写一个删除列表中重复元素的函数，要求去重后元素相对位置保持不变

1

# 无集合set
list_ = ['a','a','b','b','c','c']
list_new = []
for i in list_:
    if i not in list_new:
        list_new.append(i)

list_new

def fun_1(list_ = ['a','a','b','b','c','c']):
    list_new = []
    for i in list_:
        if i not in list_new:
            list_new.append(i)
    return list_new

fun_1()

2

# 利用集合set
list_ = ['a','a','b','b','c','c']
list_new = []
set_ = set()
for i in list_:
    if i not in set_:
        list_new.append(i)
        set_.add(i)

list_new

def fun_2(list_ = ['a','a','b','b','c','c']):
    list_new = []
    set_ = set()
    for i in list_:
        if i not in set_:
            list_new.append(i)
            set_.add(i)
    return list_new

fun_2()

3

# 无新列表
def fun_3(list_ = ['a','a','b','b','c','c']):
    set_ = set()
    for i in list_:
        if i not in set_:
            yield i
            set_.add(i)

ob = fun_3()

list_new = []
try:
    while 0<1:
        list_new.append(ob.__next__())
except:
    print(list_new)

ob.__next__()

[ i for i in fun_3() ]

ob = fun_3()
list_new = []
for i in ob:
    list_new.append(i)

list_new

list_new = []
for i in fun_3():
    list_new.append(i)

list_new