Python 3 list index out of range

Question

I am completely new to Python and I have an error I don't understand. I have a function called mult1 which I want to take a list "a" and multiply each element together and output the total so that a list of [1,2,3,4] would output 24. I am getting an error "list index is out of bounds" at line 4

def mult1(a):
   n=1
   for i in a:
      n=n*a[i-1]
   print(n)

My function works with the input [1,2,3,4] but not with [7,8,1,2]. Why is this?

`for i in a` iterates over the elements of the list, not their indices. You need `n = n*i` — John Coleman, Nov 12 '15 at 20:20

AbtPst · Answer 1 · 2015-11-12T20:27:57.790

1

you are actually iterating over the values and not the indices. hence i refers to the value in the list

def mult1(a):
   n=1
   for i in a:
      n=n*i
   print(n)

edited Nov 12 '15 at 20:27

answered Nov 12 '15 at 20:20

AbtPst

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Please, add an explanation on the behaviour for him to lear, not just post the solution code – Netwave Nov 12 '15 at 20:23
1

Thanks, I wasn't aware that "i" would be the value in "a". I assumed it acted as an index – Porteous96 Nov 12 '15 at 20:24
@Porteous96: Python is not JavaScript (fortunately). – GingerPlusPlus Nov 12 '15 at 20:28
thanks @DanielSanchez, i added some explanation. please remove the downvote. – AbtPst Nov 12 '15 at 20:28
why the downvote now? – AbtPst Nov 12 '15 at 20:30

7stud · Answer 2 · 2015-11-12T20:28:45.340

My function works with the input [1,2,3,4] but not with [7,8,1,2]. Why is this

When you read the first number 7, then 7 is assigned to i. Then you try to lookup the number at a[i-1] => a[7-1] =>a[6]--and 6 is out of bounds.

You don't need to get the index of the numbers in the list to multiply all the numbers together:

def mult1(nums):
    total = 1

    for num in nums:
        total *= num

    print(total)

mult1([7,8, 1, 2])

--output:-
112

score -1 · Accepted Answer · answered Nov 12 '15 at 20:28

-1

the python for loop relies on the iteration protocol so when you write

for i in a:
  print(a)

you are actually accessing those indices hence there's no need to use the [i] (indexing ) notation like you would do in a conventional c/c++/java for-loop.

to answer your question

def mult1(a):
  n = 1
  for i in a:
    n *= i
  return n

answered Nov 12 '15 at 20:28

Ishan Khare

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why was this answer downvoted? please provide suggestions to improve if you downvote – AbtPst Nov 12 '15 at 20:42
@7stud yes you are actually accessing those indices, if not then please provide the correct answer – Ishan Khare Nov 18 '15 at 20:21

DeepSpace · Answer 4 · 2015-11-12T20:34:10.590

I will start with explaining the behavior you have observed, and finish by showing a much better way of achieving your goal.

When you use [1, 2, 3, 4] the for loop evaluates to n=n*a[0], n=n*a[1], n=n*a[2] and n=n*a[3] and that works as you have 4 elements in the list.

However, when your list is [7, 8, 1, 2] the first iteration of the for loop evaluates to n=n*a[6] which throws an Index out of bounds exception has it tries to access the 7th element of a 4 element list.

There are many simple ways to multiply all the elements in a list. Two of them:

print (functools.reduce(lambda x,y: x*y, [7, 8, 1, 2]))
>> 112

####################

li = [7, 8, 1, 2]
mul = 1
for num in li:
    mul *= num
print (mul)
>> 112

Python 3 list index out of range

4 Answers4