सिद्धांत और अभ्यास

डेटा I/O

कार्य

शर्तें

कार्य

for लूप

कार्य

स्ट्रिंग

कार्य

while लूप

कार्य

सूचियां

कार्य

2D मैट्रिक्स

कार्य

डिक्शनरी

कार्य

सेट

कार्य

फ़ंक्शन

कार्य

पाठ 8. Dictionary

कठिनाई:

कार्य«गणना की विशिष्टता»

💻 Python

आपको नवजात शिशुओं को दिए जाने वाले नामों के डेटा का विश्लेषण करना है। आपके पास एक शब्दकोश है जहाँ कुंजी बच्चे का नाम (स्ट्रिंग) है, और मान वह संख्या है जितनी बार यह नाम प्रति वर्ष नवजात शिशुओं की सूची में पाया जाता है (एक पूर्णांक)। आपका कार्य यह निर्धारित करना है कि कितने अलग-अलग नाम (अद्वितीय कुंजियाँ) उपयोग किए गए थे और कौन सा नाम सबसे लोकप्रिय (सबसे अधिक बार पाया जाने वाला) है। यदि कई नामों की अधिकतम लोकप्रियता समान है, तो उनमें से कोई भी एक नाम प्रदर्शित करें।

इनपुट फॉर्मेट

शब्दकोश प्रस्तुत किया गया है, जो पंक्ति द्वारा दर्शाया गया है। प्रत्येक पंक्ति में नाम और मात्रा एक स्थान से अलग होते हैं। इनपुट एक खाली पंक्ति के साथ समाप्त होता है

आउटपुट फॉर्मेट

पहली पंक्ति में अद्वितीय नामों की संख्या लिखी होती है।

दूसरी पंक्ति में सबसे लोकप्रिय नाम और उसके उल्लेखों की संख्या, एक अंतराल द्वारा अलग करके प्रदर्शित की जाती है।

उदाहरण

इनपुट

एलेक्जेंडर 15
मारिया 12
इवान 15

आउटपुट

3
अलेक्जेंडर 15

संकेत

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.