python check if one before last digit is odd

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

print('even' if (-1)**i==1 else 'odd') I know how to check for a number if it's odd or even but how can I extract one before last digit — Chris, Jan 13 '13 at 23:53
See: http://www.tutorialspoint.com/python/python_strings.htm (mainly, string splicing). `i[-2]` returns '0', which is a string. We need to convert this to an integer: `int(i[-2])`. Now, to check if it is even or odd: `if int(i[-2]) % 2 == 0: return 'Even'`. `else: return 'Odd'` `%` is the modulus operator. It returns the remainder of the division of the first element and the second element. — Rushy Panchal, Jan 13 '13 at 23:57

score 3 · Accepted Answer · answered Jan 13 '13 at 23:52

3

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

answered Jan 13 '13 at 23:52

inspectorG4dget

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2

The `%10` is superfluous, I think. Just `/10%2` is what you need. – Silas Ray Jan 14 '13 at 00:04
@wim: Better maths way: `even = int(i) % 20 < 10` – Eric Jan 14 '13 at 07:46

score 2 · Answer 2 · answered Jan 14 '13 at 00:06

2

A one-liner:

f = lambda s: 'odd' if int(s[-2]) % 2 else 'even'

Tested on input:

>>> i = '12345678901'
>>> f(i)
'even'
>>> i = '12345678931'
>>> f(i)
'odd'

answered Jan 14 '13 at 00:06

ezod

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score 1 · Answer 3 · answered Jan 13 '13 at 23:53

1

This is pretty easy to achieve, you just need a tiny bit of slicing, the modulus operator(%) and you're done.

i = '12345678901'

second_last = i[-2]

if int(second_last) % 2 == 0:
    print("Even")
else:
    print("Odd")

Output:

Even

answered Jan 13 '13 at 23:53

Marius

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Volatility · Answer 4 · 2013-01-14T00:15:23.007

0

i = '12345678901'
print int(i[-2]) % 2 == 1
# False

For it to print even or odd, you need to do this:

if int(i[-2]) % 2 == 0:
    print 'Even'
else:
    print 'Odd'

You could also use the fact that 1 is interpreted as True by if statements (and 0 is False):

if int(i[2]) % 2:  #if int(i[-2]) % 2 returns 1
    print 'Odd'
else:
    print 'Even'

edited Jan 14 '13 at 00:15

answered Jan 13 '13 at 23:53

Volatility

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score 0 · Answer 5 · answered Jan 13 '13 at 23:53

0

Step 1: Get second last digit:

last_digit = i[-2]

Step 2: is it odd?

is_odd = last_digit in '13579'

answered Jan 13 '13 at 23:53

Eric

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score 0 · Answer 6 · edited May 23 '17 at 12:08

0

This starts with taking a slice of the data, see Explain Python's slice notation for a good explanation of slice. The official documentation isn't that obvious.

You can use negative numbers to slice from the end. Once you have the slice you need to check if it is a digit, find its nmeric value and see if it's even.

Coding s left as an exercise for the reader.

edited May 23 '17 at 12:08

Community

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answered Jan 13 '13 at 23:54

Peter Wooster

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score 0 · Answer 7 · answered Sep 24 '15 at 12:14

0

(i % 10) & 1

returns 1 if odd and 0 if even

code -

print "even" if not ((int(i) / 10) % 10) & 1 else ""

answered Sep 24 '15 at 12:14

Confuse

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python check if one before last digit is odd

7 Answers7