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This approach includes building a list of all the previous primes until it reaches the nth number, after which it quits the loop and return the last number in the list.
def prime(number):
if number == 0:
raise ValueError('there is no zeroth prime')
counter = 2
primes = [2]
while len(primes) < number:
counter += 1
if all(counter % test != 0 for test in range(2, counter)):
primes.append(counter)
return primes[-1]
Efficiency can be improved slightly by reducing the factor check to the square root of the number...
...
if all(counter % test != 0 for test in range(2, int(counter ** 0.5) + 1)):
...
... or even better, checking whether a new number is merely not divisible by any of the primes (in which case it's a prime itself):
...
if all(counter % test != 0 for test in primes):
...
Instead of building the list, it's more memory efficient to just save the number of primes found until now, and return the currently stored number when the nth prime is found. However, this elongates the code.
def prime(number):
if number == 0:
raise ValueError('there is no zeroth prime')
counter = 2
prime_count = 0
while True:
isprime = True
for test in range(2, int(counter ** 0.5) + 1):
if counter % test == 0:
isprime = False
break
if isprime:
prime_count += 1
if prime_count == number:
return counter
counter += 1
Tip: you can use for... else to make your code more idiomatic here.
...
for test in range(2, int(counter ** 0.5) + 1):
if counter % test == 0:
break
else:
prime_count += 1
if prime_count == number:
return counter
The else block is executed if the for loop completes normally - that is, without breaking.
Read more on [for/else][for-else]
