'String index out of range' error and I don't understand why

Question

I'm trying to take an input, drop the vowels, and print the new phrase.

Code

Here's the code I'm using.

phrase = input("Input: ")
for i in range(len(phrase)):
    if str(phrase[i]) in ["a", "A", "e", "E", "i", "I", "O", "o", "u", "U"]:  # error at this line
        phrase = str(phrase[i]).replace(phrase[i], "")
print(phrase)

Issue

On line 3, I am getting the error:

string index out of range

I don't understand how, since I thought I limited the index to the length of the phrase.

Even if there are better ways to create this script (and I would love any suggestions), I would also really love to know where what I'm doing is going wrong and how to fix it!

Answer 1

I think you don't need i as index or range(len(phrase)) to replace a character or to drop certain character in phrase string, If you're intend to take an input, drop the vowels, and print the new phrase.

phrase = input("Input: ")
for i in phrase:
    if i in ["a", "A", "e", "E", "i", "I", "O", "o", "u", "U"]:
        phrase = phrase.replace(i, "")
print(phrase)

Answer 2

If I am not wrong, it returns "IndexError: string index out of range", since you can only access one position of the string. By entering a single character or number it works perfectly.

Answer 3

Don't modify the string while you iterate over it. Just keep the letters you want:

phrase = input("Input: ")
new = []
for c in phrase:
    if c not in "aeiouAEIOU":
        new.append(c)
phrase = ''.join(new)
print(phrase)

Or:

phrase = ''.join([c for c in phrase if c not in "aeiuoAEIUO"])

Answer 4

Looks like you're updating phrase while iterating through from 0 to the original length of phrase. The second you drop a character in phrase, the max valid subscript decreases while i will still reach the original phrase length.

You could do this better with filter.

phrase = input("Input: ")
new_phrase = "".join(filter(lambda c: c not in "aAeEiIoOuU", phrase))
print(new_phrase)

Answer 5

It looks like the issue is that you are re-writing the entire phrase variable with a singular character and then you are trying to loop over it again.

In the 4th line, you set phrase equal to a single character, and then you try to loop over it again but then you have phrase[i], but phrase itself now is just length of 1. So if i is greater than 1 (it probably is) it can't do that.

Also, you can loop directly over a string and it will return each character of the string directly.

Try building a new string in line 4 instead, something like

phrase = input("Input: ")
new_string = ''
for i in phrase:
    if i not in ["a", "A", "e", "E", "i", "I", "O", "o", "u", "U"]:
        new_string += i
print(new_string)

edit: update code based off of comments

Answer 6

Spotted Issues

3 issues and 1 improvement:

1. for i in range(len(phrase)):: (a) len of string is a snapshot which will does not fit anymore as soon as the first vowel was removed. (b) indexed for-i loop can be replaced by a for-each loop for c in phrase which loops through each character of a string
2. if str(phrase[I]) in ["a", "A", "e", "E", "i", "I", "O", "o", "u", "U"]: (a) string-conversion is not needed, phrase[i] or c can be compared to string element already. (b) can directly compare if character is in a string of vowels: if c in "aAeEiIoOuU"
3. phrase = str(phrase[i]).replace(phrase[i], ""): replace parts inside a whole, like a character inside a string using phrase = phrase.replace("a", "") - don't replace the single character at a position.
4. phrase = input("Input: "): to improve towards a reproducible example, provide the input as fixed given example: phrase = "Hello World with many Vowels" instead depending on interactive user-input.

Solution

I would suggest a combination of the good answers here:

1. loop for each character in string like suggested by Tim
2. add to a new_phrase like suggested by LLSv2.0
3. drop by replace with empty-string like tried by you and suggested by ANC
4. compare agains given vowels from a string like "aAeEiIoOuU" like Tim did