如何在 python 中找出两个具有相同结构的列表？

Question

Define a procedure, same_structure, that takes two inputs. It should output True if the lists have the same structure, and False otherwise. Two values, p and q have the same structure if:

Neither p or q is a list.

Both p and q are lists, they have the same number of elements, and each
element of p has the same structure as the corresponding element of q.

EDIT: To make the picture clear the following are the expected output

same_structure([1, 0, 1], [2, 1, 2])
    ---> True
same_structure([1, [0], 1], [2, 5, 3])
    ---> False
same_structure([1, [2, [3, [4, 5]]]], ['a', ['b', ['c', ['d', 'e']]]])
    ---> True
same_structure([1, [2, [3, [4, 5]]]], ['a', ['b', ['c', ['de']]]])
    ---> False 

I thought recursion would be best to solve this problem in python I have come up with the following code but its not working.

def is_list(p):
    return isinstance(p, list)

 def same_structure(a,b):
    if not is_list(a) and not is_list(b):
        return True
    elif is_list(a) and is_list(b):
        if len(a) == len(b):
            same_structure(a[1:],b[1:])
    else:
        return False

Answer 1

Instead of same_structure(a[1:],b[1:]) , you need to check item pair of a and b one by one

def is_list(p):
    return isinstance(p, list)

def same_structure(a, b):
    if not is_list(a) and not is_list(b):
        return True
    elif (is_list(a) and is_list(b)) and (len(a) == len(b)):
        return all(map(same_structure, a, b)) # Here
    return False

Answer 2

You're missing one case, and forgot to return in the second case. Notice that is not necessary to explicitly compare the lengths of the lists, as the first case takes care of this - if one of the lists is empty and the other not, it's because one list had fewer elements than the other:

def same_structure(a, b):
    if a == [] or b == []:  # one of the lists is empty
        return a == b       # are both of the lists empty?
    elif is_list(a[0]) != is_list(b[0]):
        return False        # one of the elements is a list and the other is not
    elif not is_list(a[0]): # neither element is a list
        return same_structure(a[1:], b[1:])
    else:                   # both elements are lists
        return same_structure(a[0], b[0]) and same_structure(a[1:], b[1:])

Answer 3

Recursion would be a good idea, but not the way you've suggested it. First off (and this may be only a typo), you don't actually return anything here:

if len(a) == len(b):
    same_structure(a[1:],b[1:])

Second, you should recursively deal with each element, not each sublist. ie.:

if len(a) == len(b):
    for i in range(len(a)):
        if not same_structure(a[i], b[i]):
            return False
    return True
else:
    return False

Hope this helps.

Answer 4

Since the specification says that the input are two lists, you can iterate the lists inside your function without further checks, and only do recursive calls if you encounter sublists:

def same_structure(a, b):
    if len(a) != len(b):
        return False
    return all(is_list(x) and is_list(y) and same_structure(x, y) or
               not is_list(x) and not is_list(y)
               for x, y in zip(a, b))

Answer 5

You can try this way too also, check the type of both lists & length of both lists and do this recursively for sub-lists of both list.

def same_structure(a,b):
    return isinstance(a, list) and isinstance(b, list) and len(a) == len(b)  
    and all(same_structure_as(c, d) for c, d in zip(a, b) if isinstance(c, list))