Loop through a python list and get a 1 or 0 for a match result

Question

I have a question, I tried to search the forum and see if there is something that can help me but unfortunately I couldn't so here it is: Currently I'm using our AD system and check through that list to see if a user exists and depending on that give them access or not

users = [AD user list]
for user in users:
    if user == "user x":
        print "you are in"
     else:
        print "denied"

This works great and all but I want to know if I can make this a bit more efficient, so say if the "user x" is the 10th in the list, this code will print "denied" 9 times before it prints "you are in".

Is it possible to somehow look into the list first and if the string match is in there give a 1 and if not a 0? This way when I print this I get either a 1 or a 0.

Hope my question was clear and I provided enough details. I appreciate all the help.

Answer 1

You can do this using the in operator. For example:

users = ['john', 'adam', 'susan']
exists = 'john' in users      # will be true
exists2 = 'robert' in users     # will be false

# for 1 or 0 output
exists = int(exists)
exists2 = int(exists2)

Answer 2

Im pretty sure you can use .index here instead of a full for loop.

try:
    users.index('user x')
    print 'You are in!' # assuming you are using python 2

catch ValueError:
    print 'Denied!'

Using in is also an option.

if 'user x' in users:
    print 'You are in!'

else:
    print 'Denied!'