String Index Out of Range Error in Python

Question

I keep getting an "IndexError: string index out of range" error message when I try to execute this code:

#function countLetters(word,letter) should count the number of times
#a particular letter appears in a word.

def countLetters(word, letter):

    count=0

    wordlen=len(word)
    num=0
    wordletter=""

    while(num<=wordlen):
        wordletter=word[num]
        if(wordletter==letter):
            count=count+1
        num=num+1
    return count

print(countLetters("banana", "x"))#should print 0
print(countLetters("banana", "a"))#should print 3
print(countLetters("banana", "b"))#should print 1

The Error Message:

Traceback (most recent call last):
  File "C:\Users\Charlie Chiang\Desktop\9G.py", line 17, in <module>
    print(countLetters("banana", "x"))
  File "C:\Users\Charlie Chiang\Desktop\9G.py", line 10, in countLetters
    var=word[num]
IndexError: string index out of range

The code being run and the code you are looking into, seems to me, are different. There is no var=word[num] in your source while the Traceback seems to think there is — shaktimaan
– shaktimaan, Commented Jul 16, 2014 at 22:12
@shaktimaan: It was close enough; instead of var it is wordletter=word[num]. — Martijn Pieters
– Martijn Pieters, Commented Jul 16, 2014 at 22:13
FYI, I don't know if this is for an assigment, but you can always just use .count(). Example: "banana".count("a") returns 3. — Cory Kramer
– Cory Kramer, Commented Jul 16, 2014 at 22:16

Martijn Pieters · Accepted Answer · 2014-07-16 22:12:57Z

3

You take it one index too far:

while(num<=wordlen):

num must stay strictly below the length:

while num < wordlen:

because Python sequences are 0-based. A string of length 5 has indices 0, 1, 2, 3, and 4, not 5.

answered Jul 16, 2014 at 22:12

Martijn Pieters

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1 Comment

user3838965 Over a year ago

Thank You so much! I feel really stupid now, but at least I know what I did wrong! I can't thank you enough, I've been trying to figure this out for the past 2 days!THANK YOU!

Jan Vlcinsky · Accepted Answer · 2014-07-16 22:14:29Z

1

You are reaching one index too far:

while(num<=wordlen):

For a text "a" the len("a") is 1 and last letter can be reached by index 0. Your while condition allows trying index 1, which is not available.

Bonus: counting using Counter

Python stdlib collections provides excellent Counter:

>>> from collections import Counter
>>> Counter("banana").get("x", 0)
0
>>> Counter("banana").get("a", 0)
3

answered Jul 16, 2014 at 22:14

Jan Vlcinsky

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Comments

Rafael Barros · Accepted Answer · 2014-07-16 22:16:13Z

1

Fixing your code:

def countLetters(word, letter):

    count=0

    wordlen=len(word)
    num=0
    wordletter=""
    #here, because the index access is 0-based
    #you have to test if the num is less than, not less than or equal to length
    #last index == len(word) -1
    while(num<wordlen):
        wordletter=word[num]
        if(wordletter==letter):
            count=count+1
        num=num+1
    return count

print(countLetters("banana", "x"))#should print 0
print(countLetters("banana", "a"))#should print 3
print(countLetters("banana", "b"))#should print 1

more elegant way: str.count method

'banana'.count('x')
'banana'.count('a')
'banana'.count('b')

answered Jul 16, 2014 at 22:16

Rafael Barros

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Comments

CCKx · Accepted Answer · 2014-07-16 22:14:33Z

0

Because the index of the string starts at zero, the highest valid index will be wordlen-1.

answered Jul 16, 2014 at 22:14

CCKx

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Comments

f.rodrigues · Accepted Answer · 2014-07-16 22:15:55Z

-1

The index start at 0, not 1.

Try changing:

wordletter = word[num-1]

answered Jul 16, 2014 at 22:15

f.rodrigues

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1 Comment

Jan Vlcinsky Over a year ago

@f.rodriques The question shows code starting with index 0.

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