Python: Regex findall returns a list, why does trying to access the list element [0] return an error?

Taken from the documentation, the following is a snippet showing how the regex method findall works, and confirms that it does return a list. ``` re.findall(r"\w+ly", text) ['carefully', 'quickly'] ``` However the following code fragment generates an out of bounds error (`IndexError: list index out of range`)when trying to access the zeroth element of the list returned by findall. Relevant Code Fragment: ``` population = re.findall(",([0-9]*),",line) x = population[0] thelist.append([city,x]) ``` Why does this happen? For some more background, here's how that fragment fits into my entire script: ``` import re thelist = list() with open('Raw.txt','r') as f: for line in f: if line[1].isdigit(): city = re.findall("\"(.*?)\s*\(",line) population = re.findall(",([0-9]*),",line) x = population[0] thelist.append([city,x]) with open('Sorted.txt','w') as g: for item in thelist: string = item[0], ', '.join(map(str, item[1:])) print string ``` EDIT: Read comment below for some background on why this happened. My quick fix was: ``` if population: x = population[0] thelist.append([city,x]) ```