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Занятие 8. Dictionary

Difficulty level:

Task«Alphabetical purchase list»

You need to create a program that helps the user to make a list of purchases and sort it in alphabetical. The user introduces the name of the product and its quantity. The program should store this information in the dictionary where the name of the product is the key, and the quantity is value. After the user completes the input of the products, the program must display the list of purchases sorted by the alphabet, where each product is indicated for each product.

Input format

First, the user introduces the number of positions in the purchase list.

Output format

The program should display a list of purchases sorted according to the alphabet, where each line contains the name of the product and its amount separated by a gap

Example

Input

3
Apples 5
Banans 2
Orange 3

Output

Orange 3
Banans 2
Apples 5

Hint

Useful methods for working with dictionaries in Python

Dictionaries are one of the most powerful and flexible data structures in Python. To use them effectively, it is important to know the basic methods. Let's take a closer look at them.

The `keys() method`

Returns all dictionary keys as a special representation object dict_keys.

my_dict = {'key1': 'value1', 'key2': 'value2', 'key3': 'value3'}
keys = my_dict.keys()
print(keys)  # Output: dict_keys(['key1', 'key2', 'key3'])

A useful tip: The dict_keys object is iterable, meaning you can use it in a for loop. However, this is not a list. If you need a list of keys (for example, for index access), simply convert it using the list() function.

key_list = list(my_dict.keys())
print(key_list[0]) # Output: key1

The `values() method`

Returns all dictionary values as a representation object dict_values.

values = my_dict.values()
print(values) # Output: dict_values(['value1', 'value2', 'value3'])

A useful tip: Like dict_keys, dict_values can be iterated, but it is not a list. The conversion to list() also works if you need a list of values. This is convenient, for example, to check for the presence of a value.

The `items() method`

Returns all key-value pairs as a dict_items object, where each pair is represented as a tuple.

items = my_dict.items()
print(items) # Output: dict_items([('key1', 'value1'), ('key2', 'value2'), ('key3', 'value3')])

A useful tip: This method is incredibly convenient for iterating through a dictionary when you need both a key and a value at the same time. This is considered the most "Pythonic" way to bypass the dictionary.

for key, value in my_dict.items():
    print(f"Key: {key}, Value: {value}")

`get() method`

Returns the value by key. If there is no key, it returns the default value (or None if it is not specified).

value = my_dict.get("key1")
print(value) # Output: value1

value_nonexistent = my_dict.get("key4", "there is no such key")
print(value_nonexistent) # Output: there is no such key

A useful tip: Use get() instead of direct access through square brackets (my_dict["key4"]) when you are not sure if the key exists in the dictionary. This will avoid the KeyError error and the crash of the program. This is much safer when working with data that comes from external sources.

The `popitem() method`

Deletes and returns the last added key-value pair as a tuple.

last_item = my_dict.popitem()
print(last_item) # Output: ('key3', 'value3')
print(my_dict) # Output: {'key1': 'value1', 'key2': 'value2'}

A useful tip: This method works according to the LIFO (Last-In, First-Out) principle. It is useful when you need to process dictionary elements in the reverse order of their addition. (Guaranteed LIFO order is supported in Python 3.7+).

The `update() method`

Updates the dictionary by adding key-value pairs from another dictionary or from an iterable object with pairs.

# The original my_dict dictionary is already without 'key3'
my_dict.update({"key4": "value4", "key1": "new value1"})
print(my_dict)
# Output: {'key1': 'new value1', 'key2': 'value2', 'key4': 'value4'}

A useful tip: Note that if the key from the updating dictionary already exists in the original one, its value will be overwritten. update() is an effective way to merge two dictionaries.

Nested dictionaries

Dictionaries can contain other dictionaries as values. This allows you to create complex hierarchical data structures very similar to the JSON format.

nested_dict = {
    "key1": {"plugg1": "value1"},
"key2": {"plugg2": "value2"}
}

# Accessing the nested value
print(nested_dict["key1"]["plugg1"]) # Output: value1

An example of a multidimensional structure in dictionaries

students = {
    "Alice": {"age": 25, "grades": [88, 92, 85]},
    "Bob": {"age": 22, "grades": [79, 85, 80]},
    "Charlie": {"age": 23, "grades": [90, 87, 85]}
}

# Getting Alice's age
print(students["Alice"]["age"]) # Output: 25

# Getting the first grade Bob
print(students["Bob"]["grades"][0]) # Output: 79

# Adding a new student
students["David"] = {"age": 24, "grades": [82, 79, 88]}
print(students)

A useful tip: When working with nested dictionaries, the risk of getting a KeyError increases. For secure access to the embedded data, you can use a chain of calls get().

# Safely obtaining the age of a student, which may not be
age = students.get("Eve", {}).get("age", "Unknown")
print(age) # Output: Unknown

Here students.get("Eve", {}) will return an empty dictionary {}, since there is no student "Eve", and already this empty dictionary will be called .get("age", ...), which will prevent the error.

main.py

Test 1

Test 2

Test 3

Test 4

Test 5

Test 6

Test 7

Test 8

Test 9

Test 10

Developer’s solution

🎉 Congratulations! 🎉

You did an excellent job with the task! It was a challenging problem, but you found the correct solution. You are one step closer to mastering programming! Keep up the good work, because every stage you pass makes you even stronger.