输入:
[['a', 2], ['b',1]] (sorted by value)
[['b', 2], ['c', 1]]
输出:
[['b', 3], ['a', 2], ['c', 1]]
任何pythonic方式?当然,在Python中! (最好是2.6倍) 谢谢!
答案 0 :(得分:8)
对Python2.7 +使用collections.Counter
:
>>> from collections import Counter
>>> lis1 = [['a', 2], ['b',1]]
>>> lis2 = [['b', 2], ['c', 1]]
>>> c = Counter(dict(lis1)) + Counter(dict(lis2))
>>> c.most_common()
[('b', 3), ('a', 2), ('c', 1)]
如果列表包含重复的项目,则需要将Counter
示例修改为:
>>> lis1 = [['a', 2], ['b',1], ['b',5]]
>>> lis2 = [['b', 2], ['c', 1], ['a', 10]]
>>> from itertools import chain
>>> from collections import Counter
>>> c = sum((Counter(dict([x])) for x in chain(lis1, lis2)), Counter())
>>> c.most_common()
[('a', 12), ('b', 8), ('c', 1)]
对于2.5< = Python< = 2.6使用collections.defaultdict
:
>>> from collections import defaultdict
>>> d = defaultdict(int)
>>> for k, v in lis1 + lis2:
d[k] += v
...
>>> sorted(d.items(), key=lambda x:x[1], reverse=True)
[('b', 3), ('a', 2), ('c', 1)]
答案 1 :(得分:1)
由于python 2.6的集合中没有Counter
类,因此这个类就可以了。可以使用defautldict
,但其用法不会简化代码:
a = [['a', 2], ['b', 1]]
b = [['b', 2], ['c', 1]]
rv = {}
for k, v in a + b:
rv[k] = rv.setdefault(k, 0) + v
预期结果的输出,转换为列表列表:
>>> map(list, sorted(rv.items(), key = lambda x: x[1], reverse=True))
[['b', 3], ['a', 2], ['c', 1]]
答案 2 :(得分:0)
只是把这个答案放在这里,但你可能希望将它们组合到一个列表中,然后计算它们:
>>> a = [['a', 2], ['b',1]]
>>> b = [['b', 2], ['c', 1]]
>>> a + b
[['a', 2], ['b', 1], ['b', 2], ['c', 1]]
>>> "".join(c*n for c, n in a+b)
'aabbbc'
>>> from collections import Counter
>>> Counter("".join(c*n for c, n in a+b))
Counter({'b': 3, 'a': 2, 'c': 1})
>>> Counter("".join(c*n for c, n in a+b)).most_common()
[('b', 3), ('a', 2), ('c', 1)]
答案 3 :(得分:0)
您可以简单地从元组列表中创建dicts(在您的案例中为列表),然后添加其计数器。
>>> from collections import Counter
>>> a = [['b', 2], ['c', 1]]
>>> b = [['a', 2], ['b',1]]
>>> sorted(dict(Counter(dict(a)) + Counter(dict(b))).items(),key= lambda x:-x[1])
[('b', 3), ('a', 2), ('c', 1)]
答案 4 :(得分:0)
很好
来自集合导入计数器
lis1 = [['a',2],['b',1]]
lis2 = [['b',2],['c',1]]
c = Counter(dict(lis1))+ Counter(dict(lis2))
c.most_common()
[('b', 3), ('a', 2), ('c', 1)]
python 2.5和2.6中的collections.defaultdict以及2.7及更高版本中的collections.Counter
from collections import defaultdict
d = defaultdict(int)
表示k,v表示lis1 + lis2:
d[k] += v
排序(d.items(),key = lambda x:x [1],reverse = True)
[('b', 3), ('a', 2), ('c', 1)]
答案 5 :(得分:0)
答案:
>>> from collections import Counter >>> p = [['a', 2], ['b',1]] >>> q = [['b', 2], ['c', 1]] >>> m = Counter(dict(p)) + Counter(dict(q)) >>> sorted(m.items(), key=lambda x:x[1], reverse=True) [('b', 3), ('a', 2), ('c', 1)]