python - 遍历 *args 重置计数器

Question

我正在处理一些编造的代码以尝试理解 *args 功能，但我很难让我的 counter 变量不被重置。

在第 4 个项目之后，编号重置回 0，但我希望它在整个 args 中继续。

lst = ['apple', 'banana', 'orange', 'lemon']
lst2 = ['apple2', 'banana2', 'orange2', 'lemon2']
lst3 = ['apple3', 'banana3', 'orange3', 'lemon3']


def generateMenu(*args):
    counter = 0

    for i in args:
        def recurse(l, counter):
            for i in l:
                counter += 1
                if isinstance(i, (list, tuple)):
                    recurse(i, counter)
                else:
                    print("{}. {}.".format(counter, i))
        recurse(i, counter)


generateMenu(lst, lst2, lst3)

输出:

1. apple.
2. banana.
3. orange.
4. lemon.
1. apple2.
2. banana2.
3. orange2.
4. lemon2.
1. apple3.
2. banana3.
3. orange3.
4. lemon3.

Answer 1

你应该让counter成为一个nonlocal(nonlocal是在Python3中引入的)

lst = ['apple', 'banana', 'orange', 'lemon']
lst2 = ['apple2', 'banana2', 'orange2', 'lemon2']
lst3 = ['apple3', 'banana3', 'orange3', 'lemon3']


def generateMenu(*args):
    counter = 0

    for i in args:
        def recurse(l):
            nonlocal counter
            for i in l:
                counter += 1
                if isinstance(i, (list, tuple)):
                    recurse(i)
                else:
                    print("{}. {}.".format(counter, i))
        recurse(i)


generateMenu(lst, lst2, lst3)

或者，从 recurse 返回 counter

lst = ['apple', 'banana', 'orange', 'lemon']
lst2 = ['apple2', 'banana2', 'orange2', 'lemon2']
lst3 = ['apple3', 'banana3', 'orange3', 'lemon3']

def generateMenu(*args):
    counter = 0

    for i in args:
        def recurse(l, counter):
            for i in l:
                counter += 1
                if isinstance(i, (list, tuple)):
                    counter = recurse(i, counter)
                else:
                    print("{}. {}.".format(counter, i))
            return counter
        counter = recurse(i, counter)


generateMenu(lst, lst2, lst3)