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.