Software engineering and personal development

Tag: check

3 Ways How You Can Quickly Check Whether a Year is a Leap Year in Python

A leap year is a year that has 366 days instead of the usual 365. It occurs every four years and is the year when an extra day, February 29th, is added to the calendar. This day, known as a leap day, helps to keep the calendar aligned with the Earth’s movements around the sun. Leap years occur every four years, with the next one occurring in 2024.

We can check whether a year is a leap year in Python in a few ways.

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How to Quickly Check if 2 Strings Are Anagrams using Counter

Anagrams are strings that have the same letters but in a different order for example abc, bca, cab, acb, bac are all anagrams, since they all contain the same letters.

We can check whether two strings are anagrams in Python in different ways. One way to do that would be to use Counter from the collections module.

From the documentation:

 A Counter is a dict subclass for counting hashable objects. It is a collection where elements are stored as dictionary keys and their counts are stored as dictionary values. Counts are allowed to be any integer value including zero or negative counts. The Counter class is similar to bags or multisets in other languages.

In plain English, with Counter, we can get a dictionary that represents the frequency of elements in a list. Let us see this in an example:

 from collections import Counter
 ​
 print(Counter("Hello"))  # Counter({'l': 2, 'H': 1, 'e': 1, 'o': 1})

Now we can use Counter to quickly check whether two strings are anagrams or not:

 from collections import Counter
 ​
 ​
 def check_if_anagram(first_string, second_string):
     first_string = first_string.lower()
     second_string = second_string.lower()
     return Counter(first_string) == Counter(second_string)
 ​
 ​
 print(check_if_anagram('testinG', 'Testing'))  # True
 print(check_if_anagram('Here', 'Rehe'))  # True
 print(check_if_anagram('Know', 'Now'))  # False

We can also check whether 2 strings are anagrams using sorted():

 def check_if_anagram(first_word, second_word):
     first_word = first_word.lower()
     second_word = second_word.lower()
     return sorted(first_word) == sorted(second_word)
 ​
 print(check_if_anagram("testinG", "Testing"))  # True
 print(check_if_anagram("Here", "Rehe"))  # True
 print(check_if_anagram("Know", "Now"))  # False

That’s basically it.

I hope you find this useful.

How to Include Multiple Conditions at Once in Python

Similar to the case of using any(), we can also use a method that allows us to check whether all conditions are met. This can also greatly reduce the complexity of the code since you do not need to use multiple and checks.

Let us see this with an example.

Let us assume that we have the following conditions where we are checking whether we have more than 50 points in each school course:

 math_points = 51
 biology_points = 78
 physics_points = 56
 history_points = 72
 ​
 my_conditions = [math_points > 50, biology_points > 50,
                  physics_points > 50, history_points > 50]

Now passing all of them means that each condition should be met. To help us with that, we can simply use all(), as can be seen in the following snippet:

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How to Quickly Check Whether at Least One Condition is Met in Python

In many cases, we may have multiple conditions that we want to check, and going through each one of them can be a clutter.

First and foremost, we should keep in mind that we write code for other humans to read it. These are our colleagues that work with us, or people who use our open source projects.

As such, checking whether at least one condition is met can be done using a very quick way by using the method any().

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How to Quickly Simply “if” Statements in Python

There can be cases when we want to check whether a value is equal to one of the multiple other values.

One way that someone may start using is using or and adding multiple checks. For example, let us say that we want to check whether a number is 11, 55, or 77. Your immediate response may be to do the following:

 m = 1
 ​
 if m == 11 or m == 55 or m == 77:
     print("m is equal to 11, 55, or 77")

There is fortunately another quick way to do that.

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