在 python 列表中获取唯一的元组，而不管顺序
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Grab unique tuples in python list, irrespective of order
提问by Christopher
I have a python list:
我有一个 python 列表：
[ (2,2),(2,3),(1,4),(2,2), etc...]
What I need is some kind of function that reduces it to its unique components... which would be, in the above list:
我需要的是某种功能，将其减少到其独特的组件......这将是，在上面的列表中：
[ (2,2),(2,3),(1,4) ]
numpy unique does not quite do this. I can think of a way to do it--convert my tuples to numbers, [22,23,14,etc.], find the uniques, and work back from there...but I don't know if the complexity won't get out of hand. Is there a function that will do what I am trying to do with tuples?
numpy unique 并没有完全做到这一点。我可以想出一种方法——将我的元组转换为数字，[22,23,14,etc.]找到唯一值，然后从那里开始工作……但我不知道复杂性是否会失控。有没有一个函数可以做我想要对元组做的事情？
Here is a sample of code that demonstrates the problem:
这是演示问题的代码示例：
import numpy as np
x = [(2,2),(2,2),(2,3)]
y = np.unique(x)
returns: y: [2 3]
返回：y：[2 3]
And here is the implementation of the solution that demonstrates the fix:
这是演示修复的解决方案的实现：
x = [(2,2),(2,2),(2,3)]
y = list(set(x))
returns y: [(2,2),(2,3)]
返回 y: [(2,2),(2,3)]
采纳答案by Mehdi Golari
you could simply do
你可以简单地做
y = np.unique(x, axis=0)
z = []
for i in y:
z.append(tuple(i))
The reason is that a list of tuples is interpreted by numpy as a 2D array. By setting axis=0, you'd be asking numpy not to flatten the array and return unique rows.
原因是元组列表被 numpy 解释为二维数组。通过设置轴 = 0，您会要求 numpy 不要展平数组并返回唯一的行。
回答by timgeb
If order does not matter
如果顺序无关紧要
If the order of the result is not critical, you can convert your list to a set (because tuples are hashable) and convert the set back to a list:
如果结果的顺序不重要，您可以将列表转换为集合（因为元组是可散列的）并将集合转换回列表：
>>> l = [(2,2),(2,3),(1,4),(2,2)]
>>> list(set(l))
[(2, 3), (1, 4), (2, 2)]
If order matters
如果订单很重要
(UPDATE)
（更新）
As of CPython 3.6 (or any Python 3.7 version) regular dictionaries remember their insertion order, so you can simply issue.
从 CPython 3.6（或任何 Python 3.7 版本）开始，常规词典会记住它们的插入顺序，因此您可以简单地发出。
>>> l = [(2,2),(2,3),(1,4),(2,2)]
>>> list(dict.fromkeys(l))
[(2, 2), (2, 3), (1, 4)]
(OLD ANSWER)
（旧答案）
If the order is important, the canonical way to filter the duplicates is this:
如果顺序很重要，则过滤重复项的规范方法是：
>>> seen = set()
>>> result = []
>>> for item in l:
... if item not in seen:
... seen.add(item)
... result.append(item)
...
>>> result
[(2, 2), (2, 3), (1, 4)]
Finally, a little slower and a bit more hackish, you can abuse an OrderedDictas an ordered set:
最后，稍微慢一点，更hackish，你可以滥用 anOrderedDict作为一个有序集：
>>> from collections import OrderedDict
>>> OrderedDict.fromkeys(l).keys() # or list(OrderedDict.fromkeys(l)) if using a version where keys() does not return a list
[(2, 2), (2, 3), (1, 4)]
回答by Mureinik
Using a setwill remove duplicates, and you create a listfrom it afterwards:
使用 aset将删除重复项，然后您list从中创建一个：
>>> list(set([ (2,2),(2,3),(1,4),(2,2) ]))
[(2, 3), (1, 4), (2, 2)]
回答by zondo
set()will remove all duplicates, and you can then put it back to a list:
set()将删除所有重复项，然后您可以将其放回列表中：
unique = list(set(mylist))
Using set(), however, will kill your ordering. If the order matters, you can use a list comprehension that checks if the value already exists earlier in the list:
set()但是，使用会杀死您的订单。如果顺序很重要，您可以使用列表推导来检查该值是否已存在于列表的较早位置：
unique = [v for i,v in enumerate(mylist) if v not in mylist[:i]]
That solution is a little slow, however, so you can do it like this:
但是，该解决方案有点慢，因此您可以这样做：
unique = []
for tup in mylist:
if tup not in unique:
unique.append(tup)
