d = {'a':[1,2,3], 'b':[4,5]}
and I need
[1,2,3,4,5]
using list comprehension. How do I do it?
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1Do you want the output in any particular order? (Python dictionaries are not intrinsically sorted.)DYZ– DYZ2017-01-18 06:04:23 +00:00Commented Jan 18, 2017 at 6:04
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Order is not importantmax_max_mir– max_max_mir2017-01-18 06:08:27 +00:00Commented Jan 18, 2017 at 6:08
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4 Answers
Use a nested list comprehension:
>>> [val for lst in d.values() for val in lst]
[1, 2, 3, 4, 5]
But you may need to sort the dictionary first (because dicts are unordered) to guarantee the order:
>>> [val for key in sorted(d) for val in d[key]]
[1, 2, 3, 4, 5]
3 Comments
Mohammad Yusuf
No. Sorting the list comp is better.
MSeifert
@MYGz That depends, I assumed he wanted to get all the values depending on the (sorted) order of the keys. Not sort all the values for all keys.
Mohammad Yusuf
Thats right. Sort by key or sort by values. And 1 more possibility sort by both :)
Easy and one liner solution using list comprehension:
>>> sorted([num for val in d.values() for num in val])
[1, 2, 3, 4, 5]
Comments
sum(d.values(),[]) works but not performant because it applies a = a + b for each temp list.
Use itertools.chain.from_iterable instead:
import itertools
print(list(itertools.chain.from_iterable(d.values())))
or sorted version:
print(sorted(itertools.chain.from_iterable(d.values())))
answered Jan 18, 2017 at 6:06
4 Comments
MSeifert
chain.from_iterable is probably better and faster than unpacking (*d.values()).Jean-François Fabre
why do I always forget that?
MSeifert
Probably because it's more to type :-)
Jean-François Fabre
:) lesson learned. One should be alarmed when there's too many symbols in an answer...
If the order of elements is not important, then you cannot beat this:
sum(d.values(), [])
(Add sorted() if necessary.)
