检查项目是否在列表列表中

Question

I am making a password saver and I have a pre-determined list of lists for testing purposes which is:

passwords = [["yahoo","XqffoZeo"],["google","CoIushujSetu"]]

I give the user a number of choices. Choice #2 gives them the option to lookup a password

if choice == '2':  # Lookup a password
    print("Which website do you want to lookup the password for?")
    for keyvalue in passwords:
        print(keyvalue[0])
    passwordToLookup = input()

What I need to figure out is:

1. Setting up a loop that loops through all items in the list of lists using a FOR loop. I was told that the best way to think about a list of lists is like an Excel spreadsheet. In the case of my passwords list, I start with 2 rows and 2 columns of data (not considering any passwords that might get added by the user). So, passwords[0][0] will be equal to the item in row 1, column 1 (in my case, "yahoo"). I was told to use a combination of range() and len() so I can iterate through each "row", regardless of how long the list may be. But, I am not exactly sure how to accomplish that.
2. Once I get the FOR loop set up correctly, I need to then iterate through each "row" and compare the user's passwordToLookup input to the corresponding website name using conditional logic (eg "if..."). Since I know that the website names are stored in the first "column" that it might look something like this: passwords[i][0]. But once again, all my attempts fail and I need guidance.

Answer 1

Using filter , it is possible to select the matching element(s) in the list.

Using a list comprehension, select the second element (password) of each.

passwords = [["yahoo","XqffoZeo"],["google","CoIushujSetu"]]
website_to_lookup = input()
[e[1] for e in filter(lambda i:i[0]==website_to_lookup,passwords)]

And if you have more than one password per website, use slicing to return them all: [e[1:] for e in filter(lambda i:i[0]==website_to_lookup,passwords)]

Answer 2

Python makes many things easy and one of them is iterating through a list(or Iterator)

Suppose this is your list of list:

passwords = [["yahoo","XqffoZeo"],["google","CoIushujSetu"], ['StackOverflow','uesodsiom'], ['Facebook', 'sdhf9wk']]

You can iterate through a list like this:

for lst in passwords:
        print(lst)
['yahoo', 'XqffoZeo']
['google', 'CoIushujSetu']
['StackOverflow', 'uesodsiom']
['Facebook', 'sdhf9wk']

You did not need range or len in this case, although they could be useful in other cases.

What if you only wanted a single element of the nested list that is your website? Then you could just index:

for lst in passwords:
        print(lst[0])

yahoo
google
StackOverflow
Facebook

If you wanted the passwords, use lst[1]. The indexing starts from 0.

Note the structure of your list should remain the same: first element is the website name and the second element is the password.

Now, from my understanding you want to check if given a website, it matches a password

First method:

website = 'Facebook'
passwordToLookup = 'sdhf9wk'

for lst in passwords:
    if lst[0] == website and lst[1] == passwordToLookup:
        print(True)
True

Second Method where you are unpacking each element in the list of passwords:

for websi, password in passwords:
    if websi == website and password == passwordToLookup:
        print(True)

True

Unpacking example:

x, y =  [1, 2]
print(x)
1
print(y)
2