问题描述:

I was trying to make a program which would check a number for its greatest prime factor. I was almost done when this error message came up. `list index out of range.`

What does this mean and what is wrong with my code?

Here is my code.

`def is_prime(n):`

for i in range(3, n):

if n % i == 0:

return False

return True

def Problem3():

x = 144

n = 2

not_a_factor = []

z = []

prime = []

not_a_prime = []

while n < x:

if x%n == 0:

z.append(n)

else:

not_a_factor.append(n)

n = n + 1

for i in z:

if is_prime(z[i]) == True:

prime.append(z[i])

else:

not_a_prime.append(z[i])

print(prime)

Problem3()

You're just a bit off. for-loops in Python iterate an object and return it's entities, not a pointer/ index.

So just use the thing you get from each iteration of 'z'

(Side note: might want to check out this post, it'll help you make your is_prime function more performant)

```
def is_prime(n):
for i in range(3, n):
if n % i == 0:
return False
return True
def Problem3():
x = 144
n = 2
not_a_factor = []
z = []
prime = []
not_a_prime = []
while n < x:
if x%n == 0:
z.append(n)
else:
not_a_factor.append(n)
n =+ 1 # Python version of n++
for i in z: # Python for-loop is more like a say "for each", no need for the indexing
if is_prime(i): # no need for '=='; Python will 'truthify' your object
prime.append(i)
else:
not_a_prime.append(i)
print(prime)
Problem3()
```

`list index out of range`

- what does this mean?"The message `list index out of range`

refers to an `IndexError`

. Basically, this means that you are attempting to refer to an index in a list that doesn't exist.

Using your code as an example: you generate a list, `z`

, containing the factors of the number 144. You then iterate through each element in this list (`for i in z:`

). This means that for the:

- 1st iteration:
`i`

is the 1st element in`z`

, which is 2; - 2nd iteration:
`i`

is the 2nd element in`z`

, which is 3; - and so on.

Then, you attempt `if isprime(z[i]) == True:`

. So, as written, your program works like this:

- 1st iteration:
`if isprime(z[2]) == True:`

; - 2nd iteration:
`if isprime(z[3]) == True:`

; - ...
- 8th iteration:
`if isprime(z[16]) == True:`

At this point, your code prompts an `IndexError`

, because there are only 13 elements in `z`

.

One way to get the result that you want is to iterate through `range(len(z))`

instead of each element of `z`

. So, adjust the line `for i in z`

to `for i in range(len(z))`

.

Additionally, since `prime`

is a list, and you want to return the *greatest* prime factor, change `print(prime)`

to `print(max(prime))`

.

These two changes will give you the result you are looking for.

Overall, your program could be written much more efficiently. If you want a simple algorithm to determine the greatest prime factor of a number, here is one possibility:

```
def greatest_prime_factor(n):
greatest_prime = 1
for i in range(n + 1):
# iterate through range(n). We skip index 0.
if i == 0:
continue
# determine if the number i is a factor of n.
if n % i == 0:
# determine if the number i is prime.
for i_ in range(2,i):
if i % i_ == 0:
break
else:
# update greatest_prime.
greatest_prime = max(greatest_prime, i)
return greatest_prime
print (greatest_prime_factor(144))
```

This algorithm saves *a lot* of memory space when compared with your original program by not initializing lists to store numbers that are primes, that aren't primes, etc. If you *want* to store those values, that's up to you; there are just far more efficient possibilities for what you appear to want to achieve.

Check this link for some more info on algorithmic efficiency and how to think about time and space complexity.

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