def prime_finder(n):
“”"
Verilen bir sayıya kadar (dahil) olan tüm asal sayıları bulur.
Args:
n (int): Asal sayıların bulunacağı üst sınır.
Returns:
list: 1'den n'e (dahil) kadar olan asal sayıların listesi.
"""
primes = [] # Asal sayıları saklamak için boş bir liste oluşturun
for num in range(2, n + 1): # 2'den n'e kadar olan her sayı için döngü
# Herhangi bir sayı 2 ile karekökü arasındaki tüm sayılardan tam olarak bölünmüyorsa
if all(num % i != 0 for i in range(2, int(num**0.5) + 1)):
primes.append(num) # Asal sayıları listeye ekleyin
return primes
Post mine and ready to check cool answers from others
def prime_finder(n):
# Write your code here
prime_n = [2]
# check for exception
if type(n) != int:
return 'you have to input only one integer.'
elif n <= 1:
return 'cannot find the prime numbers for less than 2.'
elif n == 2:
return prime_n
else:
# iterate devidend from 3 to n
for d1 in range(3, n+1):
# iterate devider from 2 to d1
for d2 in range(2, d1+1):
# prime numbers only divisible by 1 and itself
if d1 % d2 == 0 and d1 != d2:
#print('{d1} % {d2} == 0 will break the loop'.format(d1=d1, d2=d2))
break
elif d1 % d2 == 0 and d1 == d2:
#print('find a prime: {d1}'.format(d1=d1))
prime_n.append(d1)
return prime_n
print(prime_finder(11))
for i in range(12, 25):
print(prime_finder(i))
print(prime_finder(1))
print(prime_finder('x'))
def prime_finder(n):
prime = []
isprime = ""
for i in range(0, n + 1):
if i == 2:
prime.append(2)
for j in range(2, i):
if i % j != 0:
isprime = "yes"
else:
isprime = "no"
break
if isprime == "yes":
prime.append(i)
return prime
print(prime_finder(11))
Just to let others compare and take it as an example if suites better:
def prime_finder(n):
prime_num = [] #Empty List
for num in range(2, n+1): #Loop through the number in n, except 1
isprime = True # Create a boolean for check point
for i in range(2, int(num ** 0.5) + 1): # Loop all possibilities of divisibles till the square root of num
if num % i == 0: # check if num is divisible by i
isprime = False # if divisible, set the breaker to False
break # the loop stops
if isprime: # if isprime is True the prime_num list will append num
prime_num.append(num)
return prime_num
print(prime_finder(11))
#create a list
primes = []
# range 2-12
for num in range(2, n+1):
for i in range(2, num):
if (num % i) == 0:
break
else:
primes.append(num)
return primes
Just checking if a number is dividable by a prime number smaller than itself.
First i add every number to the prime_numbers and just remove it, when I find out it is no prime number
def prime_finder(n):
# Write your code here
prime_numbers = []
for i in range(2, n + 1):
prime_numbers.append(i)
for number in prime_numbers:
if i != number and i % number == 0:
prime_numbers.remove(i)
break
return prime_numbers
print(prime_finder(11))
def prime_finder(n):
prime_nums = []
worked_nums = []
for i in range(1, n+1):
worked_nums = []
for j in range(1, i+1):
if i % j == 0:
worked_nums.append(j)
if worked_nums == [1, i]:
prime_nums.append(i)
return prime_nums
print(prime_finder(11))
Hi! Here is my try with this prime numbers problem.
First, I created a list with the numbers in the range we want to select the prime numbers.
Then I made the code loops through each number in this list and divide it by every number that came before. If this division is completed with no residue, it adds to a count of divisors of that number.
Finally, if the number has only 2 divisors (1 and itself), it is added to the list of prime numbers.
Made a test with large numbers and it worked really well. Tell me where you think I can do better!
def prime_finder(n):
nums_in_range = list(range(1, n+1))
prime_numbers = []
for num in nums_in_range:
new_range = list(range(1, num+1))
divisor_count = 0
for new_num in new_range:
residue = num%new_num
if residue == 0:
divisor_count += 1
else:
continue
if divisor_count == 2:
prime_numbers.append(num)
return prime_numbers
print(prime_finder(337))
Hi to everyone!! Firs time trying the challenges. Here´s my firs try to the first challenge.
