"""
Урок 41. Практикум 10 — весь код примеров одним файлом.

Источник: subjects/python-fundamentals/course/lessons/41-practice-10/examples.html
Файл собран автоматически (tools/build_lesson_examples.py): правьте страницу урока.
Запуск: python lesson-41.py
"""



# ====================================================================
# Пример 1. Counter: подсчёт, топ, арифметика
# ====================================================================
from collections import Counter

words = ["apple", "banana", "apple", "cherry", "banana", "apple"]
counts = Counter(words)
print(counts)                  # Counter({'apple': 3, 'banana': 2, 'cherry': 1})
print(counts.most_common(2))   # [('apple', 3), ('banana', 2)]
print(counts["apple"])         # 3
print(counts["missing"])       # 0 — отсутствующий ключ не бросает KeyError

more = Counter(["apple", "date"])
print(counts + more)           # Counter({'apple': 4, 'banana': 2, 'cherry': 1, 'date': 1})
print(counts - more)           # Counter({'apple': 2, 'banana': 2, 'cherry': 1})
print(list(counts.elements())[:5])   # ['apple', 'apple', 'apple', 'banana', 'banana']


# ====================================================================
# Пример 2. Задание 1 — популярные слова из нескольких текстов
# ====================================================================
def popular_words(limit, *texts):
    words = []
    for text in texts:
        words.extend(text.lower().replace(".", "").replace(",", "").split())
    return Counter(words).most_common(limit)


text1 = "This is a sample text with some repeated words."
text2 = "Another sample text with different words."
text3 = "Text processing is fun when words repeat."
print(popular_words(5, text1, text2, text3))
# [('text', 3), ('words', 3), ('is', 2), ('sample', 2), ('with', 2)]


# ====================================================================
# Пример 3. Задание 2 — defaultdict(list): группировка задач по категории
# ====================================================================
from collections import defaultdict

def group_tasks(tasks):
    result = defaultdict(list)
    for task, category in tasks.items():
        result[category].append(task)
    return dict(result)


tasks = {"task1": "работа", "task2": "учёба", "task3": "развлечения", "task4": "работа", "task5": "учёба"}
grouped = group_tasks(tasks)
print(grouped)
# {'работа': ['task1', 'task4'], 'учёба': ['task2', 'task5'], 'развлечения': ['task3']}


# ====================================================================
# Пример 4. Задание 3 — поиск задач по категории
# ====================================================================
def find_tasks(tasks_by_category, category):
    return tasks_by_category.get(category, [])


print(find_tasks(grouped, "учёба"))   # ['task2', 'task5']
print(find_tasks(grouped, "спорт"))   # [] — категории нет вовсе


# ====================================================================
# Пример 5. Задание 5 — defaultdict(int): счётчик посещений страниц
# ====================================================================
def count_visits(pages):
    counts = defaultdict(int)
    for page in pages:
        counts[page] += 1
    return dict(counts)


pages = ["home", "about", "home", "products", "home", "contact", "products"]
print(count_visits(pages))
# {'home': 3, 'about': 1, 'products': 2, 'contact': 1}


# ====================================================================
# Пример 6. Задание 6 — группировка слов по длине
# ====================================================================
def group_by_length(words):
    groups = defaultdict(list)
    for word in words:
        groups[len(word)].append(word)
    return dict(groups)


fruit_words = ["apple", "banana", "kiwi", "grape", "orange", "peach"]
print(group_by_length(fruit_words))
# {5: ['apple', 'grape', 'peach'], 6: ['banana', 'orange'], 4: ['kiwi']}


# ====================================================================
# Пример 7. Задание 4 — OrderedDict: очередь по приоритету
# ====================================================================
from collections import OrderedDict

def reorder_by_priority(tasks_dict):
    order = {"высокий": 0, "средний": 1, "низкий": 2}
    for key, _ in sorted(tasks_dict.items(), key=lambda item: order[item[1]]):
        tasks_dict.move_to_end(key)
    return tasks_dict


tasks_od = OrderedDict({"task1": "низкий", "task2": "средний", "task3": "высокий", "task4": "низкий", "task5": "высокий"})
print(reorder_by_priority(tasks_od))
# OrderedDict([('task3', 'высокий'), ('task5', 'высокий'), ('task2', 'средний'), ('task1', 'низкий'), ('task4', 'низкий')])
print(list(tasks_od.keys()))
# ['task3', 'task5', 'task2', 'task1', 'task4']


# ====================================================================
# Пример 8. Задание 7 — global и почему от него лучше уходить
# ====================================================================
counter = 0

def increment_counter():
    global counter
    counter += 1


increment_counter()
increment_counter()
print(f"Вызовов функции: {counter}")
# Вызовов функции: 2


# ====================================================================
# Пример 9. Задание 8 — LRU-очередь на OrderedDict
# ====================================================================
def update_lru_queue(tasks, new_tasks, max_size):
    queue = OrderedDict.fromkeys(tasks)
    for task in new_tasks:
        if task in queue:
            queue.move_to_end(task)
        else:
            queue[task] = None
        while len(queue) > max_size:
            queue.popitem(last=False)
    return list(queue)


result = update_lru_queue(
    ["task1", "task2", "task3", "task4", "task5", "task6"],
    ["task4", "task1", "task7", "task2"],
    4,
)
print(result)   # ['task4', 'task1', 'task7', 'task2']


# ====================================================================
# Пример 10. Когда всё же нужен именно OrderedDict, а не обычный dict
# ====================================================================
d1 = {"a": 1, "b": 2}
d2 = {"b": 2, "a": 1}
print(d1 == d2)   # True — обычный dict сравнивает только пары, порядок не важен

od1 = OrderedDict({"a": 1, "b": 2})
od2 = OrderedDict({"b": 2, "a": 1})
print(od1 == od2)   # False — OrderedDict учитывает порядок при сравнении

print(list(d1.keys()))   # ['a', 'b'] — порядок вставки сохранён и у обычного dict