String index out of range - Python Logic Based error?

Question

def double_chars(word):
    for j in range(len(word)):
        if word[j] == word[j:j+1]:
            chars_count = chars_count + 1
    return chars_count

test_word = "Hello..world!"
print(double_chars(test_word))


Error: if word[j] == word[j+1]:
IndexError: string index out of range

I keep getting string index out of range on this function. I have tried different ways of indexing and slicing.

I am trying to return the count of character in a string that has double characters for example: bell--rung --> "ll" = 1, and "--" = 1. hence count of 2.

Am I doing something wrong in the code?

How about using range(len(word) - 1)? Probably your error arises when j reaches the last character (i.e., j = len(word) - 1)) in which case j+1 is out of range. This might introduce (or not) another edge case when you provide an empty word. Deal with that separately if needed. — j1-lee
– j1-lee, Commented Nov 9, 2021 at 19:39

azro · Accepted Answer · 2021-11-09 19:52:51Z

1

The thing is that the range generates from 0 (inclusive) to the string's length (exclusive) which is ok for indexing, as string are 0-based indexed, but you don't only access current index, also the next one, so when reaching the last value of j, the [j+1] is one too far

def double_chars(word):
    chars_count = 0
    for j in range(len(word) - 1):
        if word[j] == word[j + 1]:
            chars_count += 1
    return chars_count

	0	1	2	3	4	5	6	7	8	9	10	11
`x[j]`	H	e	l	l	o	.	.	w	o	r	l	d
`x[j+1]`	e	l	l	o	.	.	w	o	r	l	d	!

One-liner proposal, and another with zip

def double_chars(word):
    return sum(1 for j in range(len(word) - 1) if word[j] == word[j + 1])

def double_chars(word):
    return sum(1 for ch_curr, ch_next in zip(word, word[1:]) if ch_curr == ch_next)

edited Nov 9, 2021 at 19:52

answered Nov 9, 2021 at 19:46

azro

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1 Comment

OlaJ Over a year ago

Thank you for the response! Your answer solves the problem and also helps me understand better. Thanks!

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