def a(n):
return max([len(n)] + [a(i) for i in n]) if isinstance(n, list) else 0
这是我最近的一个测试，我就是无法理解列表推导式。所以基本上这个函数应该返回最大列表的长度（根据正确答案的假设）。如果不是因为函数中的这部分代码，我可能已经理解了：
+ [a(i) for i in n])
这个函数计算树中最大节点的长度（树是通过列表嵌套列表实现的）。也许通过一些重命名和重新排列，它会更加清晰：
def longest_list_in_tree(tree):
if not isinstance(tree, list):
return 0 # This is a leaf-value of the tree, it has "length 0"
own_length = len(tree)
longest_descendant_of_each_subtree = [
longest_list_in_tree(subtree) for subtree in tree
]
return max([own_length] + longest_descendant_of_each_subtree)
该程序的目标是找出列表中最长列表的长度。假设您将一个列表的列表作为输入。
b = [[1], [1, 2], [3], [[1, 2, 3], [1]]]
现在，要找到内部列表的长度，我们首先必须遍历当前列表。但是，如果最初的输入本身不是一个列表怎么办？我们将该元素的长度返回为0。这个检查是由以下部分完成的：
... if isinstance(n, list) else 0
如果n不是列表的实例，我们返回0。现在我们确定当前项是一个列表，我们获取该列表的长度len(n)。然后，我们像这样迭代列表：
[a(i) for i in n]
我们取得n的每个元素并递归调用a。现在，如果该元素是一个列表，则期望a(i)返回其内最长列表的长度。因此，我们收集每个元素中最长列表的长度，将它们创建成一个列表，并将其与n的长度连接起来。这个连接是通过进行的。
[len(n)] + [a(i) for i in n]
我们通过用方括号将len(n)括起来，使其成为一个列表，然后使用+运算符将我们从所有元素中获取的长度附加到其中。因此，在每次递归函数调用中，我们得到类似于以下内容：
[length of current list,
max length of sub elements in element 1 of n,
max length of sub elements in element 2 of n,
max length of sub elements in element 3 of n,
...
]
然后我们找到所有元素中的最大值，并将其返回给调用者。
为了更好地理解这个程序，我创建了一个装饰器来记录输入和相应的输出。这可能有助于您更好地理解递归。
def logger(f):
def wrapper(args, level):
print "\t\t" * level + "Entered with input ", args
temp = f(args, level)
print "\t\t" * level + "Leaving with length", temp
return temp
return wrapper
@logger
def a(n, lev):
return max([len(n)] + [a(i, lev+1) for i in n]) if isinstance(n, list) else 0
b = [[1], [1, 2], [3], [[1, 2, 3], [1]]]
print a(b, 0)
输出结果如下：
Entered with input [[1], [1, 2], [3], [[1, 2, 3], [1]]]
Entered with input [1]
Entered with input 1
Leaving with length 0
Leaving with length 1
Entered with input [1, 2]
Entered with input 1
Leaving with length 0
Entered with input 2
Leaving with length 0
Leaving with length 2
Entered with input [3]
Entered with input 3
Leaving with length 0
Leaving with length 1
Entered with input [[1, 2, 3], [1]]
Entered with input [1, 2, 3]
Entered with input 1
Leaving with length 0
Entered with input 2
Leaving with length 0
Entered with input 3
Leaving with length 0
Leaving with length 3
Entered with input [1]
Entered with input 1
Leaving with length 0
Leaving with length 1
Leaving with length 3
Leaving with length 4
4
def b(n):
return [len(n)]+[b(i) for i in n] if isinstance(n,list) else 0
例子：
>>> x = [1,2,3,4,5,[1,2,3,4,5,6,7,8,9],7,[1,2,[1,2,3,4,5,6,7,8,9,10,11,12,13,14],4,5,6,[1,2]]]
>>> a(x)
14
>>> b(x)
[8, 0, 0, 0, 0, 0, [9, 0, 0, 0, 0, 0, 0, 0, 0, 0], 0, [7, 0, 0, [14, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], 0, 0, 0, [2, 0, 0]]]
x = [a,b,c]
y = [d,e,f]
z = x + y;
。
def a(n):
'''return the largest length of a list or sublists in that list'''
if isinstance(n, list):
# if n is a list, return the max of the len of the list
# for all subitems in the list, perform a(subitem) recursively.
# so if a sublist is longer than the list, return the largest len.
return max([len(n)] + [a(i) for i in n])
else:
# that subitem isn't a list, so return 0, which won't factor into the max calc.
return 0
def max_len_list_of_lists(a_list):
'''return the largest length of a list or sublists in that list'''
...
a([2,3,4,5]) = 4 # [2,3,4,5]
a([1, [2,3,4,5], 3]) = 4 # [2,3,4,5]
a([1, [2,3,4,5], [3, [7,8,9,0,1]]]) = 5 # [7,8,9,0,1]
a(i)。 - thefourtheye