How to extract the elements of a deep nested tuple and put them in a list in python? [closed]

-1

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:])

edited Nov 21 '18 at 13:27

asked Nov 19 '18 at 20:00

Dr.venom

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 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

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

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

-1

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:])

edited Nov 21 '18 at 13:27

asked Nov 19 '18 at 20:00

Dr.venom

closed as unclear what you're asking by Jean-François Fabre, Goyo, pushkin, Prune, lagom Nov 20 '18 at 3:29

3

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

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

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

-1

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:])

edited Nov 21 '18 at 13:27

asked Nov 19 '18 at 20:00

Dr.venom

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

edited Nov 21 '18 at 13:27

asked Nov 19 '18 at 20:00

Dr.venom

edited Nov 21 '18 at 13:27

asked Nov 19 '18 at 20:00

Dr.venom

edited Nov 21 '18 at 13:27

asked Nov 19 '18 at 20:00

Dr.venom

asked Nov 19 '18 at 20:00

Dr.venom

asked Nov 19 '18 at 20:00

Dr.venom

closed as unclear what you're asking by Jean-François Fabre, Goyo, pushkin, Prune, lagom Nov 20 '18 at 3:29

3

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

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

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 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

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

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

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

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

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

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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 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

AResem

1114

add a comment |

2 Answers
2

active

oldest

votes

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 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 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 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 Haugh

28.7k82747

answered Nov 19 '18 at 20:08

Patrick Haugh

28.7k82747

answered Nov 19 '18 at 20:08

Patrick Haugh

28.7k82747

answered Nov 19 '18 at 20:08

Patrick 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 |

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

AResem

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

AResem

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

AResem

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

AResem

1114

answered Nov 19 '18 at 20:47

AResem

1114

answered Nov 19 '18 at 20:47

AResem

1114

answered Nov 19 '18 at 20:47

AResem

1114

add a comment |

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