Find e to the Nth Digit. Enter a number and have the program generate e up to that many decimal places. Keep a limit to how far the program will go

euler.number.py

""""Find e to the Nth Digit
Enter a number and have the program generate e up to that many decimal places.
Keep a limit to how far the program will go"""


from math import e


def e_with_precision(n):
    """Return euler's number to the n-th decimal place

    :param n: number of decimal places to return
    :type n: int
    :return: euler's number with n decimal places
    :rtype: str
    """
    return '%.*f' % (n, e)

if __name__ == '__main__':
    # there is no do while loop in python, so we need to improvise
    correct_input = False
    while not correct_input:
        # ask until you get correct input
        print('Precision must be between 1 and 51')
        precision = int(raw_input('Number of decimal places: '))
        if 51 >= precision > 0:
            correct_input = True
    print(e_with_precision(precision))

return '%.*f' % (n, e)

I was looking at your problem solving and I couldn't understand what the above sentence does.
Could you explain it for me ??

return '%.*f' % (n,e) , returns the string type of nth value of e ,toned down to precision n. When the precision is not known, until before runtime, this syntax works fine in that case.

The output is incorrect for n > 15 .
https://www.math.utah.edu/~pa/math/e.html.

.....Your output for n=50 with spaces added: 2.71828 18284 59045 09079 55982 98427 64884 23347 47314 45312
Correct output for n=50 with spaces added: 2.71828 18284 59045 23536 02874 71352 66249 77572 47093 69995

return '%.*f' % (n,e).

hi, I tried to use the new formatting for this but I could not get it right, i'm new to this, so I was wondering if there is way to use the new formatting in python 3 instead of %.

thank you in advance

You can put placeholders "{ :*f}."format(n,e) in python 3 or using f string formatting in python 3.6...both works fine

Athul8raj I've tried using the placeholders in your comment but there not working.

Please provide an alternative to this return '%.*f' % (n, e) for python 3

Please provide an alternative to this return '%.*f' % (n, e) for python 3

Which is easy to read and understand as it don't know anything about python 2.

Please provide an alternative to this return '%.*f' % (n, e) for python 3

This works perfectly well on Python 3. So just learn how different formatting styles work -> https://pyformat.info/ and rewrite it yourself.

I meant can you give an alternative to this return '%.*f' % (n, e) in .format() or f-string method. As I know how these methods work.