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Count the number of zeroes, odds, and evens in a n-digits number


In our daily lives, we may need to count odd and even numbers. For instance, in the library, you are trying to know the numbers of books having odd numbers or even numbers. It can happen also that in a given digits numbers, we wanted to know the number (count) of odd numbers and even numbers or zero.


Even numbers, are numbers when you divide by 2, the remainder is 0 in opposite with odd numbers where you will get a remainder which is 1. We can also say that even numbers are numbers that are divisible by 2 and odd numbers are not divisible by 2.


We can use python programming to build an app that we can use to go through a lengthy integer to count the number of odd, even, and zero numbers that occur.


def digits_count(digits):
    """ Takes a list of single digits and returns the 
        count of zeros, evens and odds
    """
    zero = 0
    even= 0
    odd= 0

We define our function to count zero, even, and odd numbers. Our variables zero, even, and odd are assign to zero because it will help later to increment them.


    for i in digits:
        if i == 0:
            zero += 1
        elif i % 2 == 0:
            even += 1
        else:
            odd += 1
        return zero, even, odd

The for loop is to assign to i each value of in the n-digits to make the program easy and also the calculation. now, when the program will meet i=0, then it will increment(augmentation) the variable zero, if the remainder of de division is 0, it increments the variable even, else (the remainder is different than 0), the program increment our variable odd. The program will then return zero, even, and odd values.


digit_input = input("Enter a n-digits number please: ")
digits = [int(i) for i in digit_input] 

zero, even, odd = digits_count(digits)

print('zero: %d' % zero)
print('even: %d' % even)
print('odd: %d' % odd)

Our function (def digits_count(digits) ) does not give an error after runing, now the user should enter a n-digits number. the snippet digits = [int(i) for i in digit_input] is to avoid the error int object is not iterable.



Now we can try our program if it will works.


Enter a ten-digit number please: 021369495970340464277
zero: 3
even: 8
odd: 10

Clique on SOURCE CODE to have a jupyter notebook and try to play around with it and add more conditions or others functions as you want. Because programming is about doing more practice even if it is the same problem. We can solve many problems using python programming and this will make life easy. If you like share with others and follow me for more post.


Do not hesitate to send your feedback for improvement.


3 comments

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3 comentários


Data Insight
Data Insight
03 de set. de 2021

Must the number be 10 digits? Can't it be any number of digits?

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Sakayo Toadoum Sari
Sakayo Toadoum Sari
04 de set. de 2021
Respondendo a

Also I was using 10-digits everywhere because the first idea when I was writing is to use billion as an example.

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