Written by climjark

Conditionals

Requirements:
Linux Distribution
python3
A Shell Interface

My Setup:
Debian GNU/Linux for some examples, ArchLinux for others.
python3
BASH

Normally within a program, when you want to do something if a condition is met, you would use conditional statements. There are a couple conditional statements that python supports. This includes :

  1. if – “If this statement is true, do this”
  2. elif – “If the statement previous is false and this statement is true, do this”
  3. else – “If the previous statements have failed, do this”

The syntax for these are as followed:

if condition :
    #do something
elif condition :
    #do something else
else:
    #do something

Python supports the following conditionals:

  1. Equals: a == b
  2. Not Equals: a != b
  3. Less than: a < b
  4. Less than or equal to: a <= b
  5. Greater than: a > b
  6. Greater than or equal to: a >= b

These can be strung together with either “and” and “or” like followed:

if condition and condition:
    #do something
elif condition or condition:
    #do something
else:
    #do something

Lets put this in a working example :

firstNum = 10
secondNum = 30

if firstNum < secondNum and (secondNum % firstNum) == 0 :
    print("First number is smaller, and the second number is a multiple of the first.")
elif firstNum < secondNum :
    print("First number is smaller")
else:
    print("Second number is smaller")

clim@debian:~/Desktop/tests/pytests/conditionals$ python3 conditionals.py
First number is smaller, and the second number is a multiple of the first.
clim@debian:~/Desktop/tests/pytests/conditionals$

When using if statements, there needs to be atleast 1 statement within each branch. You cannot leave an if statement empty. If there is a time where nothing is to be executed within an if statement, you may use the pass keyword as such:

if condition:
  pass
else:
  pass

The most important thing to remember is the indentation. When using conditional statements, you must indent four spaces (or a tab).

Python project : Checking whether or not a year is a leap year

For this project, we input a year, and calculate whether or not it is a leap year. To determine whether or not a year is a leap year, you must divide the year by 4. If the remainder is 0, and the year is divisible by 100, it is not a leap year. However, if it is divisible by 4 or 400 , it is a leap year.

#input our year
year = int(input("Enter a year:"))
if year%4==0 and year%100!=0 or year%400==0:
    print("The year is a leap year!")
else:
    print("The year isn't a leap year!")

[clim@archnet:~/Desktop/cspcsite/PythonTutorials/examples/3]$ python3 leap.py
Enter a year:2000
The year is a leap year!
[clim@archnet:~/Desktop/cspcsite/PythonTutorials/examples/3]$ python3 leap.py
Enter a year:1900
The year isn't a leap year!
[clim@archnet:~/Desktop/cspcsite/PythonTutorials/examples/3]$ python3 leap.py
Enter a year:2020
The year is a leap year!
[clim@archnet:~/Desktop/cspcsite/PythonTutorials/examples/3]$

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