Question

我有一个元组列表：

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

我希望获得a和b值的总和。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

您可以使用collections.Counter从初始列表中创建multiset，然后根据您的情况修改结果：

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 4

与已经给出的答案非常相似。稍微长一点，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Moses Koledoye的回答复制的最后一行）

Answer 5

from collections import Counter

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

如何总结元组列表？

5 个答案: