In this post I am going to run through a function in Python that can quickly find all the Prime numbers below a given value. For example, if I passed the function a value of 1,000,000, it would find all the prime numbers below 1,000,000!
If you are not sure what a Prime number is, it is a number that can only be divided wholly by itself and one so 7 is a prime number as no other numbers apart from 7 or 1 divide cleanly into it 8 is not a prime number as while eight and one divide into it, so do 2 and 4
Let us get into it!
First let us start by setting up a variable that will act as the upper limit of numbers we want to search through. We will start with 20, so we are essentially wanting to find all prime numbers that exist that are equal to or smaller than 20
n = 20
The smallest true Prime number is 2, so we want to start by creating a list of numbers that need checking so every integer between 2 and what we set above as the upper bound which in this case was 20. We use n+1 as the range logic is not inclusive of the upper limit we set there
Instead of using a list, we are going to use a set. The reason for this is that sets have some special functions that will allow us to eliminate non-primes during our search. You will see what I mean soon…
number_range = set(range(2, n+1))
Let us also create a place where we can store any primes we discover. A list will be perfect for this job
primes_list = []
We are going to end up using a while loop to iterate through our list and check for primes, but before we construct that I always find it valuable to code up the logic and iterate manually first. This means I can check that it is working correctly before I set it off to run through everything on its own
So, we have our set of numbers (called number_range to check all integers between 2 and 20. Let us extract the first number from that set that we want to check as to whether it is a prime. When we check the value we are going to check if it is a prime…if it is, we are going to add it to our list called primes_list…if it is not a prime we do not want to keep it
There is a method which will remove an element from a list or set and provide that value to us, and that method is called pop
print(number_range)
>>> {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
If we use pop, and assign this to the object called prime it will pop the first element from the set out of number_range, and into prime
prime = number_range.pop()
print(prime)
>>> 2
print(number_range)
>>> {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
Now, we know that the very first value in our range is going to be a prime…as there is nothing smaller than it so therefore nothing else could possible divide evenly into it. As we know it is a prime, let us add it to our list of primes…
primes_list.append(prime)
print(primes_list)
>>> [2]
Now we are going to do a special trick to check our remaining number_range for non-primes. For the prime number we just checked (in this first case it was the number 2) we want to generate all the multiples of that up to our upper range (in our case, 20).
We are going to again use a set rather than a list, because it allows us some special functionality that we will use soon, which is the magic of this approach.
multiples = set(range(prime*2, n+1, prime))
Remember that when created a range the syntax is range(start, stop, step). For the starting point - we do not need our number as that has already been added as a prime, so let us start our range of multiples at 2 * our number as that is the first multiple, in our case, our number is 2 so the first multiple will be 4. If the number we were checking was 3 then the first multiple would be 6 - and so on.
For the stopping point of our range - we specify that we want our range to go up to 20, so we use n+1 to specify that we want 20 to be included.
Now, the step is key here. We want multiples of our number, so we want to increment in steps of our number so we can put in prime here
Let us have a look at our list of multiples…
print(multiples)
>>> {4, 6, 8, 10, 12, 14, 16, 18, 20}
The next part is the magic I spoke about earlier, we are using the special set functionality difference_update which removes any values from our number range that are multiples of the number we just checked. The reason we are doing this is because if a number is a multiple of anything other than 1 or itself then it is not a prime number and can remove it from the list to be checked.
Before we apply the difference_update, let us look at our two sets.
print(number_range)
>>> {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
print(multiples)
>>> {4, 6, 8, 10, 12, 14, 16, 18, 20}
difference_update works in a way that will update one set to only include the values that are different from those in a second set
To use this, we put our initial set and then apply the difference update with our multiples
number_range.difference_update(multiples)
print(number_range)
>>> {3, 5, 7, 9, 11, 13, 15, 17, 19}
When we look at our number range now, all values that were also present in the multiples set have been removed as we know they were not primes
This is amazing! We have made a massive reduction to the pool of numbers that need to be tested so this is really efficient. It also means the smallest number in our range is a prime number as we know nothing smaller than it divides into it…and this means we can run all that logic again from the top!
Whenever you can run sometime over and over again, a while loop is often a good solution.
Here is the code, with a while loop doing the hard work of updated the number list and extracting primes until the list is empty.
Let us run it for any primes below 1000…
n = 1000
# number range to be checked
number_range = set(range(2, n+1))
# empty list to append discovered primes to
primes_list = []
# iterate until list is empty
while number_range:
prime = number_range.pop()
primes_list.append(prime)
multiples = set(range(prime*2, n+1, prime))
number_range.difference_update(multiples)
Let us print the primes_list to have a look at what we found!
print(primes_list)
>>> [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]
Let us now get some interesting stats from our list which we can use to summarise our findings, the number of primes that were found, and the largest prime in the list!
prime_count = len(primes_list)
largest_prime = max(primes_list)
print(f"There are {prime_count} prime numbers between 1 and {n}, the largest of which is {largest_prime}")
>>> There are 168 prime numbers between 1 and 1000, the largest of which is 997
Amazing!
The next thing to do would be to put it into a neat function, which you can see below:
def primes_finder(n):
# number range to be checked
number_range = set(range(2, n+1))
# empty list to append discovered primes to
primes_list = []
# iterate until list is empty
while number_range:
prime = number_range.pop()
primes_list.append(prime)
multiples = set(range(prime*2, n+1, prime))
number_range.difference_update(multiples)
prime_count = len(primes_list)
largest_prime = max(primes_list)
print(f"There are {prime_count} prime numbers between 1 and {n}, the largest of which is {largest_prime}")
Now we can just pass the function the upper bound of our search and it will do the rest!
Let us go for something large, say a million…
primes_finder(1000000)
>>> There are 78498 prime numbers between 1 and 1000000, the largest of which is 999983
That is pretty cool!
I hoped you enjoyed learning about Primes, and one way to search for them using Python.
Important Note: Using pop() on a Set in Python
In the real world - we would need to make a consideration around the pop() method when used on a Set as in some cases it can be a bit inconsistent.
The pop() method will usually extract the lowest element of a Set. Sets however are, by definition, unordered. The items are stored internally with some order, but this internal order is determined by the hash code of the key (which is what allows retrieval to be so fast).
This hashing method means that we cannot 100% rely on it successfully getting the lowest value. In very rare cases, the hash provides a value that is not the lowest.
Even though here, we are just coding up something fun - it is most definitely a useful thing to note when using Sets and pop() in Python in the future!
The simplest solution to force the minimum value to be used is to replace the line…
prime = number_range.pop()
…with the lines…
prime = min(sorted(number_range))
number_range.remove(prime)
…where we firstly force the identification of the lowest number in the number_range into our prime variable, and following that we remove it.
However, because we have to sort the list for each iteration of the loop in order to get the minimum value, it is slightly slower than what we saw with pop()!