如何对元组列表进行求和？

Question

我有元组的列表：

[ (a,1), (a,2), (b,1), (b,3) ] 

我想要得到的a和b值两者的总和。结果应采用以下格式：

[ { 'key' : a, 'value' : 3 }, {'key' : b, 'value' : 4 } ] 

我该怎么做？

Answer 1

from itertools import groupby 
[{'key': k, 'value': sum(v for _,v in g)} for k, g in groupby(sorted(lst), key = lambda x: x[0])] 

# [{'key': 'a', 'value': 3}, {'key': 'b', 'value': 4}] 

Answer 2

from collections import defaultdict 

lst = [("a", 1), ("a", 2), ("b", 1), ("b", 3)] 

out = defaultdict(list) 
[out[v[0]].append(v[1]) for v in lst] 
out = [{"key": k, "value": sum(v)} for k, v in out.iteritems()] 

print out 

Answer 3

from collections import Counter 

a = [('a', 1), ('a', 2), ('b', 1), ('b', 3)] 
c = Counter() 
for tup in a: 
    c = c + Counter(dict([tup])) 

Answer 4

您可以使用collections.Counter从初始列表创建multiset，然后修改结果，以配合您的情况：

from collections import Counter 

lst = [('a', 1), ('a', 2), ('b', 1), ('b', 3)] 

part = sum((Counter({i[0]: i[1]}) for i in lst), Counter()) 
# Counter({'b': 4, 'a': 3}) 

final = [{'key': k, 'value': v} for k, v in part.items()] 
# [{'key': 'b', 'value': 4}, {'key': 'a', 'value': 3}] 

Answer 5

非常相似，已经给出答案。稍微长一些，但更容易阅读，恕我直言。

from collections import defaultdict 

lst = [('a', 1), ('a', 2), ('b', 1), ('b', 3)] 

dd = defaultdict(int) 
for name, value in lst: 
    dd[name] += value 

final = [{'key': k, 'value': v} for k, v in dd.items()] 

（从摩西Koledoye的答案复制最后一行）