I have a dictionary where the value elements are lists:
d1={'A': [], 'C': ['SUV'], 'B': []} I need to concatenate the values into a single list ,only if the list is non-empty.
Expected output:
o=['SUV'] Help is appreciated.
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4 Answers
from itertools import chain d1={'A': [], 'C': ['SUV'], 'B': []} print list(chain.from_iterable(d1.itervalues())) 
1 Comment
Sanjay T. Sharma
This answer is better than Ashwin's because it fully utilizes the fact that chain can deal with iterators and not just eagerly generated sequences.
>>> d1 = {'A': [], 'C': ['SUV'], 'B': []} >>> [ele for lst in d1.itervalues() for ele in lst] ['SUV'] 
1 Comment
Sanjay T. Sharma
+1; I like this solution though I would replace
values by itervalues.You can use itertools.chain, but the order can be arbitrary as dicts are unordered collection. So may have have to sort the dict based on keys or values to get the desired result.
>>> d1={'A': [], 'C': ['SUV'], 'B': []} >>> from itertools import chain >>> list(chain(*d1.values())) # or use d1.itervalues() as it returns an iterator(memory efficient) ['SUV'] 
1 Comment
jamylak
Please use
chain.from_iterable for these jobs>>> from operator import add >>> d1={'A': [], 'C': ['SUV'], 'B': []} >>> reduce(add,d1.itervalues()) ['SUV'] Or more comprehensive example:
>>> d2={'A': ["I","Don't","Drive"], 'C': ['SUV'], 'B': ["Boy"]} >>> reduce(add,d2.itervalues()) ['I', "Don't", 'Drive', 'SUV', 'Boy'] 5 Comments
jamylak
unfortunately this has O(N^2) behavior since you are creating a new list on each
add
HennyH
@jamylak True, this is a horribly inefficient approach (just had to put a functional approach out there)
oleg
what about
sum(d2.itervalues(), [])HennyH
f = [], map(f.extend,d2.itervalues()) is also another poor method of solving itjamylak
@HennyH the other approaches are still functional, but I see what you mean