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
– Jonathan Vanasco, Commented Oct 22, 2012 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
– Sean Vieira, Commented Oct 22, 2012 at 17:06

nneonneo · Accepted Answer · 2022-08-25 18:15:10Z

66

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, 2022 at 18:15

answered Oct 22, 2012 at 17:03

nneonneo

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5 Comments

Sam Mussmann Over a year ago

A link to the set comprehension documentation would make this perfect.

Jonathan Vanasco Over a year ago

this is really concise and nice to look at. You get the best answer out of sea of really excellent answers.

Tom Myddeltyn Over a year ago

Note: This won't work in Python3, itervalues has been removed.

Marc Over a year ago

@TomMyddeltyn substitute itervalues for values in python 3 (just like you'd substitute iteritems for items with dictionaries) since Python 3 favors lazy iteration

Elmer Over a year ago

The answer by @Jon Clements is better at leveraging ptyhon standard libs.

Elmer · Accepted Answer · 2019-07-03 11:37:05Z

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())))]

Worth pointing out: *content.values() will immediately expand the entire iterable into an argument tuple. Use chain.from_iterable(content.values()) to preserve the laziness of values() in Python 3.

Ashwini Chaudhary · Accepted Answer · 2012-10-22 17:06:24Z

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, 2012 at 17:06

answered Oct 22, 2012 at 17:00

Ashwini Chaudhary

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3 Comments

Jon Clements Over a year ago

Cool - just a couple of bits though - the list is redundant in sorted(list(x)) and *content.values() is spelt chain.from_iterable...

Ashwini Chaudhary Over a year ago

I prefer chain() as it takes less amount of characters, thanks for the sorted(list(x)) part. :)

jdi Over a year ago

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.

Sean Vieira · Accepted Answer · 2012-10-22 17:00:38Z

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.

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.

Sam Mussmann · Accepted Answer · 2012-10-22 19:22:47Z

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()))

Thanks! I was editing to add some explanation and realized and (I think) fixed that.
To my eye, a.union(set(b) is more explicit and more readable. a | set (b) certainly works as well.

georg · Accepted Answer · 2012-10-22 17:09:26Z

3

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

answered Oct 22, 2012 at 17:09

georg

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Comments

mikeviescas · Accepted Answer · 2018-08-02 16:09:24Z

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])))

This should be the most elegant solution leveraging list/dictionary comprehension.

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