为什么这个特定的函数返回“无”；我试图展示一个不断增长的列表？

Question

我想看到列表实时增长，当它完成后，我想通过索引查询附加的列表。

l=[]
def GrowingList(list):
    try:
        import random
        while True:
            rlist=lambda: print([random.randint(0,12)])
            list.append(rlist())
            print(list)
        return GrowingList(list)

    except KeyboardInterrupt:
        pass

这是一个成绩单：

Grown list
[None]
Grown list
[None, None]
Grown list
[None, None, None]
Grown list
[None, None, None, None]

当我查询新的编译列表时，它对每个索引都显示“无”：

>>> l
[None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None, None]

我是编程新手，所以冷静一下。该列表只是列表如何自动增长的练习功能。 'Grown list' 只是我想放在列表中的一个例子。

Answer 1

您在 lambda 中使用 print，它将返回 None。此外，您的 return 有一个不必要的递归。

编辑：您可以删除完整的 lambda 部分并直接将随机值附加到列表中。

您可以使用以下修改后的代码：

import random
l=[]
count=1
def GrowingList(list):
    try:
        count=1

        while True:
            rlist=random.randint(0,12)
            list.append(rlist)
            print(list)
            #below three line is added to break the code after 10 iteration you can remove them if you want to break on keyboard interrupt.  
            count+=1
            if count > 10:
                break
        return list

    except KeyboardInterrupt:
        pass

输出：

[1, 10]
[1, 10, 8]
[1, 10, 8, 11]
[1, 10, 8, 11, 2]
[1, 10, 8, 11, 2, 10]
[1, 10, 8, 11, 2, 10, 12]
[1, 10, 8, 11, 2, 10, 12, 5]
[1, 10, 8, 11, 2, 10, 12, 5, 0]
[1, 10, 8, 11, 2, 10, 12, 5, 0, 8]
[1, 10, 8, 11, 2, 10, 12, 5, 0, 8, 8]