理解 set() 函数 [英] Understanding the set() function

问题描述

在python中，set()是一个没有重复元素的无序集合.但是，我无法理解它是如何生成输出的.

例如，考虑以下内容:

<预><代码>>>>x = [1, 1, 2, 2, 2, 2, 2, 3, 3]>>>集合(x)设置([1, 2, 3])>>>y = [1, 1, 6, 6, 6, 6, 6, 8, 8]>>>设置(y)设置([8, 1, 6])>>>z = [1, 1, 6, 6, 6, 6, 6, 7, 7]>>>设置(z)设置([1, 6, 7])

set(y) 的输出不应该是:set([1, 6, 8])?我在 Python 2.6 中尝试了以上两个.

解决方案

正如你所说，集合是无序的.尽管实现集合的一种方法是使用树，但它们也可以使用哈希表实现(意味着按排序顺序获取键可能不是那么简单).

如果你想对它们进行排序，你可以简单地执行:

sorted(set(y))

将生成一个包含集合元素的排序列表.(不是集合.同样，集合是无序的.)

否则，set 唯一保证的就是它使元素唯一(不会出现多次).

希望这会有所帮助！

In python, set() is an unordered collection with no duplicate elements. However, I am not able to understand how it generates the output.

For example, consider the following:

>>> x = [1, 1, 2, 2, 2, 2, 2, 3, 3]
>>> set(x)
set([1, 2, 3])

>>> y = [1, 1, 6, 6, 6, 6, 6, 8, 8]
>>> set(y)
set([8, 1, 6])

>>> z = [1, 1, 6, 6, 6, 6, 6, 7, 7]
>>> set(z)
set([1, 6, 7])

Shouldn't the output of set(y) be: set([1, 6, 8])? I tried the above two in Python 2.6.

解决方案

Sets are unordered, as you say. Even though one way to implement sets is using a tree, they can also be implemented using a hash table (meaning getting the keys in sorted order may not be that trivial).

If you'd like to sort them, you can simply perform:

sorted(set(y))

which will produce a sorted list containing the set's elements. (Not a set. Again, sets are unordered.)

Otherwise, the only thing guaranteed by set is that it makes the elements unique (nothing will be there more than once).

Hope this helps!

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