 GFG App
Open App Browser
Continue

# Python | Count and display vowels in a string

In this program, we need to count the number of vowels present in a string and display those vowels. This can be done using various methods. In this article, we will go through few of the popular methods to do this in an efficient manner.
Examples:

```In a simple way
Input : Geeks for Geeks
Output :
5
['e', 'e', 'o', 'e', 'e']

This is in a different way
Input : Geeks for Geeks
Output : {'u': 0, 'o': 1, 'e': 4, 'a': 0, 'i': 0}```

Counting vowels: String Way

In this method, we will store all the vowels in a string and then pick every character from the enquired string and check whether it is in the vowel string or not. The vowel string consists of all the vowels with both cases since we are not ignoring the cases here. If the vowel is encountered then count gets incremented and stored in a list and finally printed.

## Python3

 `# Python code to count and display number of vowels` `# Simply using for and comparing it with a ` `# string containing all vowels` `def` `Check_Vow(string, vowels):` `    ``final ``=` `[each ``for` `each ``in` `string ``if` `each ``in` `vowels]` `    ``print``(``len``(final))` `    ``print``(final)` `    `  `# Driver Code` `string ``=` `"Geeks for Geeks"` `vowels ``=` `"AaEeIiOoUu"` `Check_Vow(string, vowels);`

Output

```5
['e', 'e', 'o', 'e', 'e']```

Time complexity: O(n)
Auxiliary Space: O(n)

Counting vowels: Dictionary Way

This also performs the same task but in a different way. In this method, we form a dictionary with the vowels and increment them when a vowel is encountered. In this method, we use the case fold method to ignore the cases, following which we form a dictionary of vowels with the key as a vowel. This is a better and efficient way to check and find the number of each vowel present in a string.

## Python3

 `# Count vowels in a different way` `# Using dictionary` `def` `Check_Vow(string, vowels):` `    `  `    ``# casefold has been used to ignore cases` `    ``string ``=` `string.casefold()` `    `  `    ``# Forms a dictionary with key as a vowel` `    ``# and the value as 0` `    ``count ``=` `{}.fromkeys(vowels, ``0``)` `    `  `    ``# To count the vowels` `    ``for` `character ``in` `string:` `        ``if` `character ``in` `count:` `            ``count[character] ``+``=` `1`    `    ``return` `count` `    `  `# Driver Code` `vowels ``=` `'aeiou'` `string ``=` `"Geeks for Geeks"` `print` `(Check_Vow(string, vowels))`

Output

`{'a': 0, 'e': 4, 'i': 0, 'o': 1, 'u': 0}`

Time complexity: O(n)
Auxiliary Space: O(n)

Counting vowels: regex way

We can also use this method to perform this task. We can use the regular expression to perform this task. We use re.findall() method to find all the vowels in string make list with them. We use len on list to find total vowels in string.

## Python3

 `import` `re` `# Count vowels in a different way` `# Using re.findall ` `def` `Check_Vow(string, vowels):` `    `  `    ``# Using re.findall in string` `    ``str_list ``=` `re.findall(f``'[{vowels}]'``, string, re.I)` `    `  `    ``# printing length of string` `    ``print``(``len``(str_list))` `    `  `    ``# Returning the list of matched element ` `    ``return` `str_list` `    `  `# Driver Code` `vowels ``=` `'aeiou'` `string ``=` `"Geeks for Geeks"` `print` `(Check_Vow(string, vowels))`

Output

```5
['e', 'e', 'o', 'e', 'e']```

Time complexity: O(n), where n is the length of the input string. The time complexity of re.findall() method is O(n) because it scans the entire string once.
Auxiliary space: O(m), where m is the number of vowels in the input string. The space complexity is proportional to the number of vowels in the input string because we are storing all the matched vowels in a list.

Another approach that could be used to count and display the vowels in a string is to use a combination of the Counter class from the collections module and the filter function.

Here is an example of how you could use this approach to solve the problem:

## Python3

 `from` `collections ``import` `Counter`   `def` `count_and_display_vowels(string):` `    ``vowels ``=` `'aeiouAEIOU'` `    ``vowels_list ``=` `filter``(``lambda` `c: c ``in` `vowels, string)` `    ``count ``=` `Counter(vowels_list)` `    ``return` `count`   `string ``=` `"Geeks for Geeks"` `print``(count_and_display_vowels(string))`

Output

`Counter({'e': 4, 'o': 1})`

This code uses the filter function to create a filter object that contains only the vowels in the input string. It then uses the Counter class to count the number of occurrences of each vowel in the filter object. The Counter class is a subclass of dict that is specifically designed for counting the occurrences of elements in a list. It returns a dictionary with the elements as keys and the number of occurrences as values.

Counting vowels: recursive function way

Here we use recursive function to get the length of vowels, vowels in a given string. we recursively call the function until we reach the base condition.

## Python3

 `#defining recursive function` `def` `Check_Vow(start,string,newlist):` `    ``if` `start``=``=``len``(string):   ``#base condition` `        ``return` `len``(newlist),newlist` `    ``if` `string[start] ``in` `[``'a'``,``'e'``,``'i'``,``'o'``,``'u'``]:   ``#check whether element is vowel or not` `        ``newlist.append(string[start])` `    ``return` `Check_Vow(start``+``1``,string,newlist)  ``#recursive calling` `#driver code` `string ``=` `"Geeks for Geeks"` `#calling recursive function` `res``=``Check_Vow(``0``,string,[])` `#printing result` `print``(``*``res,sep``=``'\n'``)`

Output

```5
['e', 'e', 'o', 'e', 'e']```

Time complexity: O(n)
Auxiliary Space: O(n)

Counting vowels: Using operator.countOf() method

## Python3

 `# Python code to count and display number of vowels` `# Simply using for and comparing it with a` `# string containing all vowels and operator.countOf() method` `import` `operator as op`     `def` `Check_Vow(string, vowels):` `    ``final ``=` `[each ``for` `each ``in` `string ``if` `op.countOf(vowels, each) > ``0``]` `    ``print``(``len``(final))` `    ``print``(final)`     `# Driver Code` `string ``=` `"Geeks for Geeks"` `vowels ``=` `"AaEeIiOoUu"` `Check_Vow(string, vowels)`

Output

```5
['e', 'e', 'o', 'e', 'e']```

Time Complexity: O(n)
Auxiliary Space: O(n)

Counting vowels: Using reduce() and lambda function:

Algorithm:

• Define a function named ‘check_vow’ which takes two arguments, a string and a string containing all vowels.
• Using the reduce function, create a list of vowels in the given string.
• Print the length of the list of vowels and the list of vowels.
• Call the function with the given string and vowel string.

## Python3

 `from` `functools ``import` `reduce`   `def` `check_vow(string, vowels):` `    ``vowel_list ``=` `reduce``(``lambda` `x, y: x ``+` `[y] ``if` `y ``in` `vowels ``else` `x, string, [])` `    ``print``(``len``(vowel_list))` `    ``print``(vowel_list)`   `string ``=` `"Geeks for Geeks"` `vowels ``=` `"AaEeIiOoUu"` `check_vow(string, vowels)` `#This code is contributed by Jyothi pinjala`

Output

```5
['e', 'e', 'o', 'e', 'e']
```

Time Complexity: O(n), where n is the length of the given string. The reduce function iterates over each character in the string, so the time complexity is linear with respect to the length of the string.

Auxiliary Space: O(m), where m is the number of vowels in the given string. The reduce function creates a list of vowels in the string, which is stored in memory. The space complexity is linear with respect to the number of vowels in the string.

My Personal Notes arrow_drop_up