Question

当条件为真时，我试图从列表中删除项目，然后从列表的最开始再次迭代。以下是我的代码：

from random import randint
x = [1,2,3,4,5]

for item in x:
    print("item is " + str(item))
    z = randint(1, 5)
    print("z is " + str(z))
    if z == item:
        x[:] = [item for item in x if item != z]
        print("Remaining list is " + str(x))
    else:
        print("Remaining list is " + str(x))

例如，我x列表中的第一个项目是1，并且假设z也是1.当z == item时，我的x列表会删除第一个项目（1）和更新到[2,3,4,5]。但我得到以下输出：

item is 1
z is 1
Remaining list is [2, 3, 4, 5]
item is 3
z is 2
Remaining list is [2, 3, 4, 5]

而不是在我最近更新的列表[2,3,4,5]中从2重新启动循环，循环从第二个位置开始，即3.如何每次列表获取时从一开始就重新启动循环更新？谢谢！

Answer 1

您可以在while循环之外添加for循环，并在删除项目时重新启动for循环。

from random import randint
x = [1,2,3,4,5]

restart = True

while restart:
    restart = False
    for item in x:
        print("item is " + str(item))
        z = randint(1, 5)
        print("z is " + str(z))
        if z == item:
            x[:] = [item for item in x if item != z]
            print("Remaining list is " + str(x))
            restart = True
            break
        else:
            print("Remaining list is " + str(x))

# results
item is 1
z is 2
Remaining list is [1, 2, 3, 4, 5]
item is 2
z is 2
Remaining list is [1, 3, 4, 5]
item is 1
z is 1
Remaining list is [3, 4, 5]
item is 3
z is 4
Remaining list is [3, 4, 5]
item is 4
z is 3
Remaining list is [3, 4, 5]
item is 5
z is 5
Remaining list is [3, 4]
item is 3
z is 4
Remaining list is [3, 4]
item is 4
z is 1
Remaining list is [3, 4]

Answer 2

For循环不会按照您预期的方式运行，如果您可以使用while进行条件检查，那么效果会更好：

from random import randint
    x = [1, 2, 3, 4, 5]
    loops = 0
    while loops<len(x):
        item = x[loops]
        print("item is " + str(item))
        z = randint(1, 5)
        print("z is " + str(z))
        if z == item:
            x[:] = [item for item in x if item != z]
            print("Remaining list is " + str(x))
            continue
        else:
            print("Remaining list is " + str(x))
        loops += 1

输出：

item is 1
z is 1
Remaining list is [2, 3, 4, 5]
item is 2
z is 2
Remaining list is [3, 4, 5]
item is 3
z is 1
Remaining list is [3, 4, 5]
item is 4
z is 2
Remaining list is [3, 4, 5]
item is 5
z is 1
Remaining list is [3, 4, 5]

从列表中有条件地删除项目并从开始迭代到Python循环

2 个答案: