【每日一题0613】删除 list 里面的重复元素 并保持顺序不变化

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,
1

删除list 里面的重复元素 并保持顺序不变化
例 [1,2,4,2,4,3] 变成 [1,2,4,3]

2

nums_list = [1,2,4,2,4,3]
res_list =
for i in nums_list:
if i not in res_list:
res_list.append(i)

4

set强转会改变顺序

5

是的 没注意到 :sweat_smile:

6

用列表推导式比较方便。
list1=[1,2,4,2,4,3]
list2=[list2.append(i) for i in list1 if i not in list2]
print(list2)

7

list3 = [1,2,4,2,4,3]
list3.reverse()
for i in list3:
for _ in range(list3.count(i)):
if list3.count(i) > 1:
list3.remove(i)
list3.reverse()

8 
"""
删除list 里面的重复元素 并保持顺序不变化
例 [1,2,4,2,4,3] 变成 [1,2,4,3]
"""


def test(test_list: list):
    i = 0
    while i < len(test_list):
        if test_list[i] in test_list[0:i]:
            test_list.pop(i)
        else:
            i += 1
    return test_list


assert test([1, 2, 4, 2, 4, 3]) == [1, 2, 4, 3]
assert test([1, 2, 2, 2, 2, 1]) == [1, 2]
assert test([]) == []
9

解法1: 列表中都是纯数字。

def single_num_list(li):
    return sorted(set(li), key=lambda x:li.index(x))
assert single_num_list([1,2,4,2,4,3]) == [1,2,4,3]

解法2: 凡是列表中的对象是可散列的都适用。例如数字，字符串和元组。

def single_hashable(li):
    existed = set()
    for x in li:
        if x not in existed:
            yield x
            existed.add(x)
    
assert list(single_hashable([1,2,4,2,4,3])) == [1,2,4,3]
1 个赞
10

“”"
每日一题：删除list 里面的重复元素 并保持顺序不变化
例 [1,2,3,2,4,3] 变成 [1,2,3,4]

“”"
alist = [1, 5, 2, 3, 5, 3, 2, 4, 3, 6]
blist =
for i in alist:
if i in blist:
pass
else:
blist.append(i)
print(blist)

11

lst_1 = [1,2,4,2,4,3]
lst_2 =
for i in lst_1:
if i not in lst_2:
lst_2.append(i)
print(lst_2)

12 
def func2(sss):
    #用字典result，将元素放入key（key不能重复）
    result = { i:'a' for i in sss }
    list_1 = result.keys()
    return list(list_1)
print(func2([1, 3, 2, 2, 3, 5, 5, 4, 4]))
13

a=[1,2,4,2,4,3]
b=list(set(a))
b.sort(key=a.index)
print(b)

1 个赞
14 
def delItem(a,b={}):
    for item in a :
        if item not in b.keys():
            b[item] = a.count(item)
    return list(b.keys())

assert delItem([1, 2, 4, 2, 4, 3]) == [1,2,4,3]
15

image
image1185×598 72.3 KB

为什么你这种不行？

16 
def solution(nums: list) -> list:
    # your code here
    result = []
    for i in nums:
        if i not in result:
            result.append(i)
    return result

assert solution([1,2,4,2,4,3]) == [1,2,4,3]
17 
def solution(num:list)->list:
    new_num = []
    for i in num:
        if i not in new_num:
            new_num.append(i)
    return new_num
18 
def solve(test_list: list):
    result = []
    for i in test_list:
        if i not in result:
            result.append(i)
    return result


assert solve([1, 2, 4, 2, 4, 3]) == [1, 2, 4, 3]
assert solve([1, 2, 2, 2, 2, 1]) == [1, 2]
assert solve([]) == []
19 
def solution(a:list)->list:
    res_list=[]
    for i in range(len(a)):
        if a[i] not in res_list:
            res_list.append(a[i])
    return res_list

assert solution([1,2,4,2,4,3])==[1,2,4,3]
20 
def test_list_info():
    list_a=[1,2,4,2,4,3]
    new_list=[]
    for i in list_a:
        if i not in new_list:
            new_list.append(i)
    print(new_list)
21

class DelectSameElement:
# 初始化定义列表
def init(self,l):
self.l = l

# 列表转成字典后，取keys再转成列表
def remove_same_elements1(self):
    # 新建空字典
    dict1 = {}
    # 遍历列表，新增字典的元素
    for i in self.l:
        dict1[i] = i
    # 返回字典的keys强转list
    return list(dict1.keys())


def remove_same_elements2(self):
    # 新建空列表
    nl=[]
    # 遍历列表
    for i in range(len(self.l)):
        # 如果元素不在新列表中
        if self.l[i] not in nl:
            # 新列表新增该元素
            nl.append(self.l[i])
    # 返回新列表
    return nl

# 强转字典后再强转列表，用dict.fromkeys()方法
def remove_same_elements3(self):
    return list(dict.fromkeys(self.l))