如何查找输入中的列表数？（python）

有人知道怎么做吗？ 我所拥有的只是

def lists(A: list) -> int:

    '''Return the total number of lists in A (including A itself).
    Each element of A and any  nested lists are either ints or other lists.

    Example:
    >>> lists([1, 2, 3])
    1
    >>> lists([[1], [2], [3]])
    4
    >>> lists([[[1, 2], [], 3]])
    4
    '''

---

这里有一种写的方法：

for i in range(0, len(A)):
    if (isinstance(A[i], list)):
        count=count+1
        return(lists(A[i]))
    else:
        B=A[i:]
return(count)

样本输出：

def numlists(lst, num = 1):
    for item in lst:
        if isinstance(item, list):
            num += numlists(item)
    return num

---

这里有一种写的方法：

for i in range(0, len(A)):
    if (isinstance(A[i], list)):
        count=count+1
        return(lists(A[i]))
    else:
        B=A[i:]
return(count)

样本输出：

def numlists(lst, num = 1):
    for item in lst:
        if isinstance(item, list):
            num += numlists(item)
    return num

---

您应该使用递归执行此操作：

print(numlists([1, 2, 3])) # 1
print(numlists([[1], [2], [3]])) # 4
print(numlists([[[1, 2], [], 3]])) # 4
print(numlists([[1,[2,3,[4,5,[6]]]],7,8,[9]])) # 6

---

您应该使用递归执行此操作：

print(numlists([1, 2, 3])) # 1
print(numlists([[1], [2], [3]])) # 4
print(numlists([[[1, 2], [], 3]])) # 4
print(numlists([[1,[2,3,[4,5,[6]]]],7,8,[9]])) # 6

---

这里有一个“肮脏”但简单的方法

def count_list(a):
    result = 0
    if isinstance(a, list):
        result += 1
    try:
        for b in a:
            result += count_list(b)
    except:
        pass
    return result

---

不需要递归

这里有一个“脏”但简单的方法

def count_list(a):
    result = 0
    if isinstance(a, list):
        result += 1
    try:
        for b in a:
            result += count_list(b)
    except:
        pass
    return result

def lists(l):
    '''
    Return the total number of lists in A (including A itself).
    Each element of A and any  nested lists are either ints or other lists.
    '''

    # convert the list to string and count the ['s
    # each [ is the start of a list, so the number of ['s equals
    # the number of lists
    nr_of_lists = str(l).count('[')

    # return the number of sublists
    return nr_of_lists

---

不需要递归

为什么要创建

B

？不需要这样做。既然您试图用递归来解决这个问题，那么您的基本情况是什么？什么是递归步骤？return（list（A[i]）是我的递归步骤，当没有更多的元素需要检查时，递归应该停止为什么要创建

B

？您不需要这样做。既然您试图用递归来解决这个问题，那么您的基本情况是什么？你的递归步骤是什么？return（list（A[i]）是我的递归步骤，当没有更多的元素需要检查时，递归应该停止。如果列表包含包含“[”？@timgeb的字符串，docstring声明它只包含子列表或intsOh，对，无需担心。尽管如此，还是相当脏；）如果列表包含包含['？@timgeb文档字符串声明它只包含子列表或intsOh，对吧，那么就不管了。尽管如此，还是相当脏；）

def lists(l):
    '''
    Return the total number of lists in A (including A itself).
    Each element of A and any  nested lists are either ints or other lists.
    '''

    # convert the list to string and count the ['s
    # each [ is the start of a list, so the number of ['s equals
    # the number of lists
    nr_of_lists = str(l).count('[')

    # return the number of sublists
    return nr_of_lists