CODEWITHSUNDEEP
python-3.xrecursion
Daniil Yefimov
Daniil Yefimov

Reputation: 1012

Subtraction(list) recursively in Python 3

I know how to add the numbers in a list recursively

def mysum(L):
    if not L:
        return 0
    else:
        return L[0] + mysum(L[1:])

Can you help me to build the subtraction recursively in the same manner(only one input - List).

mydiff([10,1,2,3])

Many thanks

Upvotes: 0

Views: 1178

Answers (2)

Nizam Mohamed
Nizam Mohamed

Reputation: 9220

if the sequence is (3,2,1), the result is 0. But subtraction is going in reverse direction, i.e. 2 -1 than 3 - (2-1) if you use return L[0] - mysum(L[1:]).

So we have to use an accumulator which keeps the difference of first and second element of a sequence. 

def mydiff(L, acc=None):
    if not L:
        return acc
    else:
        if acc is None:
            acc = L[0] - L[1]
            L = L[2:]
        else:
            acc -= L[0]
            L = L[1:]
        return mydiff(L, acc)

More simpler one;

def mysub(seq, acc=None):
    seq = iter(seq)
    if acc is None:
        item = next(seq, None)
        if item is None:
            return acc
        acc = item

    item = next(seq, None)
    if item is None:
        return acc
    return mysub(seq, acc - item)

Upvotes: 1

uselpa
uselpa

Reputation: 18917

The calculation to be done is

>>> 10-(1+(2+(3+0)))
4

so the first operation is a subtraction, but the rest of the operations remain additions. I think you need to use an inner function if you're not allowed to add a second parameter to the main function:

def mydiff(L):
    def sub(L, first=False):
        if not L:
            return 0
        else:
            if first:
                return L[0] - sub(L[1:])
            else:
                return L[0] + sub(L[1:])
    return sub(L, True)

Testing:

>>> mydiff([10,1,2,3])
4

Upvotes: 1

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