递归函数如何返回列表的元组？

Question

Question is: Define a recursive function named separate; it is passed a predicate and a list; it returns a 2-tuple whose 0 index is a list of all the values in the argument list for which the predicate returns True, and whose 1 index is a list of all the values in the argument list for which the predicate returns False. The call separate(predicate.is_positive,[1,-3,-2,4,0,-1,8]) returns ([1,4,8], [-3,-2,0,-1]) . Note 0 is not positive. Hint: like the fast version of the power function in the notes, you can define and bind (but not rebind) a local name or can write a nested function (like square in power) to help with the computation.

Here is the example of his power function:

def power(a,n):
    def square(n) : n*n
    if n == 0:
        return 1
    else:
       if n%2 == 1:
           return a*power(a,n-1)
       else:
           return square( power(a,n//2) )

My attempt:

def separate(p,l):
    l1=[]
    l2=[]
    if l == []:
        return [],[]
    else:
        if p(l[0]):
            l1=([l[0]]+map_pos(p,l[1:]))
            return l1,l2
        else:
            l2.extend([l[0]]+separate(p,l[1:]))
            return l1,l2

calling this function: print(predicate.is_positive,[1, -3, -2, 4, 0, -1, 8]) will gives me: TypeError: can only concatenate list (not "tuple") to list

Note predicate.is_positive is a function from predicate module which takes an int and return True if int is positive.

Can someone please help me with this? With actual code will be nice really appreciated.

Answer 1

This may be whay you are attempting to do

def separate(p, L):
    if L == []:
        return [], []

    l1, l2 = separate(p, L[1:])

    item = L[0]

    if p(item):
        l1.append(item)
    else:
        l2.append(item)
    return l1, l2    

It's not very efficient due to the L[1:] which creates a new list for each item

you can use a default argument to avoid making the slices

def separate(p, L, idx=0):
    if idx == len(L):
        return [], []

    l1, l2 = separate(p, L, idx + 1)

    item = L[idx]

    if p(item):
        l1.append(item)
    else:
        l2.append(item)
    return l1, l2    

It still looks clumsy. It's not really a task that calls for a recursive solution