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# Python – Remove Dictionary Key Words

• Last Updated : 16 Mar, 2023

Sometimes, while working with Python strings, we can have a problem in which we need to remove all the words from a string which are a part of key of dictionary. This problem can have application in domains such as web development and day-day programming. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using split() + loop + replace() The combination of above functions can be used to solve this problem. In this, we perform the task of converting string to list of words using split(). Then we perform a replace of word present in string with empty string using replace().

## Python3

 `# Python3 code to demonstrate working of ` `# Remove Dictionary Key Words` `# Using split() + loop + replace()`   `# initializing string` `test_str ``=` `'gfg is best for geeks'`   `# printing original string` `print``(``"The original string is : "` `+` `str``(test_str))`   `# initializing Dictionary` `test_dict ``=` `{``'geeks'` `: ``1``, ``'best'``: ``6``}`   `# Remove Dictionary Key Words` `# Using split() + loop + replace()` `for` `key ``in` `test_dict:` `    ``if` `key ``in` `test_str.split(``' '``):` `        ``test_str ``=` `test_str.replace(key, "")`   `# printing result ` `print``(``"The string after replace : "` `+` `str``(test_str)) `

Output :

```The original string is : gfg is best for geeks
The string after replace : gfg is  for ```

Time Complexity: O(n)

Space Complexity: O(n), where n is length of string.

Method #2 : Using join() + split() This is yet another way in which this task can be performed. In this, we reconstruct new string using join(), performing join by the empty string after split.

## Python3

 `# Python3 code to demonstrate working of ` `# Remove Dictionary Key Words` `# Using join() + split()`   `# initializing string` `test_str ``=` `'gfg is best for geeks'`   `# printing original string` `print``(``"The original string is : "` `+` `str``(test_str))`   `# initializing Dictionary` `test_dict ``=` `{``'geeks'` `: ``1``, ``'best'``: ``6``}`   `# Remove Dictionary Key Words` `# Using join() + split()` `temp ``=` `test_str.split(``' '``)` `temp1 ``=` `[word ``for` `word ``in` `temp ``if` `word.lower() ``not` `in` `test_dict]` `res ``=` `' '``.join(temp1)`   `# printing result ` `print``(``"The string after replace : "` `+` `str``(res)) `

Output :

```The original string is : gfg is best for geeks
The string after replace : gfg is  for ```

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

Method #3: Using keys(),split() methods and in operator

## Python3

 `# Python3 code to demonstrate working of ` `# Remove Dictionary Key Words` `# Using join() + split()`   `# initializing string` `test_str ``=` `'gfg is best for geeks'`   `# printing original string` `print``(``"The original string is : "` `+` `str``(test_str))`   `# initializing Dictionary` `test_dict ``=` `{``'geeks'` `: ``1``, ``'best'``: ``6``}`   `# Remove Dictionary Key Words` `# Using join() + split()` `temp ``=` `test_str.split(``' '``)` `temp1 ``=` `[word ``for` `word ``in` `temp ``if` `word.lower() ``not` `in` `test_dict]` `res ``=` `' '``.join(temp1)`   `# printing result ` `print``(``"The string after replace : "` `+` `str``(res)) `

Output

```The original string is : gfg is best for geeks
The string after replace : gfg is for```

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

Method #4: Using List Comprehension and Join() Method

In this method, use a list comprehension to iterate over the words in the string and check if each word is not a key in the dictionary. If it is not a key, add it to a new list. Finally, join the new list with spaces to get the modified string.

## Python

 `# Python3 code to demonstrate working of ` `# Remove Dictionary Key Words` `# Using List Comprehension and Join() Method`   `# initializing string` `test_str ``=` `'gfg is best for geeks'`   `# printing original string` `print``(``"The original string is : "` `+` `str``(test_str))`   `# initializing Dictionary` `test_dict ``=` `{``'geeks'` `: ``1``, ``'best'``: ``6``}`   `# Remove Dictionary Key Words` `# Using List Comprehension and Join() Method` `new_list ``=` `[word ``for` `word ``in` `test_str.split() ``if` `word ``not` `in` `test_dict.keys()]` `test_str ``=` `" "``.join(new_list)`   `# printing result ` `print``(``"The string after replace : "` `+` `str``(test_str)) `

Output

```The original string is : gfg is best for geeks
The string after replace : gfg is for```

Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(n), where n is the length of the string (used for creating the new list).

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