如何从列表中相似项目的集合中保留唯一项目? [英] How to keep unique items from a collection of similar items in a list?

问题描述

我有一个类似以下的列表

I have a list like the following

a = [5,5,5,5,3,3,2,2,2,2,5,5,5,2,2,2,2,2]

我希望有以下输出列表

b = [5,3,2,5,2]

请注意，我尝试使用

list(OrderedDict.fromkeys(a))

这给了我

[5, 3, 2]

请注意，由于当时2nd 5或2nd 2并不唯一，因此请注意.

Notice that it does not consider the 2nd 5 or the 2nd 2 as they are non-unique by then.

我需要一种使机器理解的方法，当它遇到一个唯一元素(例如第一个"5")时，应将其存储在列表b中.因此，当遇到第一个"3"或第一个"2"时，它也应该将它们存储在列表b中.到此为止没有问题.当它遇到第二组"5"时，应将其5作为新元素存储在列表b中.第二组2也是如此.

I need a way to make the machine understand that as and when it encounters a unique element (say the first '5'), it should store it in list b. Consequently, when it encounters the first '3' or first '2', it should store them in list b as well. No problems until this point. As it encounters the second set of '5's, it should store that 5 as a new element in list b. Same goes for the second set of 2's.

有没有Python的方法可以做到这一点?

Is there a Pythonic way to do this?

推荐答案

除了进行for循环外，您还可以使用itertools groupby:

Other than just doing a for loop you could use itertools groupby:

import itertools

a = [5,5,5,5,3,3,2,2,2,2,5,5,5,2,2,2,2,2]
b = [x[0] for x in itertools.groupby(a)]  # [5, 3, 2, 5, 2]

有关此文档的信息，请参见: https://docs .python.org/3/library/itertools.html#itertools.groupby

Documentation for this can be found here: https://docs.python.org/3/library/itertools.html#itertools.groupby

进行一些澄清.能够在列表中重复出现的次数的原因在文档(强调我的)中的这一段中得到了解释:

Some clarification. The reason this is able to count re-occurrences in the list is explained by this paragraph in the documentation (emphasis mine):

groupby()的操作类似于Unix中的uniq过滤器.它 每次键值生成一个中断或新组 功能更改 [...]

The operation of groupby() is similar to the uniq filter in Unix. It generates a break or new group every time the value of the key function changes [...]

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