Match string in python regardless of upper and lower case differences [duplicate]

Question

I'm trying to find a match value from a keyword using python. My values are stored in a list (my_list) and in the below example I'm trying to find the word 'Webcam'. I only want to return the word if it is fully matched.

Using item.find works but only if the case matches (i.e. upper and lower case must be correct). But I want to return the item regardless of the case, However, I do not wish to match all instances of the string like 'Webcamnew' so using the any() method won't work I think. Does anyone know how to do this..?

my_list = ['webcam', 'home', 'Space', 'Maybe later', 'Webcamnew']

for item in my_list:
    if item.find("Webcam") != -1:
        print item

Answer 1

my_list = ['webcam', 'home', 'Space', 'Maybe later', 'Webcamnew']

for item in my_list:
    if 'webcam' == item.lower()
        print item.lower()

Note: Strings are immutable in Python - it doesn't modify the string in the list.

Answer 2

If for some reason you REALLY need to compare case insensitive strings (rather than generating a same-case string) you can use regular expressions with the re.IGNORECASE flag. This is a terrible idea for what you seem to be trying to do, but the code is:

import re

my_list = ['webcam', 'home', 'Space', 'Maybe later', 'Webcamnew']
for item in my_list:
    if re.match("webcam$",item, flags=re.I): # re.I == re.IGNORECASE
        print item

The reason this is a Bad Idea is that using regular expressions for simple matching is kind of like using a backhoe to dig post holes. Sure, you can do it, but it's expensive, time-consuming, and has the opportunity to introduce errors you didn't think about until you accidentally swung the boom through your living room window.