The code I'm using is:
import time
def divisores(num):
for i in range(1, int(num/2+1)):
if num % i == 0:
yield i
yield num
inicio = time.time()
d = divisores(47587950)
[print(i) for i in d]
print("Fim da execução: %s segundos" % round((time.time() - inicio),2))
The return: Fim da execução: 5.94 segundos . This code is slow.
Is there a more performative way to calculate the divisors of a number using Python?
Tip:
Do not use list comprehensions to perform functions like print() , append() etc ... will create you a temporary list full of None values, and has performance implications if this is very large, you can test this by doing print(print('hey')) whose print outer will print the return of print inner (which is None ), REF .
As for performance there is not much to do and your implementation is largely well done:
Way 1:
def divisores(num):
for i in range(1, num//2+1):
if num % i == 0:
yield i
yield num
num = 47587950
print(list(divisores(num)))
Way 2:
num = 47587950
print([i for i in range(1, num//2+1) if num%i==0])
As for performance, the first way can be a bit faster, and if you do not make it into the list it's almost instantaneous:
In the above demo:
using way 1 we have a time of 1.7259 secs
using way 2 we take 1.9021 secs
Remember that your computer should run faster than in Ideone.
Using the Way 1 function, the most appropriate way to print one at a time would be:
...
num = 47587950
for d in divisores(num):
print(d)
Numpy was born for this:
import numpy as np
import time
def divisores(num):
n = np.arange(1,num)
d = num % n
zeros = d == 0
print (n[zeros])
inicio = time.time()
divisores(47587950)
print("Fim da execução: %s segundos" % round((time.time() - inicio),2))
Output:
[ 1 2 3 5 6 9 10 15
18 25 30 45 50 75 90 150
225 450 105751 211502 317253 528755 634506 951759
1057510 1586265 1903518 2643775 3172530 4758795 5287550 7931325
9517590 15862650 23793975]
Fim da execução: 0.99 segundos
It took 0.99 seconds here.