Efficient dictionary printing

Question

Here's what my task is:

Dependant on the "Logged In" user, they'll have different access rights and will see different menus.

I have tried using the .get() function.

choices = {1: "| Student Details", 2: "| Student Grades", 3: "| Finance", 4:"| Staff Details", 5:"| Staff Payments"}

print(choices.get(1))
print(choices.get(4))

GIVEN OUTPUT >>

| Student Details
| Staff Details

But it will look very inefficient with all the Access Rights. Some access rights have access to different menus e.g (Access Level 1 = Menu 1 and Menu 4). So I cannot just use a for loop to print all menus like so.

for option,num in choices.items():
    print(option,num)

I've also attempted using the itemgetter() function which is available in the operator module, but I was unable of figuring out how to: 1. Print the outcome in lines, 2. Numerate the outcome in order (1,2,3)

choices = {1: "|Student Details", 2: "| Student Grades", 3: "| Finance", 4:"| Staff Details", 5:"| Staff Payments"}

print(itemgetter(3,5)(choices))

GIVEN OUTPUT >>
('| Finance', '| Staff Payments')

REQUIRED OUTPUT >>
1 | Finance
2 | Staff Payments

Here's an example of the inefficient code I have at the moment:

if user_name == "Mike":
    print("-Managing Menu-")
    print("1 | Staff Payments. ")
    print("2 | Finance. ")
    option = int(input("Option Choice (1-2): "))
    while option not in range(1,2):
        option = int(input("Option Choice (1-2): "))
    if option == 1:
        print("Welcome to Staff Payments")
    elif option == 2:
        print("Welcome to Finance.")
    else:
        print("Error.")

I'd really appreciate any feedback and help with this task.

Answer 1

I'm not sure I understand your exact problem. But maybe you can simplify the code as follows:

class Menu(object):
    def __init__(self, title, group_name):
        self.title = title
        self.group_name = group_name

class User(object):
    def __init__(self, name, groups):
        self.name = name
        self.groups = groups

    def get_menus(self, menus):
        user_menus = []
        for menu in menus:
            if menu.group_name in self.groups:
                user_menus.append(menu)
        return user_menus

if __name__ == '__main__':
    menus = (
        Menu('Student Details', group_name='student-details'),
        Menu('Student Grades', group_name='student-grades'),
        Menu('Finance', group_name='finance')
    )

    users = (
        User('mike', groups=['finance', 'student-details', 'grades']),
        User('John', groups=['finance'])
    )

    # Actual logic

    user = users[0] # mike
    user_menus = user.get_menus(menus)

    print('-Managing Menu-')

    for index, menu in enumerate(user_menus):
        print(index+1, '|', menu.title)

    option = None
    while option not in range(1,len(user_menus)+1):
        try:
            option = int(input('Choose option: '))
        except ValueError:
            print('Invalid option')

    for index, menu in enumerate(user_menus):
        if index+1 == option:
            print('Welcome to', menu.title)

What we did here is created a menu object and a user object that hold all information relating to groups (access rights). In the menu definition we also give it a group name (or access name if you prefer). And then when we define the user we also set the groups that user can access (what menus it can access).

That simplifies the code and makes it so that we don't need to use if's for each user and each group (access right).

After defining all of this it's simply a matter of using user.get_menus(menus) to get the right menus for that user. Later on we loop through the options. We don't hardcode the choices (range(1,2)) but use the already existing user_menus list to know how to create the menu.

Answer 2

You say that using itemgetter is better than using choices.get, as you don't have to call get multiple times, however this is wrong.

If we look at itemgetter it says that it uses __getitem__. Python's function get also uses __getitem__ but allows a default value. And so the following give the same:

print(itemgetter(3, 5)(choices))
print(tuple(map(choices.get, [3, 5])))

As you say using the above is bad, as it doesn't print things on new lines, you could use str.join. As you want to join each one with a newline or '\n'.

print('\n'.join(map(choices.get, [3, 5])))

In both examples this won't give the same output as your inefficient code. As there are no numbers before the choice.

---

To improve your inefficient code, you could alternately make an options dictionary, and use options.get but with a supplied default value.

options = {
    1: "Welcome to Staff Payments",
    2: "Welcome to Finance"
}
print(options.get(option, "Error."))

---

Alternately to your current code pattern, you can make a dictionary holding users with places they are allowed to. This would be by supplying user-names, and the allowed places in a dictionary.

# Access control list
ACL = {
    "Mike": [3, 5],
    "Joe": [1]
}

places = ACL[user_name]
options = dict(enumerate(map(choices.get, places)))
print("-Managing Menu-")
for option, num in choices.items():
    print("{} {}".format(option, num))

option = int(input("Option Choice (1-2): "))
while option not in range(1,2):
    option = int(input("Option Choice (1-2): "))

if option == 1:
    print("Welcome to Staff Payments")
elif option == 2:
    print("Welcome to Finance.")
else:
    print("Error.")

---

As you can see, whilst we took advantage of a new dictionary to print the allowed places, you would still need boilerplate for all the locations. Instead you could use str.format to display the places, just like I did for the menu.

choices = {
    1: "Student Details",
    2: "Student Grades",
    3: "Finance",
    4: "Staff Details",
    5: "Staff Payments"
}

def managing_menu(user_name):
    places = ACL[user_name]
    options = dict(enumerate(map(choices.get, places)))
    print("-Managing Menu-")
    for option, place in choices.items():
        print("{} | {}".format(option, place))

    option = int(input("Where do you wish to go: "))
    while option not in options:
        option = int(input("Where do you wish to go: "))

    location = options.get(option, None)
    if location is None:
        print("Error.")
        return

    print("Welcome to {}.".format(location))

---

There are some other ways that you can improve the code:

• don't use raw int(input())'s instead use them with a try and except. This will cause the program to not crash if you enter Finance, rather than 2.

• You also are probably calling a function after printing "Welcome to Finance", and so you would need to add that functionality to the options dictionary. (It's not hard)

• You should change the place you define option to become a function, this is as you are likely to need the function in more than that one location.

Answer 3

With the itemgetter you have almost done it. You just have to enumerate it and join it with the new line as separator.

print('\n'.join("%s %s" % (str(i),s) for i,s in enumerate(itemgetter(3,5)(choices),1)))

This only prints the REQUIRED OUTPUT. Joe Wallis's answer gives you the full solution.