How can I solve the error in this algorithm code (TypeError: 'NoneType' )?

Question

I’m having trouble understanding and applying this exercise on my own with Python. I am getting this:

TypeError: '<' not supported between instances of 'int' and 'NoneType'

This is the exercise Algorithm:

Definition: An integer is said to be perfect if it is equal to the sum
of all its divisors. Examples: 6 and 28 are perfect since 6 = 1+2+3
(knowing that 1, 2 and 3 are the divisors of 6 less than 6) 28=
1+2+4+7+14 (knowing that 1, 2, 4, 7 and 14 are the divisors of 28 less
than 28)

Create a function liste_divisors(), which accepts an
integer N as a parameter and returns the list of its divisors less
than N (1 included).

Make a function is perfect(), which accepts a (positive) integer N
as a parameter and returns “True” if it is perfect and “False”
otherwise (use the function from the 1st question)

Create a Perfect List() function, which accepts a Limit parameter,
then returns a list containing the perfect numbers less than Limit

I’m having trouble with the last question

This is my attempt so far, but there are errors. How can I correct these errors?

def liste_diviseur(N):
   t = []
   for i in range(1,N):
      if(N%i == 0):
         t.append(i)
   return t

def est_parfait(M):
   s = 0
   for i in liste_diviseur(M):
      s += i 
   if(s == M):
      return True
   else:
      return False

def liste_parfait(Limite):
   t = []
   lis = print(liste_diviseur(Limite))
   if(Limite<lis):
      t.append(Limite)
   return t
m = int(input('Giving an number :'))
print(liste_parfait(m))

Asked By: user20727127

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Source

Answer 1

Your error appears to be when you set lis = print(liste_diviseur(Limite)). print() is a void function which will return null. Removing the print() should fix your problem

Answered By: justin

Answer 2

You need to create a range and then check if each number in the range is perfect.

def liste_diviseur(n):
   return [i for i in range(1,n//2+1) if n%i==0]


def est_parfait(n):
   return n == sum(liste_diviseur(n))
         

def liste_parfait(limite): 
   return [i for i in range(2, limite) if est_parfait(i)]
   

test = int(input('Giving an number: '))
print(liste_parfait(test))
# output for 8129
[6, 28, 496, 8128]

Answered By: bn_ln

Answer 3

The main part is generating the divisors. You can find some ideas for efficiency e.g. here.

For large numbers, one would use factorization of n into its factors (see e.g. an answer I wrote for another problem), and then generate all divisors by combining those factors.

Here, we’ll go for simplicity:

from math import isqrt

def divisors(n):
    m = isqrt(n)
    for i in range(1, m + 1):
        if n % i == 0:
            yield i
            j = n // i
            if i < j < n:
                yield n // i

Example:

>>> list(divisors(36))
[1, 2, 18, 3, 12, 4, 9, 6]

Note that divisors() returns a generator, not a list, and the divisors are produced out of order (more precisely: as pairs i and n // i). But this doesn’t matter to see if a number if perfect or not:

def is_perfect(n):
    return sum(divisors(n)) == n

>>> is_perfect(5)
False

>>> is_perfect(6)
True

Now for a list of perfect numbers:

>>> [n for n in range(1, 10_000) if is_perfect(n)]
[6, 28, 496, 8128]

Answered By: Pierre D

How can I solve the error in this algorithm code (TypeError: 'NoneType' )?

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