How to extract the elements of a deep nested tuple and put them in a list in python? [closed]
I've been playing with Python, and wrote the following code that finds all prime numbers in a given range as following:
def get_primes(x):
primes =
def is_prime(x):
if x == 0:
return
else:
for i in range(2, int(x)):
if (x % i) == 0:
return is_prime(x - 1)
else:
return x, is_prime(x - 1)
primes.append(is_prime(x))
return primes
print(get_primes(int(input("Enter the range: 0 - "))))
And the output is: (enter 100 for example)
Enter the range: 0 - 100
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
that looks so ugly.
So, I need a way to flatten the recursive tuple structure:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
i used the following code to do so:
x = get_primes(100)
arr =
arr.append(x[0][0])
arr.append(x[0][1][0])
arr.append(x[0][1][1][0])
arr.append(x[0][1][1][1][0])
arr.append(x[0][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
print(arr)
But of course, this is not a professional method.
So, what i want is to know how can i make this one:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
from this one:
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
I found the answer here How to flatten a tuple in python but the code there was for python 2, so, i modified it a bit.
and used this code:
def flatten(T):
if type(T) is not tuple:
return (T,)
elif len(T) == 0:
return ()
else:
return flatten(T[0]) + flatten(T[1:])
python tuples
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
closed as unclear what you're asking by Jean-François Fabre, Goyo, pushkin, Prune, lagom Nov 20 '18 at 3:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 2 more comments
I've been playing with Python, and wrote the following code that finds all prime numbers in a given range as following:
def get_primes(x):
primes =
def is_prime(x):
if x == 0:
return
else:
for i in range(2, int(x)):
if (x % i) == 0:
return is_prime(x - 1)
else:
return x, is_prime(x - 1)
primes.append(is_prime(x))
return primes
print(get_primes(int(input("Enter the range: 0 - "))))
And the output is: (enter 100 for example)
Enter the range: 0 - 100
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
that looks so ugly.
So, I need a way to flatten the recursive tuple structure:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
i used the following code to do so:
x = get_primes(100)
arr =
arr.append(x[0][0])
arr.append(x[0][1][0])
arr.append(x[0][1][1][0])
arr.append(x[0][1][1][1][0])
arr.append(x[0][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
print(arr)
But of course, this is not a professional method.
So, what i want is to know how can i make this one:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
from this one:
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
I found the answer here How to flatten a tuple in python but the code there was for python 2, so, i modified it a bit.
and used this code:
def flatten(T):
if type(T) is not tuple:
return (T,)
elif len(T) == 0:
return ()
else:
return flatten(T[0]) + flatten(T[1:])
python tuples
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
closed as unclear what you're asking by Jean-François Fabre, Goyo, pushkin, Prune, lagom Nov 20 '18 at 3:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
Why doesis_primeeven return a tuple? it should return a boolean
– DeepSpace
Nov 19 '18 at 20:03
you could post process & flatten: stackoverflow.com/questions/2158395/… and fix your indentation
– Jean-François Fabre
Nov 19 '18 at 20:04
2
I don't understand youris_prime()function. Whenif (x % i) == 0is true, you know x is not prime, so why do you continue?
– John Gordon
Nov 19 '18 at 20:05
1
Also - is there any particular reason to even attempt to use recursion here? It's not necessary and it won't scale.
– Jon Clements♦
Nov 19 '18 at 20:06
is_prime function will find all prime numbers in a specific range and return all these numbers, it checks if the current value of x is prime, if it is prime it returns this value and calls itself, otherwise it will call itself only without returning the value of x (which should be only prime number).
– Dr.venom
Nov 19 '18 at 21:01
|
show 2 more comments
I've been playing with Python, and wrote the following code that finds all prime numbers in a given range as following:
def get_primes(x):
primes =
def is_prime(x):
if x == 0:
return
else:
for i in range(2, int(x)):
if (x % i) == 0:
return is_prime(x - 1)
else:
return x, is_prime(x - 1)
primes.append(is_prime(x))
return primes
print(get_primes(int(input("Enter the range: 0 - "))))
And the output is: (enter 100 for example)
Enter the range: 0 - 100
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
that looks so ugly.
So, I need a way to flatten the recursive tuple structure:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
i used the following code to do so:
x = get_primes(100)
arr =
arr.append(x[0][0])
arr.append(x[0][1][0])
arr.append(x[0][1][1][0])
arr.append(x[0][1][1][1][0])
arr.append(x[0][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
print(arr)
But of course, this is not a professional method.
