list indices must be integers or slices, not lists error

[djwilson0495]

For the following code:

def create_ID(tmp):
    try_again = True
    in_list = False
    while try_again == True:
        New_ID = input("Please enter the new ID")
        y = 0
        for x in tmp:
            if New_ID in tmp[x][0]: # checks if ID is already in list row by row
                print("That ID is already on the list") # if it is return this error message 
                in_list = True
                y = y + 1
        if in_list == False: # if ID is not in the list close the loop
            try_again = False
    return New_ID

I'm getting this error:

Error:
Traceback (most recent call last):
  File "c:/Users/djwil/Documents/python/learning python/Chapter 19 - Chunky Challenges/Passwords.py", line 173, in <module>
    main()
  File "c:/Users/djwil/Documents/python/learning python/Chapter 19 - Chunky Challenges/Passwords.py", line 150, in main
    New_ID = create_ID(tmp) # gets New_ID from create_ID
  File "c:/Users/djwil/Documents/python/learning python/Chapter 19 - Chunky Challenges/Passwords.py", line 18, in create_ID
    if New_ID in tmp[x][0]: # checks if ID is already in list row by row
TypeError: list indices must be integers or slices, not list

But I think that I'm defining it as an integer of the a specific place in the list, can someone offer advice?

[Gribouillis]

After for x in tmp it should probably be if New_Id in x[0]

[deanhystad]

If tmp is a list of lists. for x in tmp makes x a list. You cannot use a list as an index into a list. I think you want to do this:

for x in tmp:
    if new_id == x[0]:

tmp is a bad variable name. It tells me nothing about what the variable is or how I should use it. I think that you would have fewer problems if you started using better variable names.

You should also think about what type of data structures best fit your problem. I don't know anything this challenge, but if you used a dictionary instead of a list of lists your test for unique password is:

if passwords.get(new_id):
    print("That ID is already on the list")
else:
    break;

And one last quibble. Printing is not the same as returning.

print("That ID is already on the list") # if it is return this error message 

Misleading comments are far worse than no comment at all.