问题描述
对于复制列表:shallow_copy_of_list = old_list[:].
对于复制字典:shallow_copy_of_dict = dict(old_dict).
但是对于 set，我担心类似的事情不会起作用，因为说 new_set = set(old_set) 会给出一个集合的集合?
But for a set, I was worried that a similar thing wouldn't work, because saying new_set = set(old_set) would give a set of a set?
但它确实有效.所以我在这里发布问题和答案以供参考.以防其他人有同样的困惑.
But it does work. So I'm posting the question and answer here for reference. In case anyone else has the same confusion.
推荐答案
这两个都会给出一个集合的副本:
Both of these will give a duplicate of a set:
shallow_copy_of_set = set(old_set)
或者:
shallow_copy_of_set = old_set.copy() #Which is more readable.
上面的第一种方式没有给出一个集合的集合的原因是，正确的语法应该是set([old_set]).这是行不通的，因为 sets 不能是其他 sets 中的元素，因为它们是可变的，因此它们是不可散列的.但是，对于 frozenset 来说，情况并非如此，例如frozenset(frozenset(frozenset([1,2,3]))) == frozenset([1, 2, 3]).
The reason that the first way above doesn't give a set of a set, is that the proper syntax for that would be set([old_set]). Which wouldn't work, because sets can't be elements in other sets, because they are unhashable by virtue of being mutable. However, this isn't true for frozensets, so e.g. frozenset(frozenset(frozenset([1,2,3]))) == frozenset([1, 2, 3]).
因此，在 Python 中复制任何基本数据结构(列表、字典、集合、冻结集、字符串)实例的经验法则:
So a rule of thumb for replicating any of instance of the basic data structures in Python (lists, dict, set, frozenset, string):
a2 = list(a) #a is a list
b2 = set(b) #b is a set
c2 = dict(c) #c is a dict
d2 = frozenset(d) #d is a frozenset
e2 = str(e) #e is a string
#All of the above give a (shallow) copy.
所以，如果 x 是这两种类型中的任何一种，那么
So, if x is either of those types, then
shallow_copy_of_x = type(x)(x) #Highly unreadable! But economical.
请注意，只有 dict、set 和 frozenset 具有内置的 copy() 方法.为了统一和可读性，列表和字符串也有一个 copy() 方法可能是个好主意.但他们没有，至少在我正在测试的 Python 2.7.3 中.
Note that only dict, set and frozenset have the built-in copy() method. It would probably be a good idea that lists and strings had a copy() method too, for uniformity and readability. But they don't, at least in Python 2.7.3 which I'm testing with.
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