12 at 17:06

answered Oct 22 '12 at 17:00

Ashwini Chaudhary

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Cool - just a couple of bits though - the `list` is redundant in `sorted(list(x))` and `*content.values()` is spelt `chain.from_iterable`... – Jon Clements Oct 22 '12 at 17:05
I prefer `chain()` as it takes less amount of characters, thanks for the `sorted(list(x))` part. :) – Ashwini Chaudhary Oct 22 '12 at 17:09
Isn't chain.from_iterable more efficient though? Using just chain means you are unpacking the sequence. Whereas from_iterable does it in the generator. – jdi Oct 22 '12 at 17:13

score 4 · Answer 4 · answered Oct 22 '12 at 17:00

4

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.

answered Oct 22 '12 at 17:00

Sean Vieira

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Note that this won't work in python3. Also, I think this answer explains the list comprehension a little better than the accepted answer. I like how you formatted it. – Tom Myddeltyn Dec 01 '16 at 13:58

Sam Mussmann · Answer 5 · 2012-10-22T19:22:47.527

4

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()))

edited Oct 22 '12 at 19:22

answered Oct 22 '12 at 17:01

Sam Mussmann

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This will create a bunch of temporary sets as it loops. – jdi Oct 22 '12 at 17:03
Thanks! I was editing to add some explanation and realized and (I think) fixed that. – Sam Mussmann Oct 22 '12 at 17:05
To my eye, `a.union(set(b)` is more explicit and more readable. `a | set (b)` certainly works as well. – Sam Mussmann Oct 22 '12 at 17:13

score 3 · Answer 6 · answered Oct 22 '12 at 17:09

3

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

answered Oct 22 '12 at 17:09

georg

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score 2 · Answer 7 · answered Aug 02 '18 at 16:09

2

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])))

answered Aug 02 '18 at 16:09

mikeviescas

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This should be the most elegant solution leveraging list/dictionary comprehension. – Cescante Aug 03 '20 at 22:57

Python - Flatten a dict of lists into unique values?

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