本文翻译自：Removing duplicates in lists
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 weren't duplicated/removed.我几乎需要编写一个程序来检查列表中是否有重复项，如果删除了重复项，则将其删除，并返回一个新列表，其中包含未重复/删除的项。This is what I have but to be honest I do not know what to do.这就是我所拥有的，但老实说我不知道该怎么办。
def remove_duplicates():t = ['a', 'b', 'c', 'd']t2 = ['a', 'c', 'd']for t in t2:t.append(t.remove())return t
#1楼
参考：/question/XP75/删除列表中的重复项
#2楼
>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]>>> t[1, 2, 3, 1, 2, 5, 6, 7, 8]>>> s = []>>> for i in t:if i not in s:s.append(i)>>> s[1, 2, 3, 5, 6, 7, 8]
#3楼
Another way of doing:另一种方法：
>>> seq = [1,2,3,'a', 'a', 1,2]>> dict.fromkeys(seq).keys()['a', 1, 2, 3]
#4楼
I had a dict in my list, so I could not use the above approach.我的清单上有一个字典，所以我不能使用上述方法。I got the error:我得到了错误：
TypeError: unhashable type:
So if you care aboutorderand/or some items areunhashable.因此，如果您关心订单和/或某些项目无法散列。Then you might find this useful:然后，您可能会发现这很有用：
def make_unique(original_list):unique_list = [][unique_list.append(obj) for obj in original_list if obj not in unique_list]return unique_list
Some may consider list comprehension with a side effect to not be a good solution.有些人可能认为列表理解有副作用不是一个好的解决方案。Here's an alternative:这是一个替代方案：
def make_unique(original_list):unique_list = []map(lambda x: unique_list.append(x) if (x not in unique_list) else False, original_list)return unique_list
#5楼
Here is an example, returning list without repetiotions preserving order.这是一个示例，返回不保留重复顺序的列表。Does not need any external imports.不需要任何外部进口。
def GetListWithoutRepetitions(loInput):# return list, consisting of elements of list/tuple loInput, without repetitions.# Example: GetListWithoutRepetitions([None,None,1,1,2,2,3,3,3])# Returns: [None, 1, 2, 3]if loInput==[]:return []loOutput = []if loInput[0] is None:oGroupElement=1else: # loInput[0]<>NoneoGroupElement=Nonefor oElement in loInput:if oElement<>oGroupElement:loOutput.append(oElement)oGroupElement = oElementreturn loOutput
#6楼
There are also solutions using Pandas and Numpy.也有使用Pandas和Numpy的解决方案。They both return numpy array so you have to use the function.tolist()if you want a list.它们都返回numpy数组，因此如果需要列表，则必须使用.tolist()函数。
t=['a','a','b','b','b','c','c','c']t2= ['c','c','b','b','b','a','a','a']
Pandas solution熊猫解决方案
Using Pandas functionunique():使用熊猫函数unique()：
import pandas as pdpd.unique(t).tolist()>>>['a','b','c']pd.unique(t2).tolist()>>>['c','b','a']
Numpy solution脾气暴躁的解决方案
Using numpy functionunique().使用numpy函数unique()。
import numpy as npnp.unique(t).tolist()>>>['a','b','c']np.unique(t2).tolist()>>>['a','b','c']
Note that numpy.unique() also sort the values.请注意，numpy.unique（）也对值进行排序。So the listt2is returned sorted.因此，列表t2被返回排序。If you want to have the order preserved use as in this answer :如果您想保留订单，请按照以下答案进行操作 ：
_, idx = np.unique(t2, return_index=True)t2[np.sort(idx)].tolist()>>>['c','b','a']
The solution is not so elegant compared to the others, however, compared to pandas.unique(), numpy.unique() allows you also to check if nested arrays are unique along one selected axis.与其他解决方案相比，该解决方案并不那么优雅，但是与pandas.unique（）相比，numpy.unique（）还可让您检查嵌套数组在一个选定轴上是否唯一。
