我正在寻找重构下面Python代码的最佳方法。我认为在2或3行代码中有一种Pythonic方法可以做到这一点,但无法弄明白。我搜索过Stackoverflow但找不到类似的问题和解决方案。非常感谢!
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
答案 0 :(得分:5)
您正在寻找collections.defaultdict
:
from collections import defaultdict
from itertools import chain
merged = defaultdict(int)
for key, value in chain(list1, list2):
merged[key] += value
如果您需要list
tuple
s:
list3 = merged.items()
如果你想在没有chain
的情况下这样做,你可以这样做:
from collections import defaultdict
merged = defaultdict(int)
merged.update(list1)
for key, value in list2:
merged[key] += value
编辑:正如Beni在评论中指出的那样,在2.7 / 3.2 +上,您可以这样做:
from collections import Counter
merged = Counter(dict(list1))
merged.update(dict(list2))
这要求您将列表转换为dict
,但其他方面则完美无缺。