在2.6[重复]上合并N字典和添加值的最快方法

Question

我有一个字典列表，我想把它们组合成一个字典，并在列表中添加每个字典的值。例如：

ds = [{1: 1, 2: 0, 3: 0}, {1: 2, 2: 1, 3: 0}, {1: 3, 2: 2, 3: 1, 4: 5}]

merged = {1: 6, 2: 3, 3: 1, 4: 5}

from collections import defaultdict

merged = defaultdict(int)

for d in ds:
    for k, v in d.items():
        merged[k] += v

在Python 2.6中有没有更快的方法来实现这一点？

Answer 1

DefaultDict仍然是最快的，我找到了几种通过缓存函数名来加快速度的方法，现在又找到了另一种方法，它可以显著加快速度，只需在d中为k迭代，而不是使用d.items()或d.iteritems()

到目前为止的一些时间安排：

from random import randrange
ds = [dict((randrange(1, 1000), randrange(1, 1000)) for i in xrange(500))
      for i in xrange(10000)]

# 10000 dictionaries of approx. length 500

from collections import defaultdict

def merge1(dicts, defaultdict=defaultdict, int=int):
    merged = defaultdict(int)
    for d in dicts:
        for k in d:
            merged[k] += d[k]
    return merged

def merge2(dicts):
    merged = {}
    merged_get = merged.get
    for d in dicts:
        for k in d:
            merged[k] = merged_get(k, 0) + d[k]
    return merged


def merge3(dicts):
    merged = {}
    for d in dicts:
        for k in d:
            merged[k] = merged[k] + d[k] if k in merged else 0
    return merged


from timeit import timeit
for func in ('merge1', 'merge2', 'merge3'):
    print func, timeit(stmt='{0}(ds)'.format(func),
                       setup='from __main__ import merge1, merge2, merge3, ds',
                       number=1)

merge1 0.992541510164
merge2 1.40478747997
merge3 1.23502204889