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

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

taken documentation, next snippet showing how regex method findall works, , confirms homecoming list.

re.findall(r"\w+ly", text) ['carefully', 'quickly']

however next code fragment generates out of bounds error (indexerror: list index out of range)when trying access zeroth element of list returned findall.

relevant code fragment:

population = re.findall(",([0-9]*),",line) x = population[0] thelist.append([city,x])

why happen?

for more background, here's how fragment fits entire script:

import re thelist = list() open('raw.txt','r') f: 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]) open('sorted.txt','w') g: item in thelist: string = item[0], ', '.join(map(str, item[1:])) print string

edit: read comment below background on why happened. quick prepare was:

if population: x = population[0] thelist.append([city,x])

re.findall homecoming empty list if there no matches:

>>> re.findall(r'\w+ly', 'this not work') []

python regex