Another things that could be included is a filter by type of input, like a text output if you input a text, or a float or a negative. I thik it could be easy adding some if, elif, else statements with a type() function. What do you think?
def prime_finder(n):
prime = []
for i in range(1, n +1):
divisores = []
for j in range(1, i+1):
if i % j == 0:
divisores.append(j)
if len(divisores) == 2:
prime.append(i)
return prime
print(prime_finder(11))
def prime_finder(n):
result = []
for i in range(2, n+1):
count = 0
for s in range(2, i+1):
if i%s==0:
count+=1
if count==1:
result.append(i)
return result
print(prime_finder(100))
Here is my answer to the challenge. Any feedback would be great.
One area I am trying to get better at is adding better comments to my code
def prime_finder(n):
# Write your code here
#define variables
prime_numbers = [] #Empty list prime_numbers to be return at end of function
num_list = list(range(3,n+1)) #list of all numbers from 3 up to n, inclusive
#check if n is 1 or zero, return empty prime_number list, else append list with 2
if n <= 1:return prime_numbers
else: prime_numbers.append(2)
#loop through all integers, i, in num_list
for i in num_list:
#loop through identified prime numbers
for prime in prime_numbers:
if prime <= i**(1/2):#check if prime number is less than square root of i
#if prime number is factor of i break loop, goes to next i
if i%prime == 0:break
#if loop through all prime numbers less than sqrt of i and no factors if i
#were identified then i is a prime number and we append prime_numbers list
else:
prime_numbers.append(i)
break
return prime_numbers
print(prime_finder(11))
def prime_finder(n): # function
prime_numbers = [] # create a list to store the numbers
for number in range(2, n + 1): # First for loop to loop through each number
count = 0 # Creating a counter
for test in range(1, n): # Second one to test if any other numbers divide to 0
if number % test == 0: # number divided by test with a remainder
count += 1 # count only if the remainder is 0
if count <= 2: # check if the amount of numbers it can divide by is 2 or less
prime_numbers.append(number) # add number to the list like a boss
return(prime_numbers) # returning the list created at the start
print(prime_finder(11)) # call the epic function
Here is my solution. I am nesting a function, is_prime. If the number is less than 2, the is_prime function returns False. If the number is 2, the is_prime function returns True. If the number is divisible by 2 and is greater than 2, the is_prime function returns False. Now concentrate on the odd numbers up to and including the square root of the number (to the integer). In the for loop of the is_prime function, if the number is divisible by the iterator, return False. If the odd number is not divisible by any of the iterator (3, 5, 7, 9, …) the is_prime function returns ‘True’. Now going back to prime_finder function. Define a list named primes. Iterate over the loop from 1 to n inclusive. If the iterator is a prime number, add the iterator to the list primes. Return the list primes.
def prime_finder(n):
# Write your code here
def is_prime(num):
if num<2:
return False
if num == 2:
return True
if num%2 == 0 and num > 2:
return False
for i in range(3,int(num**0.5)+1,2):
if num%i == 0:
return False
return True
primes = []
for i in range(1,n+1):
if is_prime(i):
primes.append(i)
return primes
print(prime_finder(11))
Here is my attempt at solving the prime number finder challenge. I did this by using 2 for loops. One ran through all of the numbers, and the other ran through all of the possible division equations. If the modulus equation had no remainder, the counter z was incremented. All iterations where z = 2, are appended to the list of prime numbers.
def prime_finder(n):
z = 0
results = []
for i in range (1, n+1, 1):
for j in range (1, n+1, 1):
if i % j == 0:
z = z + 1
if z > 3:
z = 0
continue
elif z == 2:
results.append(i)
z = 0
else:
z = 0
continue
return(results)
print(prime_finder(11))
def prime_finder(n):
# Write your code here
prime_list = []
for num in range(1, n+1):
prime = True
for i in range(2,num):
if num % i == 0:
prime = False
if prime and num != 1:
prime_list.append(num)
return prime_list
print(prime_finder(11))
def prime_finder(n):
list_nums = range(2, n + 1)
list_primes = []
for x in list_nums:
list_val = []
for y in list_nums:
if x % y == 0:
list_val.append(x)
if len(list_val) == 1:
list_primes.append(x)
return list_primes
print(prime_finder(11))
def prime_finder(n):
prime=[] #primes list creation
for i in range(2,n+1):
if i==2:
prime.append(i) #first prime number
else:
a=i-1
while (i%a)!=0: #succession of divisions to search even only by itself, stops if result even by another number thzn itself
if a==2:
prime.append(i)
break
else:
a=a-1
return(prime)
print(prime_finder(11))
def prime_finder(n):
prime_list = []
for number in range(2, n+1):
count = 0
for i in range(1, number+1):
if number % i == 0:
count += 1
if count == 2:
prime_list.append(number)
return prime_list
print(prime_finder(11))
