function to sum range gives: 'int' object is not iterable

Question

I'm trying to write a function where the user gives the start and end of a range and the output is that range summed. This is my best try:

def sum_range (start, end):
    output = 0
    userange = range(start, end)
    for i in userange :
        sum(i, output)

    return output

I get the following error:

TypeError: 'int' object is not iterable

Try replacing `sum(i,output)` with `output += i`. It will add `i` to the current value of `ouput` — Nuageux, Aug 25 '17 at 12:39
Tip: virtually no function in Python accepts an output argument. Results are always *returned*. — deceze, Aug 25 '17 at 12:41

Moses Koledoye · Answer 1 · 2017-08-25T13:30:12.060

3

You probably don't need to build a range object. Simply take the sum of an arithmetic progression:

def sum_range(start, end):
    return (end - start) * (start + end - 1) // 2


print(sum_range(4, 30))
# 429.0
assert sum_range(4, 30) == sum(range(4, 30))

edited Aug 25 '17 at 13:30

answered Aug 25 '17 at 12:42

Moses Koledoye

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the only thing that's different is the datatype of the result returned (`float` vs `int`) – Ma0 Aug 25 '17 at 12:45
This presupposes that `start <= end`, which `range` handles implicitly. – chepner Aug 25 '17 at 12:45
@Ev.Kounis I added `2.` to handle integer division in Python 2. Nothing to it – Moses Koledoye Aug 25 '17 at 12:47
@chepner True. I seem to have assumed that from the OP since they're not passing a stride parameter – Moses Koledoye Aug 25 '17 at 12:47
@StefanPochmann I don't see how your point has to do with integer division. Rewriting the same expression as `((start + end - 1)/2) * (end - start)` will lead to problems only in Python 2. – Moses Koledoye Aug 25 '17 at 13:06
I'm somewhat surprised that `sum` isn't optimized to recognize when it gets a `range` object to compute the sum directly, instead of iterating. – chepner Aug 25 '17 at 13:06
@MosesKoledoye I think the point is, one of `(end - start)` and `(start + end - 1)` is guaranteed to be even, so since the product is computed first, there's no problem with simply dividing by 2. – chepner Aug 25 '17 at 13:07
@chepner Right. It's not immediately obvious for someone glancing through tho. – Moses Koledoye Aug 25 '17 at 13:08
1

@MosesKoledoye That's what comments are for :) – chepner Aug 25 '17 at 13:10

score 1 · Answer 2 · answered Aug 25 '17 at 12:40

1

You either need:

def sum_range(start, end):
    output = 0
    userange = range(start, end)
    for i in userange:
        output += i

    return output

or

def sum_range(start, end):
    return sum(range(start, end))

answered Aug 25 '17 at 12:40

quamrana

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The first option ends up just returning the value given for start. Not sure what that's about. – Jenna Maree Fae Aug 25 '17 at 12:52
@JennaMareeFae Are you sure you typed it correctly? Sounds like you put the `return` statement in the body of the `for` loop. – chepner Aug 25 '17 at 12:59
Yup, you're totally right I did. I seriously cannot wait to be less noobish at this. Thanks for you help! – Jenna Maree Fae Aug 25 '17 at 13:09

score 1 · Accepted Answer · answered Aug 25 '17 at 12:40

1

you can do :

def sum_range(start, end):
    return sum(range(start, end))

example :

>>> def sum_range(start, end):
...     return sum(range(start, end))
... 
>>> sum_range(10,14)
46

answered Aug 25 '17 at 12:40

Dadep

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Of course, once you boil it down like that… what's the point of `sum_range()` when you can write `sum(range())`…? – deceze Aug 25 '17 at 12:42

function to sum range gives: 'int' object is not iterable

3 Answers3