How to count the number of sub-sequences in a string?












0















This program is to count the number of sub-strings in a string.



Test case:

    string ABCDCDC

    pattern CDC



The answer should be 2 but I am getting 0.



def count_substring(string, sub_string):

  for i in range(0,len(string)-len(sub_string)+1):

    count=0

    for j in range(0,len(sub_string)):

       if(string[i+j]!=sub_string[j]):

           break;

    if j==len(sub_string):

       count=count+1

    return count





python string substring







share|improve this question

























  • Make sure your indentation is correct when you post...

    – Julien
    Nov 21 '18 at 0:05
















0















This program is to count the number of sub-strings in a string.



Test case:

    string ABCDCDC

    pattern CDC



The answer should be 2 but I am getting 0.



def count_substring(string, sub_string):

  for i in range(0,len(string)-len(sub_string)+1):

    count=0

    for j in range(0,len(sub_string)):

       if(string[i+j]!=sub_string[j]):

           break;

    if j==len(sub_string):

       count=count+1

    return count





python string substring







share|improve this question

























  • Make sure your indentation is correct when you post...

    – Julien
    Nov 21 '18 at 0:05














0












0








0








This program is to count the number of sub-strings in a string.



Test case:

    string ABCDCDC

    pattern CDC



The answer should be 2 but I am getting 0.



def count_substring(string, sub_string):

  for i in range(0,len(string)-len(sub_string)+1):

    count=0

    for j in range(0,len(sub_string)):

       if(string[i+j]!=sub_string[j]):

           break;

    if j==len(sub_string):

       count=count+1

    return count





python string substring







share|improve this question
















This program is to count the number of sub-strings in a string.



Test case:

    string ABCDCDC

    pattern CDC



The answer should be 2 but I am getting 0.



def count_substring(string, sub_string):

  for i in range(0,len(string)-len(sub_string)+1):

    count=0

    for j in range(0,len(sub_string)):

       if(string[i+j]!=sub_string[j]):

           break;

    if j==len(sub_string):

       count=count+1

    return count





python string substring




python string substring






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 '18 at 1:28









martineau

68.9k1091186




68.9k1091186










asked Nov 20 '18 at 23:43







user10628441




















  • Make sure your indentation is correct when you post...

    – Julien
    Nov 21 '18 at 0:05



















  • Make sure your indentation is correct when you post...

    – Julien
    Nov 21 '18 at 0:05

















Make sure your indentation is correct when you post...

– Julien
Nov 21 '18 at 0:05





Make sure your indentation is correct when you post...

– Julien
Nov 21 '18 at 0:05












2 Answers
2






active

oldest

votes


















0














At the end of the for j in range(0,len(sub_string)) loop j can be at most len(sub_string)-1 (if you don't break early). So if j==len(sub_string): is always False. You want if j==len(sub_string)-1: or simply else: (meaning you've reached the end of the for loop) instead.



Also count = 0 should be initialized at the start, not in the first for loop:



def count_substring(string, sub_string):

    count=0

    for i in range(0,len(string)-len(sub_string)+1):

        for j in range(0,len(sub_string)):

            if(string[i+j]!=sub_string[j]):

                break;

        else:

            count=count+1

    return count





share|improve this answer































    0














    Strings are sequences of 1-character substrings in Python, so you can use the built-in count() method — in other words, you don't have to re-invent the wheel and write your own function to do it.



    test_string = 'ABCDCDC'
    
sub_string = 'CDC'
    

    
print(test_string.count(sub_string))  # -> 2





    share|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      At the end of the for j in range(0,len(sub_string)) loop j can be at most len(sub_string)-1 (if you don't break early). So if j==len(sub_string): is always False. You want if j==len(sub_string)-1: or simply else: (meaning you've reached the end of the for loop) instead.



      Also count = 0 should be initialized at the start, not in the first for loop:



      def count_substring(string, sub_string):
      
    count=0
      
    for i in range(0,len(string)-len(sub_string)+1):
      
        for j in range(0,len(sub_string)):
      
            if(string[i+j]!=sub_string[j]):
      
                break;
      
        else:
      
            count=count+1
      
    return count





      share|improve this answer




























        0














        At the end of the for j in range(0,len(sub_string)) loop j can be at most len(sub_string)-1 (if you don't break early). So if j==len(sub_string): is always False. You want if j==len(sub_string)-1: or simply else: (meaning you've reached the end of the for loop) instead.



        Also count = 0 should be initialized at the start, not in the first for loop:



        def count_substring(string, sub_string):
        
    count=0
        
    for i in range(0,len(string)-len(sub_string)+1):
        
        for j in range(0,len(sub_string)):
        
            if(string[i+j]!=sub_string[j]):
        
                break;
        
        else:
        
            count=count+1
        
    return count





        share|improve this answer


























          0












          0








          0







          At the end of the for j in range(0,len(sub_string)) loop j can be at most len(sub_string)-1 (if you don't break early). So if j==len(sub_string): is always False. You want if j==len(sub_string)-1: or simply else: (meaning you've reached the end of the for loop) instead.



          Also count = 0 should be initialized at the start, not in the first for loop:



          def count_substring(string, sub_string):
          
    count=0
          
    for i in range(0,len(string)-len(sub_string)+1):
          
        for j in range(0,len(sub_string)):
          
            if(string[i+j]!=sub_string[j]):
          
                break;
          
        else:
          
            count=count+1
          
    return count





          share|improve this answer













          At the end of the for j in range(0,len(sub_string)) loop j can be at most len(sub_string)-1 (if you don't break early). So if j==len(sub_string): is always False. You want if j==len(sub_string)-1: or simply else: (meaning you've reached the end of the for loop) instead.



          Also count = 0 should be initialized at the start, not in the first for loop:



          def count_substring(string, sub_string):
          
    count=0
          
    for i in range(0,len(string)-len(sub_string)+1):
          
        for j in range(0,len(sub_string)):
          
            if(string[i+j]!=sub_string[j]):
          
                break;
          
        else:
          
            count=count+1
          
    return count






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 20 '18 at 23:56









          JulienJulien

          7,69831637




          7,69831637

























              0














              Strings are sequences of 1-character substrings in Python, so you can use the built-in count() method — in other words, you don't have to re-invent the wheel and write your own function to do it.



              test_string = 'ABCDCDC'
              
sub_string = 'CDC'
              

              
print(test_string.count(sub_string))  # -> 2





              share|improve this answer




























                0














                Strings are sequences of 1-character substrings in Python, so you can use the built-in count() method — in other words, you don't have to re-invent the wheel and write your own function to do it.



                test_string = 'ABCDCDC'
                
sub_string = 'CDC'
                

                
print(test_string.count(sub_string))  # -> 2





                share|improve this answer


























                  0












                  0








                  0







                  Strings are sequences of 1-character substrings in Python, so you can use the built-in count() method — in other words, you don't have to re-invent the wheel and write your own function to do it.



                  test_string = 'ABCDCDC'
                  
sub_string = 'CDC'
                  

                  
print(test_string.count(sub_string))  # -> 2





                  share|improve this answer













                  Strings are sequences of 1-character substrings in Python, so you can use the built-in count() method — in other words, you don't have to re-invent the wheel and write your own function to do it.



                  test_string = 'ABCDCDC'
                  
sub_string = 'CDC'
                  

                  
print(test_string.count(sub_string))  # -> 2






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 21 '18 at 1:20









                  martineaumartineau

                  68.9k1091186




                  68.9k1091186






























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