So, what i want is to know how can i make this one:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
from this one:
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
I found the answer here How to flatten a tuple in python but the code there was for python 2, so, i modified it a bit.
and used this code:
def flatten(T):
if type(T) is not tuple:
return (T,)
elif len(T) == 0:
return ()
else:
return flatten(T[0]) + flatten(T[1:])
python tuples
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
I've been playing with Python, and wrote the following code that finds all prime numbers in a given range as following:
def get_primes(x):
primes =
def is_prime(x):
if x == 0:
return
else:
for i in range(2, int(x)):
if (x % i) == 0:
return is_prime(x - 1)
else:
return x, is_prime(x - 1)
primes.append(is_prime(x))
return primes
print(get_primes(int(input("Enter the range: 0 - "))))
And the output is: (enter 100 for example)
Enter the range: 0 - 100
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
that looks so ugly.
So, I need a way to flatten the recursive tuple structure:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
i used the following code to do so:
x = get_primes(100)
arr =
arr.append(x[0][0])
arr.append(x[0][1][0])
arr.append(x[0][1][1][0])
arr.append(x[0][1][1][1][0])
arr.append(x[0][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
arr.append(x[0][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][1][0])
print(arr)
But of course, this is not a professional method.
So, what i want is to know how can i make this one:
[97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
from this one:
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
I found the answer here How to flatten a tuple in python but the code there was for python 2, so, i modified it a bit.
and used this code:
def flatten(T):
if type(T) is not tuple:
return (T,)
elif len(T) == 0:
return ()
else:
return flatten(T[0]) + flatten(T[1:])
python tuples
python tuples
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
edited Nov 21 '18 at 13:27
Dr.venom
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
asked Nov 19 '18 at 20:00
Dr.venomDr.venom
13
13
closed as unclear what you're asking by Jean-François Fabre, Goyo, pushkin, Prune, lagom Nov 20 '18 at 3:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Jean-François Fabre, Goyo, pushkin, Prune, lagom Nov 20 '18 at 3:29
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
Why doesis_primeeven return a tuple? it should return a boolean
– DeepSpace
Nov 19 '18 at 20:03
you could post process & flatten: stackoverflow.com/questions/2158395/… and fix your indentation
– Jean-François Fabre
Nov 19 '18 at 20:04
2
I don't understand youris_prime()function. Whenif (x % i) == 0is true, you know x is not prime, so why do you continue?
– John Gordon
Nov 19 '18 at 20:05
1
Also - is there any particular reason to even attempt to use recursion here? It's not necessary and it won't scale.
– Jon Clements♦
Nov 19 '18 at 20:06
is_prime function will find all prime numbers in a specific range and return all these numbers, it checks if the current value of x is prime, if it is prime it returns this value and calls itself, otherwise it will call itself only without returning the value of x (which should be only prime number).
– Dr.venom
Nov 19 '18 at 21:01
|
show 2 more comments
3
Why doesis_primeeven return a tuple? it should return a boolean
– DeepSpace
Nov 19 '18 at 20:03
you could post process & flatten: stackoverflow.com/questions/2158395/… and fix your indentation
– Jean-François Fabre
Nov 19 '18 at 20:04
2
I don't understand youris_prime()function. Whenif (x % i) == 0is true, you know x is not prime, so why do you continue?
– John Gordon
Nov 19 '18 at 20:05
1
Also - is there any particular reason to even attempt to use recursion here? It's not necessary and it won't scale.
– Jon Clements♦
Nov 19 '18 at 20:06
is_prime function will find all prime numbers in a specific range and return all these numbers, it checks if the current value of x is prime, if it is prime it returns this value and calls itself, otherwise it will call itself only without returning the value of x (which should be only prime number).
– Dr.venom
Nov 19 '18 at 21:01
3
3
Why does
is_prime even return a tuple? it should return a boolean– DeepSpace
Nov 19 '18 at 20:03
Why does
is_prime even return a tuple? it should return a boolean– DeepSpace
Nov 19 '18 at 20:03
you could post process & flatten: stackoverflow.com/questions/2158395/… and fix your indentation
– Jean-François Fabre
Nov 19 '18 at 20:04
you could post process & flatten: stackoverflow.com/questions/2158395/… and fix your indentation
– Jean-François Fabre
Nov 19 '18 at 20:04
2
2
I don't understand your
is_prime() function. When if (x % i) == 0 is true, you know x is not prime, so why do you continue?– John Gordon
Nov 19 '18 at 20:05
I don't understand your
is_prime() function. When if (x % i) == 0 is true, you know x is not prime, so why do you continue?– John Gordon
Nov 19 '18 at 20:05
1
1
Also - is there any particular reason to even attempt to use recursion here? It's not necessary and it won't scale.
