python字典的比较_python-将字典与其他字典进行比较,并仅选择...

您可以使用字典理解.这将创建一个新的字典,但与您的for循环具有相同的复杂性：

d = {k: v for k, v in packaged.items() if k in union}

这是一些现实的基准测试,以及两个稍微更有效的变体：

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()) # 1.94 ms

%timeit dct_cmp_from_key(union, packaged.copy()) # 1.8 ms

%timeit dct_cmp_from_key_intersect(union, packaged.copy()) # 1.8 ms

%timeit loopy(union, packaged.copy()) # 2.75 ms