比较字典忽略特定键

如何在不考虑某些键的情况下测试两个字典是否相等。例如，

equal_dicts(
    {'foo':1, 'bar':2, 'x':55, 'y': 77 },
    {'foo':1, 'bar':2, 'x':66, 'z': 88 },
    ignore_keys=('x', 'y', 'z')
)

应该返回 True。

UPD：我正在寻找一种高效、快速的解决方案。

UPD2。我最终得到了这段代码，它似乎是最快的：

def equal_dicts_1(a, b, ignore_keys):
    ka = set(a).difference(ignore_keys)
    kb = set(b).difference(ignore_keys)
    return ka == kb and all(a[k] == b[k] for k in ka)

时间：https://gist.github.com/2651872 https://gist.github.com/2651872

def equal_dicts(d1, d2, ignore_keys):
    d1_filtered = {k:v for k,v in d1.items() if k not in ignore_keys}
    d2_filtered = {k:v for k,v in d2.items() if k not in ignore_keys}
    return d1_filtered == d2_filtered

编辑：这可能更快并且更节省内存：

def equal_dicts(d1, d2, ignore_keys):
    ignored = set(ignore_keys)
    for k1, v1 in d1.iteritems():
        if k1 not in ignored and (k1 not in d2 or d2[k1] != v1):
            return False
    for k2, v2 in d2.iteritems():
        if k2 not in ignored and k2 not in d1:
            return False
    return True