Given a String, Split the String on all the punctuations.
Input : test_str = 'neveropen! is-best' Output : ['neveropen', '!', 'is', '-', 'best'] Explanation : Splits on '!' and '-'. Input : test_str = 'geek-sfo, rLazyroar! is-best' Output : ['geek', '-', 'sfo', ', ', 'rLazyroar', '!', 'is', '-', 'best'] Explanation : Splits on '!', ', ' and '-'.
Method : Using regex + findall()
This is one way in which this problem can be solved. In this, we construct appropriate regex and task of segregating and split is done by findall().
Python3
# Python3 code to demonstrate working of # Split String on all punctuations# using regex + findall()import re# initializing stringtest_str = 'neveropen ! is-best, for @Lazyroar'# printing original Stringprint("The original string is : " + str(test_str))# using findall() to get all regex matches. res = re.findall( r'\w+|[^\s\w]+', test_str)# printing result print("The converted string : " + str(res)) |
Output
The original string is : neveropen! is-best, for @Lazyroar The converted string : ['neveropen', '!', 'is', '-', 'best', ', ', 'for', '@', 'Lazyroar']
Time Complexity: O(n)
Auxiliary Space: O(n)
Method : Using replace() and split() methods
Python3
# Python3 code to demonstrate working of# Split String on all punctuations# initializing stringtest_str = 'neveropen ! is-best, for @Lazyroar'# printing original Stringprint("The original string is : " + str(test_str))# import string library functionimport string # Storing the sets of punctuation in variable resultresult = string.punctuationfor i in test_str: if i in result: test_str=test_str.replace(i,"*"+i+"*")res=test_str.split("*")# printing resultprint("The converted string : " + str(res)) |
Output
The original string is : neveropen ! is-best, for @Lazyroar The converted string : ['neveropen ', '!', ' is', '-', 'best', ',', ' for ', '@', 'Lazyroar']
Method #3: Using re.sub() method with lambda function
Steps:
- Import the required module ‘re‘ and initialize the input string “test_str” and print the original string “test_str“.
- Apply the ‘re.sub()’ method to add space before and after each punctuation symbol.
- Assign the result to the variable “res” and split the modified string into a list of words.
- Print the resulting list of words.
Python3
# Python3 code to demonstrate working of # Split String on all punctuations# Using re.sub() method with lambda function:import re# initializing stringtest_str = 'neveropen ! is-best, for @Lazyroar'# printing original Stringprint("The original string is : " + str(test_str))res = re.sub(r'(\W+)', lambda x: ' '+x.group(0)+' ', test_str).split()# printing result print("The converted string : " + str(res)) |
Output
The original string is : neveropen ! is-best, for @Lazyroar The converted string : ['neveropen', '!', 'is', '-', 'best', ',', 'for', '@', 'Lazyroar']
Time complexity: O(n), as ‘re.sub()’ method has a time complexity of O(n) where n is the length of the input string “test_str”
Space complexity: O(n), where n is the length of the input string “test_str”.
