(Python) List Index out of range error

Question

This function returns a ListIndex error on line 8. I try to calculate the sum of all squared values in the nList:

def oneVolume(n):
  j = 0
  Sum = 0
  nList = []
  for i in range (n):
    nList.append(2*random()-1)
    while j <= n:
      Sum = Sum +nList[j] # line with list index error
      j = j+1
  return Sum

i want to square each value in the list and then sum all the squared values — user3349164
– user3349164, Commented Mar 1, 2014 at 2:08
@user3349164 or in similar flavor as @loki, sum(map(lambda x: x**2, nList)) — Nitish
– Nitish, Commented Mar 1, 2014 at 2:13

mhlester · Accepted Answer · 2014-03-01 02:10:55Z

2

while j <= n:

Should be

while j < n;

range(5) == [0, 1, 2, 3, 4]. There is no 5 element (but there are five elements, because there's a 0)

But this isn't the pythonic approach. Iterate over the list instead of a range the size of the list:

def oneVolume(n):
    Sum = 0
    nList = []
    for i in range(n):
        nList.append(2*random()-1)
    for j in nList:
        Sum = Sum + j
    return Sum

But you can go a whole lot further in simplifying, using a generator and the built-in sum function:

def oneVolume(n):
    return sum(2*random()-1 for _ in range(n))

The generator is a lot like a list, [2*random()-1 for _ in range(n)], only more transient. That gets passed to the sum, which automatically does what you intended by the second half of your function.

answered Mar 1, 2014 at 2:10

mhlester

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Comments

Gary Walker · Accepted Answer · 2014-03-01 02:10:40Z

0

You have an off by 1 error.

l = []
l.append(1)
l.append(2)
l.append(3)

l will that 3 elements. They will be l[0] (containing 0), l[1], and l[2]

len(l) will be 3

referencing l[3] will be an index out of bounds.

answered Mar 1, 2014 at 2:10

Gary Walker

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