Remove all duplicates from a given string in Python

We are given a string and we need to remove all duplicates from it? What will be the output if the order of character matters?
Examples:

Input : geeksforgeeks
Output : efgkos

This problem has existing solution please refer Remove all duplicates from a given string.
Method 1:

filter_none

edit
close

play_arrow

link
brightness_4
code

from collections import OrderedDict 
  
# Function to remove all duplicates from string 
# and order does not matter 
def removeDupWithoutOrder(str): 
  
    # set() --> A Set is an unordered collection 
    #        data type that is iterable, mutable, 
    #        and has no duplicate elements. 
    # "".join() --> It joins two adjacent elements in 
    #            iterable with any symbol defined in 
    #            "" ( double quotes ) and returns a 
    #            single string 
    return "".join(set(str)) 
  
# Function to remove all duplicates from string 
# and keep the order of characters same 
def removeDupWithOrder(str): 
    return "".join(OrderedDict.fromkeys(str)) 
  
# Driver program 
if __name__ == "__main__": 
    str = "geeksforgeeks"
    print ("Without Order = ",removeDupWithoutOrder(str)) 
    print ("With Order = ",removeDupWithOrder(str)) 

chevron_right






                        filter_none








Output:


Without Order =  egfkosr
With Order    =  geksfor



Method 2:













                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















def removeDuplicate(str): 


    s=set(str) 


    s="".join(s) 


    print("Without Order:",s) 


    t="" 


    for i in str: 


        if(i in t): 


            pass


        else: 


            t=t+i 


        print("With Order:",t) 


      


str="geeksforgeeks"


removeDuplicate(str) 



















                        chevron_right







                        filter_none








Output:


Without Order: rofgeks
With Order: geksfor


What do OrderedDict and fromkeys() do ?


An OrderedDict is a dictionary that remembers the order of the keys that were inserted first. If a new entry overwrites an existing entry, the original insertion position is left unchanged.


 For example see below code snippet :











                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















from collections import OrderedDict 


  


ordinary_dictionary = {} 


ordinary_dictionary['a'] = 1


ordinary_dictionary['b'] = 2


ordinary_dictionary['c'] = 3


ordinary_dictionary['d'] = 4


ordinary_dictionary['e'] = 5


  


# Output = {'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4} 


print (ordinary_dictionary)     


  


ordered_dictionary = OrderedDict() 


ordered_dictionary['a'] = 1


ordered_dictionary['b'] = 2


ordered_dictionary['c'] = 3


ordered_dictionary['d'] = 4


ordered_dictionary['e'] = 5


  


# Output = {'a':1,'b':2,'c':3,'d':4,'e':5} 


print (ordered_dictionary)       



















                        chevron_right







                        filter_none








fromkeys() creates a new dictionary with keys from seq and values set to value and returns list of keys, fromkeys(seq[, value]) is the syntax for fromkeys() method.


Parameters :


    

  • seq : This is the list of values which would be used for dictionary keys preparation.
    • 

  • value : This is optional, if provided then value would be set to this value.
    • 



For example see below code snippet :











                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















from collections import OrderedDict 


seq = ('name', 'age', 'gender') 


dict = OrderedDict.fromkeys(seq) 


  


# Output = {'age': None, 'name': None, 'gender': None} 


print (str(dict))  


dict = OrderedDict.fromkeys(seq, 10) 


  


# Output = {'age': 10, 'name': 10, 'gender': 10} 


print (str(dict))        



















                        chevron_right







                        filter_none








This article is contributed by Shashank Mishra (Gullu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.




To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course.





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:

Improved By :  vs164758


					
		

    Article Tags : 
Python
Strings

           	
Practice Tags : 
Strings

        

         

         
 
                2

                

        



        
        
        
        
        

    

            

            2
            

            

             
                 Based on 2 vote(s)
                    

            
            
            
            
            
        

        
        

        

        

        

        

        

        


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.