Question

I have a string:

i = '12345678901'

How to check (with python) if one before last digit in a string is even or odd ?

In a string above I want function to return or print “even”.
I found some examples but they are in C and i don’t know how to use it in Python

Answer 1

There are (at least) two ways of doing this:

Method 1:

Get the character at index -2 and intify it and check its modulus with 2

In [119]: i = '12345678901'

In [120]: int(i[-2])%2
Out[120]: 0

In [121]: int(i[-1])%2
Out[121]: 1

Method 2:

intify i, and divide by 10 and mod by 10 to get at the second last digit and nod by 2 get determine if it’s even or odd

In [122]: i = 12345678901

In [123]: (i/10)%10
Out[123]: 0

In [124]: ((i/10)%10)%2
Out[124]: 0