Reputation: 738
I am looking for the best way to refactor the Python code below. I think there is a Pythonic way to do this in 2 or 3 lines of code, but couldn't figure it out. I have searched Stackoverflow but couldn't find similar problems and solutions. Many thanks!
list1 = [(Python, 5), (Ruby, 10), (Java, 15), (C++, 20)]
list2 = [(Python, 1), (Ruby, 2), (Java, 3), (PHP, 4), (Javascript, 5)]
# I want to make an unsorted list3 like this
# list3 = [(Python, 6), (Ruby, 12), (Java, 18), (PHP, 4), (Javasript, 5), (C++, 20)]
common_keys = list(set(dict(list1).keys()) & set(dict(list2).keys()))
if common_keys:
common_lst = [(x, (dict(list1)[x] + dict(list2)[x])) for x in common_keys]
rest_list1 = [(x, dict(list1)[x]) for x in dict(list1).keys() if x not in common_keys]
rest_list2 = [(x, dict(list2)[x]) for x in dict(list2).keys() if x not in common_keys]
list3 = common_lst + rest_list1 + rest_list2
else:
list3 = list1 + list2
Upvotes: 1
Views: 1278
Reputation: 176910
You're looking for collections.defaultdict
:
from collections import defaultdict
from itertools import chain
merged = defaultdict(int)
for key, value in chain(list1, list2):
merged[key] += value
If you want a list
of tuple
s:
list3 = merged.items()
If you want to do it without chain
, you can do it as:
from collections import defaultdict
merged = defaultdict(int)
merged.update(list1)
for key, value in list2:
merged[key] += value
Edit: As Beni points out in a comment, on 2.7/3.2+, you can do:
from collections import Counter
merged = Counter(dict(list1))
merged.update(dict(list2))
Which requires you convert the lists to dict
s but is otherwise perfect.
Upvotes: 5