CODEWITHSUNDEEP
pythonpython-3.xrecursionsum
cedricroux
cedricroux

Reputation: 21

list index out of range (Python 3)

Trying to experiment around with calculating the sum of a list of integers or floating-points using recursion. However, I am getting a 'list index out of range' error. Forgive me if this is silly but I am very new to this and still just playing around.

def sum(listOfNumbers):
    listOfNumbers == int or float
    if len(listOfNumbers) == 1:
        return listOfNumbers[0]
    else:
        return listOfNumbers[0] + sum(listOfNumbers[1:])


for (input, output) in [ ([1.0], 1), ([1,2], 3), ([1,2.0,3], 6) ]:
    result = 'PASSED' if sum(input) == output else 'FAILED'
    print('Test', input, result)
    print(sum([]))

Upvotes: 2

Views: 88

Answers (2)

hamiltoncw
hamiltoncw

Reputation: 23

You are passing an empty list as a parameter to the print(sum()) call you are making. Perhaps try passing input to see the result of your sum() function printed.

Upvotes: 1

Kevin
Kevin

Reputation: 76194

Your sum function does not work properly when the list has no elements in it. len(listOfNumbers) would equal zero, so the else clause executes, and tries to access listOfNumbers[0]. But a list with no elements doesn't have a zeroth element, so listOfNumbers[0] crashes with a "list index out of range" error.

Change your function so it handles lists of length zero. The simplest way to implement this change would be:

def sum(listOfNumbers):
    if len(listOfNumbers) == 0:
        return 0
    if len(listOfNumbers) == 1:
        return listOfNumbers[0]
    else:
        return listOfNumbers[0] + sum(listOfNumbers[1:])

... But you really only need one base case, so you can remove the middle clause:

def sum(listOfNumbers):
    if len(listOfNumbers) == 0:
        return 0
    else:
        return listOfNumbers[0] + sum(listOfNumbers[1:])

Upvotes: 0

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