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osanseviero Group Title

I need to create a program that counts the string "bob" inside another string, given by the user. The bobs can overlap. What is wrong with my code?

  • one year ago
  • one year ago

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  1. osanseviero Group Title
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    http://dpaste.com/1431048/ How to make it work?

    • one year ago
  2. e.mccormick Group Title
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    Well, you are looking at characters, but trying to test for a string.

    • one year ago
  3. osanseviero Group Title
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    Changing to i wont make it work

    • one year ago
  4. osanseviero Group Title
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    http://dpaste.com/1431056/

    • one year ago
  5. e.mccormick Group Title
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    string.find(s, sub[, start[, end]]) Return the lowest index in s where the substring sub is found such that sub is wholly contained in s[start:end]. Return -1 on failure. Defaults for start and end and interpretation of negative values is the same as for slices. Loop or recurse until you find all copies. As you find one, advance how far you start in the string, or pass from just beyond if recursing.

    • one year ago
  6. osanseviero Group Title
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    I know I need to use slicing and ranges, but I dont know how to apply them in this

    • one year ago
  7. e.mccormick Group Title
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    All stringgs are arrays. Slicing is s[start:stop] basically. But you can use - numbers to go backwards.

    • one year ago
  8. osanseviero Group Title
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    s=raw_input("Enter a word: ") Bobs= 0 for i in range(len(s)): if s[i]=="b": if s[i+1]=="o": if s[i+2]=="b": Bobs +=1 print("Number of bobs: " + str(Bobs)) Why doesnt that work?

    • one year ago
  9. e.mccormick Group Title
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    The end is not inclided in the slice. ``` >>> s="This is a test" >>> s[1:] 'his is a test' >>> s[:-5] 'This is a' >>> s[2:3] 'i' >>> ``` OK, let me look.

    • one year ago
  10. e.mccormick Group Title
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    You can end up testing more than exists. Your s[i+2]=="b": can go past the end of the string.

    • one year ago
  11. osanseviero Group Title
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    So how can i control it?

    • one year ago
  12. osanseviero Group Title
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    String index is out of range :/

    • one year ago
  13. e.mccormick Group Title
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    Yes. And how far are you going last the end in the middle of the loop?

    • one year ago
  14. e.mccormick Group Title
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    ``` for i in range(len(s)): if s[i]=="b": if s[i+1]=="o": if s[i+2]=="b": # <== How much too far? Bobs +=1 ```

    • one year ago
  15. osanseviero Group Title
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    can this work? (and remember me with what can I paste code, please) s=raw_input("Enter a word: ") Bobs= 0 for i in range(len(s)): if i+2<=len(s): if s[i]=="b": if s[i+1]=="o": if s[i+2]=="b": Bobs +=1 print("Number of bobs: " + str(Bobs))

    • one year ago
  16. osanseviero Group Title
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    Forget it, same mistake :/

    • one year ago
  17. e.mccormick Group Title
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    Use ``` above and below the code block. I don't know if that can work, but there is a simpler way to solve it.

    • one year ago
  18. osanseviero Group Title
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    Yep...this worked...But how could I do this simpler?

    • one year ago
  19. e.mccormick Group Title
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    Look at what I asked about how much too far you are going. How far is it past the end of the string?

    • one year ago
  20. osanseviero Group Title
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    2

    • one year ago
  21. e.mccormick Group Title
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    Right. So, how could you change the range so that it goes 2 less?

    • one year ago
  22. osanseviero Group Title
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    so i could make the end of the range the leng - 2

    • one year ago
  23. e.mccormick Group Title
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    Exactly. ``` #s=raw_input("Enter a word: ") s="This is a bob test for bob find bob" Bobs= 0 for i in range(len(s)-2): if s[i]=="b": if s[i+1]=="o": if s[i+2]=="b": Bobs +=1 print("Number of bobs: " + str(Bobs)) ``` Number of bobs: 3

    • one year ago
  24. osanseviero Group Title
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    Oh... now I see :D

    • one year ago
  25. osanseviero Group Title
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    Thanks a lot :D Next problem set looks much harder

    • one year ago
  26. e.mccormick Group Title
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    You can also see one other good thing there for testing. Rather than enter a string each time, I just make a test string. That makes the trying out code faster.

    • one year ago
  27. osanseviero Group Title
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    Thanks :)

    • one year ago
  28. osanseviero Group Title
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    Can you help me with the start of another one?

    • one year ago
  29. e.mccormick Group Title
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    Sure.

    • one year ago
  30. osanseviero Group Title
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    I will open a new question, thanks :)

    • one year ago
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