检查字典中是否所有值都作为键存在

您的方法将起作用,但是,从转换到 set .

另一个具有相同时间复杂性的解决方案是:

all(i in bar for i in foo)

两者都有时间复杂性 O(len(foo))

bar = {str(i): i for i in range(100000)}
foo = [str(i) for i in range(1, 10000, 2)]

%timeit all(i in bar for i in foo)
462 µs ± 14.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit set(foo) - set(bar)
14.6 ms ± 174 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

# The overhead is all the difference here:

foo = set(foo)
bar = set(bar)

%timeit foo - bar
213 µs ± 1.48 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

这里的开销很大,所以我会选择 all 在这里。

试试看是否有任何未映射的项:

has_unmapped = all( (x in bar) for x in foo   )

要查看未映射的项:

unmapped_items = [ x for x in foo if x not in bar ]