Python Slicing 'bob' in s

Question

    s = 'gfdhbobobyui'
    bob = 0
    for x in range(len(s)):
         if x == 'bob':
             bob += 1
    print('Number of times bob occurs is: ' + str(bob))

Attempting to write a code that will count the amount of times 'bob' appears in s, but for some reason this always outputs 0 for number of 'bob'.

Ahh why noone uses regex? `len(re.findall("(?=bob)", "gfdhbobobyui"))` — JadedTuna, Oct 27 '13 at 11:27
@Vik2015 I don't know why, but i studied regex for a few pages and i found out that i hate it.. — aIKid, Oct 27 '13 at 15:29

aIKid · Accepted Answer · 2013-10-29T15:53:24.000

3

Here, try this, handcrafted :)

for i, _ in enumerate(s): #i here is the index, equal to "i in range(len(s))"
    if s[i:i+3] == 'bob': #Check the current char + the next three chars.
        bob += 1
print('Number of times bob occurs is: ' + str(bob))

Demo

>>> s = 'gfdhbobobyui'
>>> bob = 0
>>> for i, v in enumerate(s): #i here is the index, equal to "i range(len(s))"
    if s[i:i+3] == 'bob': #Check the current char + the next two chars.
        bob += 1


>>> bob
2

Hope this helps!

edited Oct 29 '13 at 15:53

answered Oct 27 '13 at 10:42

aIKid

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Thank you, what is enumerate? – Patrick Hayes Oct 27 '13 at 10:53
It returns a tuple of the index and the value. So i'm catching it with two variables, `v` is for the current value, and `i` is for the current index. – aIKid Oct 27 '13 at 10:54
For details, see [the docs](http://docs.python.org/3.3/library/functions.html#enumerate) :) – aIKid Oct 27 '13 at 10:55
I'll check that out appreciate it! – Patrick Hayes Oct 27 '13 at 11:01
2

I'd suggest you to write `for i, _ in enumerate(s)` since you don't use `v`. Also, `enumerate` might be slightly slower than using `range(len(s))` and you also do 2 extra unnecessary iterations. – Maxime Chéramy Oct 27 '13 at 11:02

score 3 · Answer 2 · answered Oct 27 '13 at 10:52

3

x is a number, it can't be equal to 'bob'. That's why it always output 0.

You should use x in order to get a substring of s:

bob = 0
for x in range(len(s) - 2):
    if s[x:x+3] == 'bob':
        bob += 1

You can also use enumerate.

answered Oct 27 '13 at 10:52

Maxime Chéramy

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OK thank you that makes since! – Patrick Hayes Oct 27 '13 at 10:58

Games Brainiac · Answer 3 · 2013-10-27T10:48:06.977

2

s = 'gfdhbobobyui'
bob = 0
for x in range(len(s)):
     if s[x:x+3] == 'bob':  # From x to three 3 characters ahead.
         bob += 1
print('Number of times bob occurs is: ' + str(bob))

Working example.

But the best way is this, however it does not work with overlapping strings:

s.ount('bob')

edited Oct 27 '13 at 10:48

answered Oct 27 '13 at 10:41

Games Brainiac

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actually, since you're only comparing the string, `len(s)` is sufficient. – aIKid Oct 27 '13 at 10:48
Also, `str.count()` doesn't work with overlapping occurences :) – aIKid Oct 27 '13 at 10:50
@aIKid You'll gen an index error. – Games Brainiac Oct 27 '13 at 10:50
Nope. I didn't. See my example, i copy pasted it. – aIKid Oct 27 '13 at 10:51
@aIKid Well thats strange, it seems that slicing out of bounds raises no exception. – Games Brainiac Oct 27 '13 at 10:53
Yep. I'm curious of that too actually HAHAHA. But it just works. – aIKid Oct 27 '13 at 10:53
@GamesBrainiac Yep, there is no index error. However, why stopping at `len(s)-1` ? – Maxime Chéramy Oct 27 '13 at 10:54
@Maxime since if you're on the -1th index, it's impossible to have a three character word, isn't it? =) – aIKid Oct 27 '13 at 10:57
@alKid exactly. So you can stop even earlier ;). That's what I'm saying. – Maxime Chéramy Oct 27 '13 at 10:58
@GamesBrainiac You have updated your answer but forgot to change the code on Ideone which was incorrect (didn't work with gfdhbobob because of the `-3` instead of `-2`). – Maxime Chéramy Oct 27 '13 at 11:08

Python Slicing 'bob' in s

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