IndexError string index out of range

Question

s="(8+(2+4))"
def checker(n):
if len(n) == 0:
    return True
if n[0].isdigit==True:
    if n[1].isdigit==True:
        return False
    else:
        checker(n[1:])
else:
    checker(n[1:])

This is what I have so far. Simple code, trying to see if a string meets the following conditions. However when i perform checker(s) i get:

True
IndexError: string index out of range

Any help? Thanks in advance Edit: The function's purpose is to produce true if the string contains only single digit numbers, and false if 2 or more-figured digits exist in the string.

Answer 1

When the length of n is 0, the n[0] part is going to raise an error because the string in empty. You should add a return statement there instead of print.

def checker(n):
    if len(n) < 2:
        return True
    if n[0] in x:

Note that the conditions must be len(n) < 2 otherwise you'll get an error on n[1] when the length of string is 1.

Secondly you're trying to match characters to a list which contains integers, so the in checks are always going to be False. Either convert the list items to string or better use str.isdigit.

>>> '1'.isdigit()
True
>>> ')'.isdigit()
False
>>> '12'.isdigit()
True

---

Update:

You can use regex and all for this:

>>> import re
def check(strs):
    nums = re.findall(r'\d+',strs)
    return all(len(c) == 1 for c in nums)
... 
>>> s="(8+(2+4))"
>>> check(s)
True
>>> check("(8+(2+42))")
False

---

Working version of your code:

s="(8+(2+4))"
def checker(n):
    if not n:           #better than len(n) == 0, empty string returns False in python
        return True
    if n[0].isdigit():  #str.digit is a method and it already returns a boolean value   
        if n[1].isdigit():   
            return False
        else:
            return checker(n[1:])  # use return statement for recursive calls
                                   # otherwise the recursive calls may return None  
    else:
        return checker(n[1:])        

print checker("(8+(2+4))")
print checker("(8+(2+42))")

output:

True
False