删除列表中的重复项 [英] Removing duplicates in lists

问题描述

很多我需要编写一个程序来检查一个列表是否有任何重复的，如果它删除它们，并返回一个没有被重复/删除的项目的新列表。这是我所说的，但老实说，我不知道该怎么做。

Pretty much I need to write a program to check if a list has any duplicates and if it does it removes them and returns a new list with the items that werent duplicated/removed. This is what I have but to be honest I do not know what to do.

def remove_duplicates():
    t = ['a', 'b', 'c', 'd']
    t2 = ['a', 'c', 'd']
    for t in t2:
        t.append(t.remove())
    return t

推荐答案

获取唯一的项目集合的常见方法是使用 设置 。集合 要从任何可迭代创建一个集合，您只需将其传递给内置的 set（） 功能。如果您以后需要一个真实的列表，您可以同样将该集合传递到 list（） 功能。

The common approach to get a unique collection of items is to use a set. Sets are unordered collections of distinct objects. To create a set from any iterable, you can simply pass it to the built-in set() function. If you later need a real list again, you can similarly pass the set to the list() function.

以下示例应涵盖您尝试执行的任何操作： / p>

The following example should cover whatever you are trying to do:

>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> t
[1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> list(set(t))
[1, 2, 3, 5, 6, 7, 8]
>>> s = [1, 2, 3]
>>> list(set(t) - set(s))
[8, 5, 6, 7]

从示例结果可以看出，原始顺序不被维护。如上所述，集合本身是无序集合，因此订单丢失。将集合转换为列表时，将创建任意顺序。

As you can see from the example result, the original order is not maintained. As mentioned above, sets themselves are unordered collections, so the order is lost. When converting a set back to a list, an arbitrary order is created.

如果订单对您很重要，则必须使用其他机制。 此问题更详细地介绍了该主题。

If order is important to you, then you will have to use a different mechanism. This question covers that topic in more detail.

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