Question

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?

Answer 1

http://dpaste.com/1431048/ How to make it work?

Answer 2

Well, you are looking at characters, but trying to test for a string.

Answer 3

Changing to i wont make it work

Answer 4

http://dpaste.com/1431056/

Answer 5

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.

Answer 6

I know I need to use slicing and ranges, but I dont know how to apply them in this

Answer 7

All stringgs are arrays. Slicing is s[start:stop] basically. But you can use - numbers to go backwards.

Answer 8

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?

Answer 9

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.

Answer 10

You can end up testing more than exists. Your s[i+2]=="b": can go past the end of the string.

Answer 11

So how can i control it?

Answer 12

String index is out of range :/

Answer 13

Yes. And how far are you going last the end in the middle of the loop?

Answer 14

```
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
```

Answer 15

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))

Answer 16

Forget it, same mistake :/

Answer 17

Use ``` above and below the code block. I don't know if that can work, but there is a simpler way to solve it.

Answer 18

Yep...this worked...But how could I do this simpler?

Answer 19

Look at what I asked about how much too far you are going. How far is it past the end of the string?

Answer 20

2

Answer 21

Right. So, how could you change the range so that it goes 2 less?

Answer 22

so i could make the end of the range the leng - 2

Answer 23

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

Answer 24

Oh... now I see :D

Answer 25

Thanks a lot :D Next problem set looks much harder

Answer 26

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.

Answer 27

Thanks :)

Answer 28

Can you help me with the start of another one?

Answer 29

Sure.

Answer 30

I will open a new question, thanks :)