Python - Flatten a dict of lists into unique values?
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())))]
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.
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]
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})