def prime_finder(n):
prime = []
num_list = []
for num in range(2, n + 1):
for x in range(1, num + 1):
if str(num / x)[-2:] == '.0':
num_list.append(x)
if len(num_list) == 2:
prime.append(num)
num_list = []
return prime
print(prime_finder(11))
I was nervous to post my code to the forum because I assumed that my answer was going to be overly complicated compared to others but it looks very similar to most of them!
I realize now I could’ve just used modulus to find the whole numbers from the division but oh well, this way is a tad bit different than most.
Thanks for highlighting the discrepancies in the prime_finder function and sharing your valuable insights. I’ll definitely investigate and work on finding an improved solution.
def prime_finder(n):
# Write your code here
prime_list = []
for i in range(2,n+1):
# if i == 2:
# prime_list.append(2)
# else:
# for j in range(2,n):
# if i % j == 0:
# break
# else:
# if j == i - 1:
# prime_list.append(i)
# return prime_list
flag = 0
for j in range(1, i+1):
if i % j == 0:
flag += 1
if flag == 2:
prime_list.append(i)
return prime_list
print(prime_finder(11))
def prime_finder(n):
nums = [i for i in range(1, n+1)]
primes = []
for num in nums[1:]:
counter = 0
for i in nums:
if num % i == 0:
counter += 1
if counter == 2:
primes.append(num)
return primes
I used a pretty simple solution. Considering I never want 1 or 0 in my finished list, I just started the counter at 2. I used two for loops giving me a time complexity of n^2 that can probably be streamlined a bit. My code runs each number separately and if it is ever evenly divisible by j (2 through that number minus one) it trips a flag and calls it true, which then doesn’t allow it to be added to the primes list.
def prime_finder(n):
primes = [];
for i in range(2, n+1):
flag = False;
for j in range(2, i):
if i % j == 0:
flag = True
if flag == False:
primes.append(i)
return primes
print(prime_finder(11))
This took me quite some time. My approach was as follows:
def prime_finder(n):
answer=[i for i in range(2,n+1) if is_prime(i)]
return answer
def is_prime(x):
for l in range(2,x):
if x==l:
continue
elif x%l==0:
return False
else:
continue
return True
print(prime_finder(11))
Following is a version of a solution for the prime finder challenge:
def find_primes(num):
# Create the list of numbers to check
primes = []
for n in range(2, num + 1):
# Check the number in the prime list
is_prime = True
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
is_prime = False
break
if is_prime:
primes.append(n)
return primes
print(find_primes(11))
def prime_finder(n):
# Write your code here
lst = [] # Empty list to hold the prime numbers
for i in range(2, (n + 1)): # Loop through the number. It is inclusive say 11
# print(i, end=" ")
# Output
# 2 3 4 5 6 7 8 9 10 11
# Loop through 2 3 4 5 6 7 8 9 10 11
for j in range(2, i): # i is the input number for inner loop from the outer loop
if i % j == 0:
# Not prime number
# Do nothing
break
else:
# Prime number
lst.append(i)
return lst
print(prime_finder(11))
def prime_finder(n):
prime_numbers = []
for i in range(2, n+1):
if [j for j in range(2, i) if i%j == 0] == []:
prime_numbers.append(i)
return prime_numbers
print(prime_finder(11))
hello guys, this is mine solution of this challenge:
def prime_finder(n):
primes = list(range(2, n+1))
for x in range(1, n+1):
for num in range(2, x):
if x!=2 and x%num == 0:
primes.remove(x)
break
return primes
print(prime_finder(11))
start by iterating through numbers to find what is divisible by one and itself
for numbers in list_of_numbers:
if numbers == 0 or numbers == 1:
continue
if numbers > 2 and numbers%2 == 0 :
continue #print(numbers)
if numbers > 3 and numbers%3 == 0 :
continue
if numbers > 5 and numbers%5 == 0:
continue
if numbers > 7 and numbers%7 == 0:
continue
if numbers > 11 and numbers%11 == 0:
continue
prime_numbers.append(numbers)
return prime_numbers
Hi, This was my first challenge and I created this code.
def prime_finder(n):
Write your code here
prime_numbers =
for i in range(2, n + 1):
x = 2
prime_bool = True
while(prime_bool == True and x < i):
if i % x == 0:
prime_bool = False
x += 1
if prime_bool == True:
prime_numbers.append(x)
return prime_numbers
remove all numbers that are divisible by any number between (excluding) 1 and itself
def prime_finder(n):
listOfPrimes = list(range(2,n+1))
for potentialPrime in range(2,n+1):
for divider in range(2, potentialPrime):
if potentialPrime % divider == 0:
listOfPrimes.remove(potentialPrime)
break
return listOfPrimes
unfortunately you can’t simply you the same list for the for loop. It skips some numbers, because IMHO the .remove is probably messing with the iterator. ↩︎
def prime_finder(n):
# Write your code here
new_list = []
for number in range(2, n+1):
pieces = 0
for i in range(1, number+1):
if number % i == 0: pieces +=1
if pieces == 2:
new_list.append(number)
return new_list
print(prime_finder(1000))
Here’s my solution (hard-coded empty lists for any number below 2, then a conditional comprehension within a comprehension to handle the prime-checking: if any int smaller than the current number but larger than 1 can divide evenly into it- i.e., modulo equals 0, which is false-y, then it is not prime):
def prime_finder(n):
if n < 2:
return []
else:
return [
x for x in range(2, n+1)
if all([x % i for i in range(x-1, 1, -1)])
]
I’m not very good at recursion and this problem is always my goto to refresh myself how to implement it
def prime_finder(n):
# Write your code here
primes = []
candidates = range(2, n+1)
count = n
for candidate in candidates:
if (prime(candidate)):
primes.append(candidate)
return primes
def prime(n, i=2):
if n == i:
return True
elif n % i == 0:
return False
return prime(n, i + 1)
print(prime_finder(11))
