在两个不同的列表中查找相同索引号以比较值的最有效方法

Question

I have the following code and it is the login feature that I need help with. I have two lists - usernames and passwords . The login feature asks the user to enter a username and password. If the entered username is in the usernames list AND corresponds to the same index number in the passwords list, THEN, return "Access granted", else "Denied".

I'd be interested in two things for teaching purposes: a) a simple fix to the problem using the two lists as specified. b) suggestions as to the best way to solve this problem. (eg dictionaries, 2darrays, or anything else).

The issue is needing to iterate through both lists simulatenously and look up the same corresponding index number.

Example:

username1 and pass1 = access granted but username1 and pass2 =access denied

CODE:

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

def main():
   mainmenu()


def mainmenu():
   print("****MAIN MENU****")
   print("=======Press L to login :")
   print("=======Press R to register :")
   choice1=input()
   if choice1=="L" or choice1=="l":
      login()
   elif choice1=="R" or choice1=="r":
      register()
   else:
      print("please make a valid selection")

def login():
   print("*****LOGIN SCREEN******")
   username=input("Username: ")
   password=input("Password: ")
   if username in usernames and password in passwords:
      print("yes")
   else:
      print("denied")


def register():
   print("*****REGISTRATION****")
   username=input("Enter a username:")
   password=input("Enter a password:")
   usernames.append(username)
   passwords.append(password)
   answer=input("Do you want to make another registration?")
   if answer=="y":
      register()
   else:
      registration_details()

def registration_details():
   print(usernames)
   print(passwords)

main()

Note: I am aware that storing the lists in a 2d array would be an obvious solution/suggestion, but this fix is necessary for pedagogical reasons - ie students have not yet covered arrays at all. Looking at simple solutions first, but also stackoverflow users would benefit from suggestions to alternate/more efficient methods to solve this problem as well.

UPDATE:

As someone has commented below ...I thought I'd clarify. I'm aware that what is needed is to get at the index numbers of the said values in the lists. My question is - what is the best solution, or some of the solutions. Enumerate. zip. simply using a for loop? It is quite difficult to know how to start in python as there is not just one way ...any comments as to which would be the most idiomatic (pythonic) would also be useful.

BEST ANSWER:

This is possibly the best answer, presented below by Damian Lattenero The indentation, a common error, below is off. Is it possible to also just make a quick comment on why? How to fix it?

def login():
   print("*****LOGIN SCREEN******")
   username=input("Username: ")
   password=input("Password: ")
   for ind, user in enumerate(usernames):
     if username == user and passwords[ind] == password:
       print("correct login")
     else:
       print("invalid username or password")

OUTPUT

*****LOGIN SCREEN******
Username: user3
Password: pass3
invalid username or password
invalid username or password
correct login
>>> 

Answer 1

If you want to teach python foundations...

zip(usernames, passwords)

leads to

dict(zip(usernames, passwords))

but you could also do...

for (idx, username) in enumerate(usernames):
   valid_password = passwords[idx]

Answer 2

I suggest to use dictionary in this case, look I'll show you how:

users_pass = {"user1" : "pass1", "user2":"pass2", "user3":"pass3"}

def login():
   print("*****LOGIN SCREEN******")
   username=input("Username: ")
   password=input("Password: ")
   if username not in users_pass:
      print("The user doesnt exist")
   elif users_pass[username] == password:
      print("password ok")


def register():
   print("*****REGISTRATION****")
   username=input("Enter a username:")
   password=input("Enter a password:")
   users_pass[username] = password
   answer=input("Do you want to make another registration?")
   if answer=="y":
      register()
   else:
      registration_details()

if you only want to use lists:

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      print("correct login")
    else:
      print("invalid username or password")

def register():
  print("*****REGISTRATION****")
  username=input("Enter a username:")
  password=input("Enter a password:")
  users_pass[username] = password
  answer=input("Do you want to make another registration?")
  if answer=="y":
    register()
  else:
    registration_details()

Answer 3

An easy fix to your code, but not recommended, is by using zip() .

You need to replace this if statement:

if username in usernames and password in passwords:
    print("yes")
else:
    print("denied")

by:

if (username, password) in zip(usernames, passwords):
    print("yes")
else:
    print("denied")

However, you can use a dict where you store your unique usernames ans password and then check if the username is in this current dict then check if the password is correct or not.

Answer 4

Here are a couple more methods, neither of which I'd particularly recommend, but most other decent ways have been covered in previous answers.

These methods might be better for teaching some general programming basics, but not necessarily for teaching Python...

# Both methods assume usernames are unique

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

username = "user2"
password = "pass2"


# Method 1, with try-catch

try:
  idx = usernames.index(username)
except ValueError:
  idx = None

if idx is not None and password == passwords[idx]:
  print "yes1"
else:
  print "denied1"


# Method 2, no try-catch

idx = None
if username in usernames:
  idx = usernames.index(username)

  if password != passwords[idx]:
    idx = None

if idx is not None:
  print "yes2"
else:
  print "denied2"

Answer 5

This is a great scenario for the zip and enumerate functions. If I read your question correctly, you want to

• Iterate across both usernames and passwords simultaneously (zip)
• Keep track of index (enumerate)

Given your two lists (usernames and passwords), you will want to do the following

for i, (username, password) in enumerate(zip(usernames, passwords)):
    print(i, username, password)

Here's a description on what's going on.

1) The zip function is taking your usernames and passwords lists and creates a new list (an iterable zip object to be precise) where each username and password is appropriately paired.

>>> zip(usernames, passwords)
<zip object at 0x_________> # hmm, cant see the contents

>>> list(zip(usernames, passwords))
[("user1", "pass1"), ("user2", "pass2"), ("user3","pass3")]

2) The enumerate function is taking a list, and creating a new list (actually an iterable enumerate object) where each item is now paired with an index.

>>> enumerate(usernames)
<enumerate object 0x_________> # Lets make this printable

>>> list(enumerate(usernames))
[(0, "user1"), (1, "user2"), (2, "user3")]

3) When we combine these, we get the following.

>>> list(enumerate(zip(usernames, passwords))
[(0, ("user1", "pass1")), (1, ("user2", "pass2")), (2, ("user3", "pass3"))]

This gives us a list where each element is of the form (index, (username, password)) . Which is super easy to use with a loop!

4) Setup your loop with the above!

for i, (username, password) in enumerate(zip(usernames, passwords)):
    # Freely use i, username and password!