Andy K Andy K - 1 year ago 60
Python Question

Why this way for the recursion and what does it mean?

I'm trying to understand this piece of code

def listSum(ls):
if not ls:
return 0

return ls[0] + listSum(ls[1:])

Two things bug me a bit

  1. if not ls:
    - What does it mean? As nothing has been specified, what is it looking for?

  2. ls[0] + listSum(ls[1:])
    - I've tried using
    only and I have running error. Why? Why should I stick to
    ls[0] + listSum(ls[1:])

Answer Source

Imagine it's called...

listSum([3, 5, 3, 7])

It get down to...

return ls[0] + listSum(ls[1:])

...which is evaluated as...

return 3 + listSum([5, 3, 7])

Then listSum([5, 3, 7]) as 5 + listsum([3, 7]).

Then listSum([3, 7]) as 3 + listsum([7]).

Then listSum([7]) as 7 + listsum([]), which is where the not ls kicks in and returns 0.

So, the complete thing is then evaluated as...

3 + 5 + 3 + 7 + 0


  • listSum(ls[1:]) is shortening the list each time - reducing the complexity of the remaining problem, until an invocation of listSum receives an empty list: the trivial case that if not ls: return 0 handles

  • recursive solutions are often made up of two parts: a solution for the simplest possible input (or sometimes, a couple very simple cases), and something saying how to take an arbitrarily complicated input and reduce it to a formula making one or more recusive calls that are each at least a little bit simpler, always moving towards that simplest possible input

  • "I've tried using listSum(ls[0:])" - ls[0:] returns a copy of ls - it would ask for the same list to be processed ad infinitum

  • it's arguably more intuitive to explicitly handle a list of one element - if len(ls) == 1: return ls[0] - but then if anyone called listSum([]) you'd try to access [0] and raise an exception; handling the empty list makes the function safer to use (if "0" is meaningful as the sum of an empty list in your specific application), and happily the len(ls) == 1 case then "just works"

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