Question

像

这样的东西

{A: 3, 45, 34, 4, 2, 5, 94, 2139, 230345, 283047, 230847}, {B: 92374, 324, 345, 345, 45879, 34857987, 3457938457), {C: 23874923874987, 2347}

如何将其减少到

{A: 2304923094820398}, {B: 2374923784897}, {C: 29348239847239847}

值显然不准确。我只想以非耗时的方式添加所有值。

Answer 1

d = [{'A': [3, 45, 34, 4, 2, 5, 94, 2139, 230345, 283047, 230847]}, {'B': [92374, 324, 345, 345, 45879, 34857987, 3457938457]}, {'C': [23874923874987, 2347]}]
[{x.keys()[0]:sum(x.values()[0])} for x in d]

Answer 2

这样的事情：

a={'a':range(100), 'b':range(200)}
b={}

for k,v in a.iteritems():
    b[k]=sum(v)

print b

给出

{'a': 4950, 'b': 19900}

Answer 3

假设值在列表中，例如：

d = {A: [3, 45, 34, 4, 2, 5, 94, 2139, 230345, 283047, 230847]}, {B: [92374, 324, 345, 345, 45879, 34857987, 3457938457]), {C: [23874923874987, 2347]}

然后：

new_d = []
for i in d:
    new_d.append(dict([(k,sum(v)) for k,v in i.itervalues()]))

Answer 4

如果你有很多很多很多值，这样的东西可能会起作用（假设“词典列表”你的意思是“列表字典”，因为有一个单键/值对词典的大列表会是愚蠢的，因此值存储在dict {'A': [3, 45, 34, 4, 2, 5, 94, 2139, 230345, 283047, 230847], 'B': [92374, 324, 345, 345, 45879, 34857987, 3457938457], ...}}中：

from multiprocessing import Process, Lock

def sumfunc(lock, key, values):
    total = sum(values)

    lock.acquire()
    print(key + ':', total)   # The actual code would be better off utilizing shared multiprocessing.Array objects, but I haven't used them before so am not sure quite how you set them up so decided to leave that aside for now as this is just a theoretical example.
    lock.release()

if __name__ == '__main__':
    lock = Lock()

    for key, values in dict_of_values_to_sum.items():
        Process(target=sumfunc, args=(lock, key, values)).start()

Answer 5

如果你有很多值，我可能会使用numpy并添加一个向量。

所以，像

 a = numpy.arange(10, 2000000)
 total = numpy.sum(a)

没有更多，没有更少：）

Answer 6

d = [{'A': [3, 45, 34, 4, 2, 5, 94, 2139, 230345, 283047, 230847]}, {'B': [92374, 324, 345, 345, 45879, 34857987, 3457938457]}, {'C': [23874923874987, 2347]}]
abc = {}
for x in d:
    for k, v in x.iteritems():
        abc[k] = reduce(lambda x,y: x+y, v)

我有一个非常大的字典列表，我想总结内部

6 个答案: