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Python: Matches a word at the end of a string, with optional punctuation

Python Regular Expression: Exercise-11 with Solution

Write a Python program that matches a word at the end of a string, with optional punctuation.

Sample Solution:-

Python Code:

import re
def text_match(text):
        patterns = '\w+\S*$'
        if re.search(patterns,  text):
                return 'Found a match!'
        else:
                return('Not matched!')

print(text_match("The quick brown fox jumps over the lazy dog."))
print(text_match("The quick brown fox jumps over the lazy dog. "))
print(text_match("The quick brown fox jumps over the lazy dog "))

Sample Output:

Found a match!                                                                                                
Not matched!                                                                                                  
Not matched!

Pictorial Presentation:

Python: Regular Expression - Matches a word at the end of a string, with optional punctuation.

Flowchart:

Flowchart: Regular Expression - Matches a word at the end of a string, with optional punctuation.

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Python: Tips of the Day

Getting the last element of a list:

some_list[-1] is the shortest and most Pythonic.

In fact, you can do much more with this syntax. The some_list[-n] syntax gets the nth-to-last element. So some_list[-1] gets the last element, some_list[-2] gets the second to last, etc, all the way down to some_list[-len(some_list)], which gives you the first element.

You can also set list elements in this way. For instance:

>>> some_list = [1, 2, 3]
>>> some_list[-1] = 5 # Set the last element
>>> some_list[-2] = 3 # Set the second to last element
>>> some_list
[1, 3, 5]

Note that getting a list item by index will raise an IndexError if the expected item doesn't exist. This means that some_list[-1] will raise an exception if some_list is empty, because an empty list can't have a last element.

Ref: https://bit.ly/3d8TfFP