Finding prime numbers up till a number - prime-factoring

Question

I am trying to list down all the prime numbers up till a specific number e.g. 1000. The code gets slower as the number increase. I am pretty sure it is because of the for loop where (number -1) is checked by all the prime_factors. Need some advise how I can decrease the processing time of the code for larger numbers. Thanks
import time
t0 = time.time()
prime_list = [2]
number = 0
is_not_prime = 0
count = 0
while number < 1000:
print(number)
for i in range (2,number):
count = 0
if (number%i) == 0:
is_not_prime = 1
if is_not_prime == 1:
for j in range (0,len(prime_list)):
if(number-1)%prime_list[j] != 0:
count += 1
if count == len(prime_list):
prime_list.append(number-1)
is_not_prime = 0
count = 0
break
number += 1
print(prime_list)
t1 = time.time()
total = t1-t0
print(total)

Answer 1

Your solution, on top of being confusing, is very inefficient - O(n^3). Please, use the Sieve of Eratosthenes. Also, learn how to use booleans.
Something like this (not optimal, just a mock-up). Essentially, you start with a list of all numbers, 1-1000. Then, you remove ones that are the multiple of something.
amount = 1000
numbers = range(1, amount)
i = 1
while i < len(numbers):
n = i + 1
while n < len(numbers):
if numbers[n] % numbers[i] == 0:
numbers.pop(n)
else:
n += 1
i += 1
print(numbers)
Finally, I was able to answer because your question isn't language-specific, but please tag the question with the language you're using in the example.