How to remove duplicate elements from a list in Python
Delete by traversing the list directly
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] for el in l1: if l1.count(el) > 1: l1.remove(el) print(l1)# 会漏删,因为删除一个元素后,后面的元素向前补位,导致紧跟的一个元素被跳过.
Delete by traversing the index
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] for el in range(len(l1)): # 此时len(l1)已经确定,不会随着l1后面的变化而变化 if l1.count(l1[el]) > 1: l1.remove(l1[el]) print(l1) # 会报错,因为删除元素后导致l1的长度变短了,但是for遍历的还是之前的索引长度,会导致索引超过范围而报错
Delete the original list by traversing the created slice
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] for el in l1[:]: if l1.count(el) > 1: l1.remove(el) # 没有问题,可以去重,但是无法保留原有的顺序 print(l1)
Use the new list Record the elements that need to be retained
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] lst = [] for el in l1: if lst.count(el) < 1: lst.append(el) print(lst) # 没有问题,也能保留原有顺序,但是创建了新列表
Delete through index backwards
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] for el in range(len(l1)-1, -1, -1): if l1.count(l1[el]) > 1: l1.pop(el) # 没有问题,且保留原顺序 # l1.remove(l1[el]) # 没有问题,但是不能保留原有顺序 # del l1[el] # 这样则会保留原有顺序,小伙伴可以想一想为什么 print(l1)
Delete through recursive function
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] def set_lst(lst): for el in lst: if lst.count(el) > 1: lst.remove(el) set_lst(lst) # 每次开辟一个新函数,判断上次被删除了一个元素后的列表 else: # 直到最后,列表里的元素都是一个,运行了else return lst print(set_lst(l1)) # 因为是从前面开始删除的,所以不保留原有顺序 ''' [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] [1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] [1, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] [1, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] [1, 3, 3, 3, 6, 6, 5, 5, 2, 2] [1, 3, 3, 6, 6, 5, 5, 2, 2] [1, 3, 6, 6, 5, 5, 2, 2] [1, 3, 6, 5, 5, 2, 2] [1, 3, 6, 5, 2, 2] [1, 3, 6, 5, 2] return lst = [1, 3, 6, 5, 2] '''
There is no doubt that set() is the most convenient
l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2] lst = list(set(l1)) print(lst)
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