– Jon Clements♦
Nov 19 '18 at 20:06
Also - is there any particular reason to even attempt to use recursion here? It's not necessary and it won't scale.
– Jon Clements♦
Nov 19 '18 at 20:06
is_prime function will find all prime numbers in a specific range and return all these numbers, it checks if the current value of x is prime, if it is prime it returns this value and calls itself, otherwise it will call itself only without returning the value of x (which should be only prime number).
– Dr.venom
Nov 19 '18 at 21:01
is_prime function will find all prime numbers in a specific range and return all these numbers, it checks if the current value of x is prime, if it is prime it returns this value and calls itself, otherwise it will call itself only without returning the value of x (which should be only prime number).
– Dr.venom
Nov 19 '18 at 21:01
|
show 2 more comments
2 Answers
2
active
oldest
votes
You can adapt your function to be a generator pretty easily:
def is_prime(x):
if x != 0:
for i in range(2, int(x)):
if (x % i) == 0:
yield from is_prime(x - 1)
return
else:
yield x
yield from is_prime(x - 1)
Then to get all of the primes up to 100 you can pass that generator to list to get a list containing the values from that generator:
print(list(is_prime(100)))
# [97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
Recursion isn't a great approach to this though, I would suggest looking up the Sieve of Eratosthenes for a better approach.
The yield from syntax is only available in newer versions of Python (>= 3.3). You can replace
yield from x
with
for y in x:
yield y
if necessary.
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
– Dr.venom
Nov 19 '18 at 21:19
add a comment |
If you create a nested tuple, then the only way to flatten it is by unpacking. Best way to not deal with that is just not creating one.
Here goes an alternative version of your code:
def get_primes(limit):
def is_prime(x):
if x in (2, 3):
return True
if (x % 2 == 0) or (x % 3 == 0):
return False
i, w = 5, 2
while i**2 <= x:
if x % i == 0:
return False
i += w
w = 6 - w
return True
return [x for x in range(2, limit + 1) if is_prime(x)]
print(get_primes(int(input("Enter the range: 0 - "))))
answered Nov 19 '18 at 20:47
AResemAResem
1114
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can adapt your function to be a generator pretty easily:
def is_prime(x):
if x != 0:
for i in range(2, int(x)):
if (x % i) == 0:
yield from is_prime(x - 1)
return
else:
yield x
yield from is_prime(x - 1)
Then to get all of the primes up to 100 you can pass that generator to list to get a list containing the values from that generator:
print(list(is_prime(100)))
# [97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
Recursion isn't a great approach to this though, I would suggest looking up the Sieve of Eratosthenes for a better approach.
The yield from syntax is only available in newer versions of Python (>= 3.3). You can replace
yield from x
with
for y in x:
yield y
if necessary.
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
– Dr.venom
Nov 19 '18 at 21:19
add a comment |
You can adapt your function to be a generator pretty easily:
def is_prime(x):
if x != 0:
for i in range(2, int(x)):
if (x % i) == 0:
yield from is_prime(x - 1)
return
else:
yield x
yield from is_prime(x - 1)
Then to get all of the primes up to 100 you can pass that generator to list to get a list containing the values from that generator:
print(list(is_prime(100)))
# [97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
Recursion isn't a great approach to this though, I would suggest looking up the Sieve of Eratosthenes for a better approach.
The yield from syntax is only available in newer versions of Python (>= 3.3). You can replace
yield from x
with
for y in x:
yield y
if necessary.
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
– Dr.venom
Nov 19 '18 at 21:19
add a comment |
You can adapt your function to be a generator pretty easily:
def is_prime(x):
if x != 0:
for i in range(2, int(x)):
if (x % i) == 0:
yield from is_prime(x - 1)
return
else:
yield x
yield from is_prime(x - 1)
Then to get all of the primes up to 100 you can pass that generator to list to get a list containing the values from that generator:
print(list(is_prime(100)))
# [97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
Recursion isn't a great approach to this though, I would suggest looking up the Sieve of Eratosthenes for a better approach.
The yield from syntax is only available in newer versions of Python (>= 3.3). You can replace
yield from x
with
for y in x:
yield y
if necessary.
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
You can adapt your function to be a generator pretty easily:
def is_prime(x):
if x != 0:
for i in range(2, int(x)):
if (x % i) == 0:
yield from is_prime(x - 1)
return
else:
yield x
yield from is_prime(x - 1)
Then to get all of the primes up to 100 you can pass that generator to list to get a list containing the values from that generator:
print(list(is_prime(100)))
# [97, 89, 83, 79, 73, 71, 67, 61, 59, 53, 47, 43, 41, 37, 31, 29, 23, 19, 17, 13, 11, 7, 5, 3, 2, 1]
Recursion isn't a great approach to this though, I would suggest looking up the Sieve of Eratosthenes for a better approach.
