list1=[1,2,3,3,4,5,6,1,3,4,5]
 n=int(input('enter  number'))
 while n in list1:
    list1.remove(n)
 print(list1)

这里的许多答案都涉及到创建一个新列表。这涉及到将所有数据从旧列表复制到新列表(删除的项除外)。如果你的清单很长，你可能负担不起(或者你不应该想要)。

在这些情况下，就地更改列表要快得多。如果你必须从列表中删除一个以上的元素，这可能会很棘手。假设您循环遍历列表，并删除了一个项，那么列表将更改，标准for循环将不考虑这一点。循环的结果可能不是您所期望的。

例子:

a = [0, 1, 2, 3, 4, 5]
for i in a:
    a.remove(i)  # Remove all items
print(a)

Out: [1, 3, 5]

一个简单的解决方案是以相反的顺序遍历列表。在这种情况下，你会得到:

a = [0, 1, 2, 3, 4, 5]
for i in reversed(a):
    a.remove(i)  # Remove all items
print(a)

Out: []

然后，如果你只需要删除一些特定值的元素，你可以简单地在循环中放入一个if语句，结果是:

a = [0, 1, 2, 3, 4, 5]
for i in reversed(a):
    if i == 2 or i == 3:  # Remove all items having value 2 or 3.
        a.remove(i)
print(a)

Out: [0, 1, 4, 5]

一句话:

a.remove('b') if 'b' in a else None

有时它很有用。

更简单:

if 'b' in a: a.remove('b')

要删除一个元素的第一次出现，使用list.remove:

>>> xs = ['a', 'b', 'c', 'd']
>>> xs.remove('b')
>>> print(xs)
['a', 'c', 'd']

要移除所有出现的元素，使用列表推导式:

>>> xs = ['a', 'b', 'c', 'd', 'b', 'b', 'b', 'b']
>>> xs = [x for x in xs if x != 'b']
>>> print(xs)
['a', 'c', 'd']

arr = [1, 1, 3, 4, 5, 2, 4, 3]

# to remove first occurence of that element, suppose 3 in this example
arr.remove(3)

# to remove all occurences of that element, again suppose 3
# use something called list comprehension
new_arr = [element for element in arr if element!=3]

# if you want to delete a position use "pop" function, suppose 
# position 4 
# the pop function also returns a value
removed_element = arr.pop(4)

# u can also use "del" to delete a position
del arr[4]

这就是我的答案，用while和for就可以了

def remove_all(data, value):
    i = j = 0
    while j < len(data):
        if data[j] == value:
            j += 1
            continue
        data[i] = data[j]
        i += 1
        j += 1
    for x in range(j - i):
        data.pop()