python - 从字典列表中优化平均计算

Question

我有一个字典列表，键为“a”、“n”、“o”、“u”。 有没有办法加快这个计算，例如 NumPy ？列表中有数万个项目。

数据是从数据库中提取的，所以我必须接受它最初是以字典列表的形式存在的。

x = n = o = u = 0
for entry in indata:
    x += (entry['a']) * entry['n']  # n - number of data points
    n += entry['n']
    o += entry['o']
    u += entry['u']

    loops += 1

average = int(round(x / n)), n, o, u

Answer 1

我怀疑这会快得多，但我想它是 timeit 的候选者...

from operator import itemgetter
x = n = o = u = 0
items = itemgetter('a','n','o','u')
for entry in indata:
    A,N,O,U = items(entry)
    x += A*N  # n - number of data points
    n += N
    o += O    #don't know what you're doing with O or U, but I'll leave them
    u += U

average = int(round(x / n)), n, o, u

至少，它保存了对 entry['n'] 的查找，因为我现在已经将它保存到一个变量中