python-3.x - Counter 的 most_common 方法的底层实现是什么？

Question

我找到了一个 pyi 文件，它具有以下定义

def most_common(self, n: Optional[int] = ...) -> List[Tuple[_T, int]]: ...

这怎么可能发生？列表没有定义，没有执行？

---

在这里为追随者强调一些有价值的建议：

列表是从打字模块导入的；它与列表不同。.pyi 文件不需要导入它，因为存根文件永远不会被执行；它们只需要在语法上是有效的 Python

如果您使用 from future import annotations，您将不必导入类型来使用 List 等。在 .py 文件中的函数注释中也是如此，因为函数注释将被视为字符串文字。（从 Python 4 开始，这将是默认行为。有关详细信息，请参阅 PEP 563。）

标签： python-3.xcounterpyi

Answer 1

您正在查看pyi仅用于注释的文件。它永远不会由 Python 解释器执行。pyi您可以通过阅读PEP484了解有关文件的更多信息。

使用调试器，在您调用的行上放置一个断点，most_common然后单步执行该方法。

Python 3.7 实现。

...\Lib\collections\__init__.py：

def most_common(self, n=None):
    '''List the n most common elements and their counts from the most
    common to the least.  If n is None, then list all element counts.

    >>> Counter('abcdeabcdabcaba').most_common(3)
    [('a', 5), ('b', 4), ('c', 3)]

    '''
    # Emulate Bag.sortedByCount from Smalltalk
    if n is None:
        return sorted(self.items(), key=_itemgetter(1), reverse=True)
    return _heapq.nlargest(n, self.items(), key=_itemgetter(1))

_heapq.nlargest(in ...\Lib\heapq.py) 实施：

def nlargest(n, iterable, key=None):
    """Find the n largest elements in a dataset.

    Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]
    """

    # Short-cut for n==1 is to use max()
    if n == 1:
        it = iter(iterable)
        sentinel = object()
        if key is None:
            result = max(it, default=sentinel)
        else:
            result = max(it, default=sentinel, key=key)
        return [] if result is sentinel else [result]

    # When n>=size, it's faster to use sorted()
    try:
        size = len(iterable)
    except (TypeError, AttributeError):
        pass
    else:
        if n >= size:
            return sorted(iterable, key=key, reverse=True)[:n]

    # When key is none, use simpler decoration
    if key is None:
        it = iter(iterable)
        result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)]
        if not result:
            return result
        heapify(result)
        top = result[0][0]
        order = -n
        _heapreplace = heapreplace
        for elem in it:
            if top < elem:
                _heapreplace(result, (elem, order))
                top, _order = result[0]
                order -= 1
        result.sort(reverse=True)
        return [elem for (elem, order) in result]

    # General case, slowest method
    it = iter(iterable)
    result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)]
    if not result:
        return result
    heapify(result)
    top = result[0][0]
    order = -n
    _heapreplace = heapreplace
    for elem in it:
        k = key(elem)
        if top < k:
            _heapreplace(result, (k, order, elem))
            top, _order, _elem = result[0]
            order -= 1
    result.sort(reverse=True)
    return [elem for (k, order, elem) in result]