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)# 會(huì)漏刪,因?yàn)閯h除一個(gè)元素后,后面的元素向前補(bǔ)位,導(dǎo)致緊跟的一個(gè)元素被跳過(guò).

l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2]
for el in range(len(l1)):  # 此時(shí)len(l1)已經(jīng)確定,不會(huì)隨著l1后面的變化而變化
    if l1.count(l1[el]) > 1:
        l1.remove(l1[el])
print(l1)  # 會(huì)報(bào)錯(cuò),因?yàn)閯h除元素后導(dǎo)致l1的長(zhǎng)度變短了,但是for遍歷的還是之前的索引長(zhǎng)度,會(huì)導(dǎo)致索引超過(guò)范圍而報(bào)錯(cuò)

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) # 沒有問(wèn)題,可以去重,但是無(wú)法保留原有的順序
print(l1)   

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)   # 沒有問(wèn)題,也能保留原有順序,但是創(chuàng)建了新列表

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)  # 沒有問(wèn)題,且保留原順序
        # l1.remove(l1[el]) # 沒有問(wèn)題,但是不能保留原有順序
        # del l1[el] # 這樣則會(huì)保留原有順序,小伙伴可以想一想為什么
print(l1)   

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)    # 每次開辟一個(gè)新函數(shù),判斷上次被刪除了一個(gè)元素后的列表
    else:  # 直到最后,列表里的元素都是一個(gè),運(yùn)行了else
        return lst
print(set_lst(l1))  # 因?yàn)槭菑那懊骈_始刪除的,所以不保留原有順序
'''
[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]
'''

l1 = [1, 1, 2, 2, 3, 3, 3, 3, 6, 6, 5, 5, 2, 2]
lst = list(set(l1))
print(lst)