Lesson 026: Dictionary: when the index is not easy to use 2 | after class test questions and answers

Test questions ：

0. Python Does your dictionary support one click （Key） Multivalued （Value）？

answer ： I won't support it , Assigning another value to the same key will directly overwrite the previous value .

1. In the dictionary , If you try to create a key that doesn't exist （Key） What happens to the assignment ？

answer ： The corresponding key is automatically created （Key） And add the corresponding value （Value） go in .

2. Membership operator （in and not in） You can check whether an element exists in the sequence , Of course, it can also be used to check a key （Key） Whether it exists in the dictionary , What kind of inspection is more efficient ？ Why? ？

answer ： Check the key in the dictionary （Key） Existence is more efficient than checking the existence of a specified element in a sequence . Because the principle of dictionary is to use hash algorithm to store , One step in place , There is no need to use a lookup algorithm for matching , So the time complexity is O（1）, Very efficient .

3. Python Key matching （Key） And the value （Value） There are no type restrictions ？

answer ：Python The requirements for keys are relatively strict , They must be hashable （Hash） The object of , Cannot be of variable type （ Including variable 、 list 、 Dictionary itself, etc ）.

however Python There is no limit to the value , They can be arbitrary Python object .

4. Please visually check that the following code is executed , Dictionaries dict1 What is the content of ？

>>> dict1.fromkeys((1, 2, 3), ('one', 'two', 'three')) 
>>> dict1.fromkeys((1, 3), ' Numbers ')

answer ： After execution , Dictionaries dict1 The content is ：{1: ‘ Numbers ’, 3: ‘ Numbers ’}
What we should pay attention to here is ,fromkeys The method is to create a new dictionary directly , Don't try to use it to modify an existing dictionary , Because it will cover the whole dictionary directly and ruthlessly .

5. If you need a dictionary dict1 = {1: ‘one’, 2: ‘two’, 3: ‘three’} copy to dict2, What should you do ？

answer ： You can use the dictionary copy() Method ：dict2 = dict1.copy(), Transfer to... In other languages Python At the beginning, the partners may habitually use the method of assignment directly （dict2 = dict1）, Do this in Python Just copy the reference of the object in .

>>> a = {
    1:'one', 2:'two', 3:'three'}
>>> b = a.copy()
>>> c = a
>>> c[4] = 'four'
>>> c
{
    1: 'one', 2: 'two', 3: 'three', 4: 'four'}
>>> a
{
    1: 'one', 2: 'two', 3: 'three', 4: 'four'}
>>> b
{
    1: 'one', 2: 'two', 3: 'three'}

use one's hands ：

0. Try to write a user login program （ This is an attempt to encapsulate functionality into functions ）, The program implementation is shown in the figure ：

user_data = {
    }

def new_user():
    prompt = ' Please enter a user name ：'
    while True:
        name = input(prompt)
        if name in user_data:
            prompt = ' This user name is already in use , Please re-enter ：'
            continue
        else:
            break

    passwd = input(' Please input a password ：')
    user_data[name] = passwd
    print(' Registered successfully , Try logging in ^_^')

def old_user():
    prompt = ' Please enter a user name ：'
    while True:
        name = input(prompt)
        if name not in user_data:
            prompt = ' The user name you entered does not exist , Please re-enter ：'
            continue
        else:
            break
    
    passwd = input(' Please input a password ：')
    pwd = user_data.get(name)
    if passwd == pwd:
        print(' Welcome to XXOO System , Please click... In the upper right corner X End procedure ！')
    else:
        print(' Wrong password ！')

def showmenu():
    prompt = ''' |---  A new user ：N/n ---| |---  Login account ：E/e ---| |---  Launch program ：Q/q ---| |---  Please enter the instruction code ：'''

    while True:
        chosen = False
        while not chosen:
            choice = input(prompt)
            if choice not in 'NnEeQq':
                print(' The command code you entered is incorrect , Please re-enter ：')
            else:
                chosen = True

        if choice == 'q' or choice == 'Q':
            break
        if choice == 'n' or choice == 'N':
            new_user()
        if choice == 'e' or choice == 'E':
            old_user()

showmenu()