你可以使用字典推导式。这会创建一个新的字典,但其复杂度与你的
for
循环相同:
d = {k: v for k, v in packaged.items() if k in union}
下面是一些真实的基准测试,以及两个稍微更高效的变体:
union = {'a':'gamma','b':'beta','d':'theta', **dict.fromkeys(range(5000, 10000))}
packaged = {'a':'alpha', 'b':'gamma', 'c':'alpha', **dict.fromkeys(range(1, 15000))}
def dct_cmp(union, packaged):
return {k: v for k, v in packaged.items() if k in union}
def dct_cmp_from_key(union, packaged):
return {k: packaged[k] for k in packaged if k in union}
def dct_cmp_from_key_intersect(union, packaged):
return {k: packaged[k] for k in set(packaged) & set(union)}
def loopy(union, packaged):
for k, v in list(packaged.items()):
if k not in union:
del packaged[k]
return packaged
assert dct_cmp(union, packaged.copy()) == loopy(union, packaged.copy())
assert dct_cmp(union, packaged.copy()) == dct_cmp_from_key(union, packaged.copy())
assert dct_cmp(union, packaged.copy()) == dct_cmp_from_key_intersect(union, packaged.copy())
%timeit dct_cmp(union, packaged.copy())
%timeit dct_cmp_from_key(union, packaged.copy())
%timeit dct_cmp_from_key_intersect(union, packaged.copy())
%timeit loopy(union, packaged.copy())
d = {k: packaged[k] for k in packaged if k in union}
,因为它不会从items()
创建副本。 - RoadRunner