Python:如何检查一个项目是否被添加到一个集合中,没有 2x(散列,查找)
问题描述
我想知道是否有一种清晰/简洁的方法可以将某些内容添加到集合中,并检查它是否是在没有 2x 哈希的情况下添加的.查找.
I was wondering if there was a clear/concise way to add something to a set and check if it was added without 2x hashes & lookups.
这是你可能会做的,但它有 2 倍的项目哈希
this is what you might do, but it has 2x hash's of item
if item not in some_set: # <-- hash & lookup
some_set.add(item) # <-- hash & lookup, to check the item already is in the set
other_task()
这适用于单个哈希和查找,但有点难看.
This works with a single hash and lookup but is a bit ugly.
some_set_len = len(some_set)
some_set.add(item)
if some_set_len != len(some_set):
other_task()
有没有更好的方法使用 Python 的 set api 来做到这一点?
Is there a better way to do this using Python's set api?
解决方案
我认为没有内置的方法可以做到这一点.当然,您可以编写自己的函数:
I don't think there's a built-in way to do this. You could, of course, write your own function:
def do_add(s, x):
l = len(s)
s.add(x)
return len(s) != l
s = set()
print(do_add(s, 1))
print(do_add(s, 2))
print(do_add(s, 1))
print(do_add(s, 2))
print(do_add(s, 4))
或者,如果您更喜欢神秘的单线:
Or, if you prefer cryptic one-liners:
def do_add(s, x):
return len(s) != (s.add(x) or len(s))
(这取决于从左到右的评估顺序以及 set.add()
总是返回 None
的事实,这是错误的.)
(This relies on the left-to-right evaluation order and on the fact that set.add()
always returns None
, which is falsey.)
除此之外,我只会在双重哈希/查找明显是性能瓶颈时才考虑这样做并且如果使用函数明显更快.
All this aside, I would only consider doing this if the double hashing/lookup is demonstrably a performance bottleneck and if using a function is demonstrably faster.
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