Sometimes, while working with Python Strings, we can have a problem in which we need to extract indices of vowels in it. This kind of application is common in day-day programming. Lets discuss certain ways in which this task can be performed.
Method #1 : Using loop This is one way in which this task can be performed. In this we use brute force to perform this task. In this we iterate for each string element and test for vowel.
Python3
# Python3 code to demonstrate working of # Vowel indices in String # Using loop # initializing string test_str = "neveropen"# printing original string print("The original string is : " + test_str) # Vowel indices in String # Using loop res = [] for ele in range(len(test_str)): if test_str[ele] in "aeiou": res.append(ele) # printing result print("The vowel indices are : " + str(res)) |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(N)
Auxiliary Space: O(N)
Method #2 : Using enumerate() + list comprehension The combination of above methods can also be used to perform this task. In this, we access the index using enumerate() and list comprehension is used to check for vowels.
Python3
# Python3 code to demonstrate working of # Vowel indices in String # Using list comprehension + enumerate() # initializing string test_str = "neveropen"# printing original string print("The original string is : " + test_str) # Vowel indices in String # Using list comprehension + enumerate() res = [idx for idx, ele in enumerate(test_str) if ele in "aeiou"] # printing result print("The vowel indices are : " + str(res)) |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
The Time and Space Complexity for all the methods are the same:
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : Using ord() method
Python3
# Python3 code to demonstrate working of# Vowel indices in String# Using loop# initializing stringtest_str = "neveropen"# printing original stringprint("The original string is : " + test_str)# Vowel indices in String# Using loopres = []vow=[97,101,105,111,117]for ele in range(len(test_str)): if ord(test_str[ele]) in vow: res.append(ele)# printing resultprint("The vowel indices are : " + str(res)) |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(N)
Auxiliary Space: O(N)
Method #4 : Using operator.countOf() method
Python3
# Python3 code to demonstrate working of# Vowel indices in Stringimport operator as opvowels = "aeiou"# initializing stringtest_str = "neveropen"# printing original stringprint("The original string is : " + test_str)# Vowel indices in String# Using loopres = []for ele in range(len(test_str)): if op.countOf(vowels, test_str[ele]) > 0: res.append(ele)# printing resultprint("The vowel indices are : " + str(res)) |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method#5: Using Recursive method.
Python3
# Python3 code to demonstrate working of # Vowel indices in String # Using recursive methoddef vowel_indices(string, idx=0, result=None): if result is None: result = [] if idx == len(string): return result if string[idx] in "aeiou": result.append(idx) return vowel_indices(string, idx + 1, result)# initializing string test_str = "neveropen"# printing original string print("The original string is : " + test_str) res=vowel_indices(test_str)# printing result print("The vowel indices are : " + str(res)) #this code contributed by tvsk |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(n)
Space Complexity: O(n)
Method #6 : Using Regex
Python3
# Python3 code to demonstrate working of# Vowel indices in String# Using regeximport re # initializing stringtest_str = "neveropen" # printing original stringprint("The original string is : " + test_str) # Vowel indices in String# Using regexvowel_regex = re.compile("[aeiou]")res = [i.start() for i in vowel_regex.finditer(test_str)] # printing resultprint("The vowel indices are : " + str(res))#This code is contributed by Edula Vinay Kumar Reddy |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(n) where n is the length of the string
Auxiliary Space: O(1)
Explanation: In this method, we use the re.compile method to compile a regular expression pattern “[aeiou]” which matches the vowels in the string. The re.finditer method returns an iterator yielding match objects for all non-overlapping matches. The start method of each match object returns the starting position of the match. The result of the code is the list of starting positions of all the matches of vowels in the string.
Method #7 : Using Filter() and lambda():
Python3
# Initialize the stringtest_str = "neveropen"# Specify the vowelsvowels = "aeiou"# printing original stringprint("The original string is : " + test_str)# Find the indices of vowels using filter() function and lambda functionresult = list(filter(lambda x: test_str[x] in vowels, range(len(test_str))))# Print the resultprint("The vowel indices are : " + str(result))#This code is contributed by Jyothi pinjala |
The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #8 : Using numpy:
Algorithm:
- Create a string ‘test_str’ and initialize it with some value.
- Convert the string ‘test_str’ to a list of characters using the built-in list() function.
- Create a numpy array using the converted list of characters.
- Create another numpy array using the list of vowels in the order ‘aeiou’.
- Use the built-in np.isin() function to create a Boolean array indicating whether each character in the ‘test_str’ array is in the ‘aeiou’ array.
- Use the built-in np.where() function to get the indices of the True values in the Boolean array returned in step 5.
- Access the first array of the tuple returned in step 6 using index 0 and convert it to a list.
- Print the list of vowel indices.
Python3
import numpy as nptest_str = "neveropen"# printing original stringprint("The original string is : " + test_str) vowel_indices = np.where(np.isin(np.array(list(test_str)), np.array(list("aeiou")))) # returns a tuple of arraysvowel_indices = vowel_indices[0].tolist() # get the first array and convert to list# printing resultprint("The vowel indices are : " + str(vowel_indices))#This code is contributed by Rayudu. |
Output: The original string is : neveropen The vowel indices are : [1, 2, 6, 9, 10]
Time Complexity: O(n), where n is the length of the input string ‘test_str’. The time complexity is dominated by the np.isin() function, which has a time complexity of O(n).
Auxiliary Space: O(n), where n is the length of the input string ‘test_str’. The space complexity is dominated by the two numpy arrays created in steps 3 and 4, each of which has a size of n. The remaining space used by the algorithm is constant.
