4 
list=['a','a','x','c','e','e','f','f','f']

i=0
count = 0

while count < len(list)-2:
    if list[i] == list[i+1]:
        if list [i+1] != list [i+2]:
            print list[i]
            i+=1
            count +=1
        else:print "no"
    else:   
        i +=1
        count += 1

I'm getting:

    else:print "no"
   ^
 IndentationError: unexpected indent

I'm trying to only print the elts that match the following element, but not the element following that. I'm new to Python, and I'm not sure why this isn't working.

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6
  • have you pasted your code exactly ? Because there is no error in this piece of code. Commented May 19, 2013 at 3:49
  • 4
    make sure you do not mix in tabs with spaces. Also, have the if and else at the same indentation level. Also, dont use list as a variable name, as it is a builtin type Commented May 19, 2013 at 3:51
  • 2
    Avoid the use of list, since that's a defined internal function (and overriding it will cause problems). Commented May 19, 2013 at 3:51
  • 1
    Also, using a for-loop instead of a while-loop will make all of this a little easier. Commented May 19, 2013 at 4:01
  • @Makoto, that's wrong in three aspects: list is a built-in type and not a function; you don't override it but simply shadow the name in the scope of the variable; and it doesn't directly cause problems unless you also want to reference the type in the same scope. The actual problem with using list as a variable name is that it is a very indescriptive name, that it could confuse people reading the code if they don't see it is shadowed, and that it stops python and tools like pyflakes from warning you about undefined variables and also could make it harder to find bugs. Commented May 19, 2013 at 22:08

2 Answers 2

Reset to default
7

Here is the fixed-up code (added a count += 1 after the else-clause to make sure it terminates):

list=['a','a','x','c','e','e','f','f','f']

i=0
count = 0

while count < len(list)-2:
    if list[i] == list[i+1]:
        if list [i+1] != list [i+2]:
            print list[i]
            i+=1
            count +=1
        else:
            print "no"
            count += 1
    else:   
        i +=1
        count += 1

A more advanced solution using itertools is more compact and easier to get right:

from itertools import groupby

data = ['a','a','x','c','e','e','f','f','f']
for k, g in groupby(data):
    if len(list(g)) > 1:
        print k
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3 Comments

This also shows why you shouldn't use list as a variable name
It also appears that the itertools version catches more conditions (example the 'f' isn't printed using the while loop). Also, the original loop prints nothing if the list is modified to contain three 'a's at once (i is never incremented).
Wow itertools is a great solution. Just replacing the ">" with "==" gives me exactly what I want. Thanks!
2

The code works for me without error (although you get stuck in a loop). Make sure you are not mixing tabs and spaces.

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2 Comments

I'd argue that getting stuck in a loop constitutes an error of some kind.
The loop has nothing to do with the question and I've pointed it out. The question at hand is regarding the unexpected indent.

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