Python：列表列表的唯一性
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/3724551/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Python: Uniqueness for list of lists
提问by Hellnar
I am curious what would be an effient way of uniquefying such data objects:
我很好奇什么是唯一化此类数据对象的有效方法：
testdata =[ ['9034968', 'ETH'], ['14160113', 'ETH'], ['9034968', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15724032', 'ETH'], ['15481740', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['10307528', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['15481740', 'ETH'], ['15379365', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15379365', 'ETH']
]
For each data pair, left numeric string PLUS the type at the right tells the uniqueness of a data element. And it returns a list of lists as same as the testdata, but only uniques are existing.
对于每个数据对，左边的数字字符串加上右边的类型说明数据元素的唯一性。它返回一个与测试数据相同的列表列表，但只有唯一的存在。
Regards
问候
采纳答案by Mark Byers
回答by Manoj Govindan
I tried @Mark's answer and got an error. Converting the list and each elements into a tuple made it work. Not sure if this the best way though.
我尝试了@Mark 的回答，但出现错误。将列表和每个元素转换为元组使其工作。不确定这是否是最好的方法。
list(map(list, set(map(lambda i: tuple(i), testdata))))
Of course the same thing can be expressed using a list comprehension instead.
当然，同样的事情可以用列表推导来表达。
[list(i) for i in set(tuple(i) for i in testdata)]
I am using Python 2.6.2.
我正在使用 Python 2.6.2。
Update
更新
@Mark has since changed his answer. His current answer uses tuples and will work. So will mine :)
@Mark 从那以后改变了他的答案。他目前的答案使用元组并且会起作用。我的也会这样:)
Update 2
更新 2
Thanks to @Mark. I have changed my answer to return a list of lists rather than a list of tuples.
感谢@Mark。我已更改答案以返回列表列表而不是元组列表。
回答by pyfunc
import sets
testdata =[ ['9034968', 'ETH'], ['14160113', 'ETH'], ['9034968', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15724032', 'ETH'], ['15481740', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['10307528', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['15481740', 'ETH'], ['15379365', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15379365', 'ETH']]
conacatData = [x[0] + x[1] for x in testdata]
print conacatData
uniqueSet = sets.Set(conacatData)
uniqueList = [ [t[0:-3], t[-3:]] for t in uniqueSet]
print uniqueList
回答by Sam Morgan
Expanding a bit on @Mark Byerssolution, you can also just do one list comprehension and typecast to get what you need:
对@Mark Byers解决方案进行一些扩展，您也可以只做一个列表理解和类型转换来获得您需要的内容：
testdata = list(set(tuple(x) for x in testdata))
Also, if you don't like list comprehensions as many find them confusing, you can do the same in a for loop:
此外，如果您不喜欢列表推导式，因为很多人认为它们令人困惑，您可以在 for 循环中执行相同的操作：
for i, e in enumerate(testdata):
testdata[i] = tuple(e)
testdata = list(set(testdata))
回答by Khan
if you have a list of objects than you can modify @Mark Byers answer to:
如果您有一个对象列表，您可以将@Mark Byers 的答案修改为：
unique_data = [list(x) for x in set(tuple(x.testList) for x in testdata)]
where testdata is a list of objects which has a list testList as attribute.
其中 testdata 是具有列表 testList 作为属性的对象列表。
