Testing user input against a list in python

I need to test if the user input is the same as an element of a list, right now I’m doing this:

cars = ("red", "yellow", "blue")
guess = str(input())

if guess == cars[1] or guess == cars[2]:
        print("success!")

But I’m working with bigger lists and my if statement is growing a lot with all those checks, is there a way to reference multiple indexes something like:

if guess == cars[1] or cars[2]

or

if guess == cars[1,2,3]

Reading the lists docs I saw that it’s impossible to reference more than one index like, I tried above and of course that sends a syntax error.

Answers:

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Method 1

The simplest way is:

if guess in cars:
    ...

but if your list was huge, that would be slow. You should then store your list of cars in a set:

cars_set = set(cars)
....
if guess in cars_set:
    ...

Checking whether something is present is a set is much quicker than checking whether it’s in a list (but this only becomes an issue when you have many many items, and you’re doing the check several times.)

(Edit: I’m assuming that the omission of cars[0] from the code in the question is an accident. If it isn’t, then use cars[1:] instead of cars.)

Method 2

Use guess in cars to test if guess is equal to an element in cars:

cars = ("red","yellow","blue")
guess = str(input())

if guess in cars:
        print ("success!")

Method 3

Use in:

if guess in cars:
    print( 'success!' )

See also the possible operations on sequence type as documented in the official documentation.

Method 4

@Sean Hobbs:
First you’d have to assign a value to the variable index.

index = 0

You might want to use while True to create the infinite loop, so your code would be like this:

while True:

    champ = input("Guess a champion: ")
    champ = str(champ)

    found_champ = False
    for i in listC:
        if champ == i:
            found_champ = True

    if found_champ:
        print("Correct")
    else:
        print("Incorrect")

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0