w3resource

Python: Find the number of divisors of a given integer is even or odd

Python Basic - 1: Exercise-24 with Solution

Write a Python program to find the number of divisors of a given integer is even or odd.

Sample Solution:

Python Code:

def divisor(n):
  x = len([i for i in range(1,n+1) if not n % i])
  return x
print(divisor(15))
print(divisor(12))
print(divisor(9))
print(divisor(6))
print(divisor(3))

Sample Output:

4
6
3
4
2

Pictorial Presentation:

Python: Find the number of divisors of a given integer is even or odd

Flowchart:

Flowchart: Python - Find the number of divisors of a given integer is even or odd

Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Python Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous: Write a Python program that accept a positive number and subtract from this number the sum of its digits and so on. Continues this operation until the number is positive.
Next: Write a Python program to find the digits which are absent in a given mobile number.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



Python: Tips of the Day

How to make a flat list out of list of lists?

Given a list of lists l

flat_list = [item for sublist in l for item in sublist]

which means:

flat_list = []
for sublist in l:
    for item in sublist:
        flat_list.append(item)

is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:

flatten = lambda l: [item for sublist in l for item in sublist]

As evidence, you can use the timeit module in the standard library:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Ref: https://bit.ly/3dKsNTR