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Python | Count and display vowels in a string
• Difficulty Level : Basic
• Last Updated : 25 Feb, 2020

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}
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

 `# Python code to count and display number of vowels ` `# Simply using for and comparing it with a  ` `# string containg 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']
```

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.

 `# 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:

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

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