I have this function py def get_fast_primes(): primes = [] for i in range(2, 1000): for prime in primes: if i%prime == 0: continue primes.append(i)
My intention was for the continue to continue the outer for loop. Is there a way to do that with a different statement such as end or exit?
#🔒 How can I improve my function?
shell sinew
vestal zealotBOT
I have this function ```py
def get_fast_primes():
primes = []
for i in r...
@shell sinew
Python help channel opened
solid canyon
def get_fast_primes():
primes = []
for i in range(2, 1000):
for prime in primes:
if i%prime == 0:
break
else:
primes.append(i)
This does what you want
When the if statement is True, it will end the inner loop with break.
The else is on the same line as the inner for. for/else means that when for completes without breaking, the else will execute.
@shell sinew
You could also use a flag:
def get_fast_primes():
primes = []
for i in range(2, 1000):
is_prime = True # Assume the number is prime
for prime in primes:
if i%prime == 0:
is_prime = False # Turns out, not prime ..
break
if is_prime:
primes.append(i)
Oh wow that was a fast answer, I'll look at it and I'm sure I'll have some questions 1 minute
shell sinew
```py
def get_fast_primes():
primes = []
for i in range(2, 1000):
...
I've never seen a for else block before, so the else executes when the for loop is not broken? py for prime in primes: if i%prime == 0: break else: primes.append(i)
solid canyon
Yes, it is a Python gimmick. You should probably use the second code snippet.
shell sinew
Maybe but I love the idea of this for else block 😂
The 2nd snippet is also very clean
The point of this was I wanted to optimise a function that I saw on youtube just for my own curiosity so I'll test both and tell you the results 😛
solid canyon
Sure, you can also go up untill the sqrt of i
You don't need to check all primes up until i
solid canyon
Seems like it, but Python should not be used if you want speed 😛
