CODEWITHSUNDEEP
python-3.7
peregrinefalcon
peregrinefalcon

Reputation: 45

Python - Only Odd Numbers

I need to find out if n is made up of odd numbers. If a number is made up of odd numbers, I should return True, else I should return False.

n = 1357 should return True because all four values are odd. n = 1356 should return False because 6 is an even number.

How would I go about solving this? My first steps included looping over the "stringified" number but then I felt stuck because I was unable to use modulo to check whether or not the number is even or odd.

Upvotes: 1

Views: 1633

Answers (3)

Mark Tolonen
Mark Tolonen

Reputation: 177481

If the set of odd digits is subtracted from the set of str(n) and the result is empty, then it was all odd digits:

>>> def odd(n):
...   return set(str(n))-set('13579')==set()
...
>>> odd(123)
False
>>> odd(113355)
True

It's fast, too, compared to checking each digit numerically:

C:\>py -m timeit -s "odd=set('13579');n=111333555777999" "all([int(x)%2 for x in str(n)])"
50000 loops, best of 5: 4.69 usec per loop

C:\>py -m timeit -s "odd=set('13579');n=111333555777999" set(str(n))-odd==set()"
200000 loops, best of 5: 1.36 usec per loop

Upvotes: 1

Andreas
Andreas

Reputation: 9197

n = 1357
print(all([int(x)%2!=0 for x in str(n)]))
#True

You essentially explained a possible way already. To check if a number is even use:

number % 2

Upvotes: 0

Countour-Integral
Countour-Integral

Reputation: 1154

def isMadeofEven(num):
    for number in str(num):
        if (int(number) % 2 !=0):
            return False
    return True
  • Basically loops through each characters of the converted string number and if it finds an odd returns false

Upvotes: 0

Related Questions