The yield from syntax is only available in newer versions of Python (>= 3.3). You can replace
yield from x
with
for y in x:
yield y
if necessary.
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
answered Nov 19 '18 at 20:08
Patrick HaughPatrick Haugh
28.7k82747
28.7k82747
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
– Dr.venom
Nov 19 '18 at 21:19
add a comment |
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]
– Dr.venom
Nov 19 '18 at 21:19
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]– Dr.venom
Nov 19 '18 at 21:19
Thanks for clarification, this is definitely a better way to generate these prime numbers in a specific range, but my question is about how can i get these values out from the returned list:
[(97, (89, (83, (79, (73, (71, (67, (61, (59, (53, (47, (43, (41, (37, (31, (29, (23, (19, (17, (13, (11, (7, (5, (3, (2, (1, None))))))))))))))))))))))))))]– Dr.venom
Nov 19 '18 at 21:19
add a comment |
If you create a nested tuple, then the only way to flatten it is by unpacking. Best way to not deal with that is just not creating one.
Here goes an alternative version of your code:
def get_primes(limit):
def is_prime(x):
if x in (2, 3):
return True
if (x % 2 == 0) or (x % 3 == 0):
return False
i, w = 5, 2
while i**2 <= x:
if x % i == 0:
return False
i += w
w = 6 - w
return True
return [x for x in range(2, limit + 1) if is_prime(x)]
print(get_primes(int(input("Enter the range: 0 - "))))
answered Nov 19 '18 at 20:47
AResemAResem
1114
add a comment |
If you create a nested tuple, then the only way to flatten it is by unpacking. Best way to not deal with that is just not creating one.
Here goes an alternative version of your code:
def get_primes(limit):
def is_prime(x):
if x in (2, 3):
return True
if (x % 2 == 0) or (x % 3 == 0):
return False
i, w = 5, 2
while i**2 <= x:
if x % i == 0:
return False
i += w
w = 6 - w
return True
return [x for x in range(2, limit + 1) if is_prime(x)]
print(get_primes(int(input("Enter the range: 0 - "))))
answered Nov 19 '18 at 20:47
AResemAResem
1114
add a comment |
If you create a nested tuple, then the only way to flatten it is by unpacking. Best way to not deal with that is just not creating one.
Here goes an alternative version of your code:
def get_primes(limit):
def is_prime(x):
if x in (2, 3):
return True
if (x % 2 == 0) or (x % 3 == 0):
return False
i, w = 5, 2
while i**2 <= x:
if x % i == 0:
return False
i += w
w = 6 - w
return True
return [x for x in range(2, limit + 1) if is_prime(x)]
print(get_primes(int(input("Enter the range: 0 - "))))
answered Nov 19 '18 at 20:47
AResemAResem
1114
If you create a nested tuple, then the only way to flatten it is by unpacking. Best way to not deal with that is just not creating one.
Here goes an alternative version of your code:
def get_primes(limit):
def is_prime(x):
if x in (2, 3):
return True
if (x % 2 == 0) or (x % 3 == 0):
return False
i, w = 5, 2
while i**2 <= x:
if x % i == 0:
return False
i += w
w = 6 - w
return True
return [x for x in range(2, limit + 1) if is_prime(x)]
print(get_primes(int(input("Enter the range: 0 - "))))
answered Nov 19 '18 at 20:47
AResemAResem
1114
answered Nov 19 '18 at 20:47
AResemAResem
1114
answered Nov 19 '18 at 20:47
AResemAResem
1114
answered Nov 19 '18 at 20:47
AResemAResem
1114
1114
add a comment |
add a comment |
3
Why does
is_primeeven return a tuple? it should return a boolean– DeepSpace
Nov 19 '18 at 20:03
you could post process & flatten: stackoverflow.com/questions/2158395/… and fix your indentation
– Jean-François Fabre
Nov 19 '18 at 20:04
2
I don't understand your
is_prime()function. Whenif (x % i) == 0is true, you know x is not prime, so why do you continue?– John Gordon
Nov 19 '18 at 20:05
1
Also - is there any particular reason to even attempt to use recursion here? It's not necessary and it won't scale.
– Jon Clements♦
Nov 19 '18 at 20:06
is_prime function will find all prime numbers in a specific range and return all these numbers, it checks if the current value of x is prime, if it is prime it returns this value and calls itself, otherwise it will call itself only without returning the value of x (which should be only prime number).
– Dr.venom
Nov 19 '18 at 21:01