Python: Expected an indented block [duplicate]

Question

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I'm getting an IndentationError. How do I fix it? (5 answers)

Closed last year.

I thought everything was properly indented here but I am getting an IndentationError: expected an indented block at the else: statement. Am I making an obvious mistake here?

def anti_vowel(text):
    new_string = ""
    vowels = "aeiou"
    for letter in text:
       for vowel in vowels:
           if (lower(letter) == vowel):
               #do nothing
           else:
               #append letter to the new string
               new_string += letter
    return new_string

Answer 1

Do nothing translates to using the pass keyword to fill an otherwise empty block (which is not allowed). See the official documentation for more information.

def anti_vowel(text):
    new_string = ""
    vowels = "aeiou"
    for letter in text:
       for vowel in vowels:
           if (lower(letter) == vowel):
               #do nothing
               pass
           else:
               #append letter to the new string
               new_string += letter
    return new_string

Answer 2

You need to put something inside the if block. If you don't want to do anything, put pass.

Alternatively, just reverse your condition so you only have one block:

if lower(letter) != vowel:
    new_string += letter

Incidentally, I don't think your code will do what you intend it to do, but that's an issue for another question.

Answer 3

You can't do this.

if (lower(letter) == vowel):
    #do nothing

Try:

if (lower(letter) == vowel):
    #do nothing
    pass

Answer 4

This can be caused by mixing tabs and spaces. However, this is probably caused by having nothing inside your if statement. You can use "pass" as a placeholder (see example below).

    if condition:
        pass

Information on pass: http://docs.python.org/release/2.5.2/ref/pass.html