Python如何实现字典？

Question

我想知道python字典如何在后台运行，尤其是动态方面？创建字典时，其初始大小是多少？如果我们用很多元素更新它，我想我们需要扩大哈希表。我想我们需要重新计算散列函数以适应新的更大的散列表的大小，同时又与先前的散列表保持某种逻辑？

如您所见，我不完全了解此结构的内部。

Answer 1

（部分）以下答案来自“ 升级Python技能”：检查字典。有关Python哈希表的更多信息，请参见The
Hood的Python Hash Tables
：

当我们创建字典时，它的初始大小是多少？

1. 从源代码中可以看出：

    /* PyDict_MINSIZE is the starting size for any new dict.
     * 8 allows dicts with no more than 5 active entries; experiments suggested
     * this suffices for the majority of dicts (consisting mostly of usually-small
     * dicts created to pass keyword arguments).
     * Making this 8, rather than 4 reduces the number of resizes for most
     * dictionaries, without any significant extra memory use.
     */
    #define PyDict_MINSIZE 8

假设我们使用许多键值对进行更新，我想我们需要消除散列表。我想我们需要重新计算哈希函数，以适应新的更大哈希表的大小，同时保持与先前哈希表的某种逻辑关系。

每当我们添加键时，CPython都会检查哈希表的大小。如果表已满三分之二，它将调整哈希表的大小GROWTH_RATE（当前设置为3），并插入所有元素：

    /* GROWTH_RATE. Growth rate upon hitting maximum load.
     * Currently set to used*3.
     * This means that dicts double in size when growing without deletions,
     * but have more head room when the number of deletions is on a par with the
     * number of insertions.  See also bpo-17563 and bpo-33205.
     *
     * GROWTH_RATE was set to used*4 up to version 3.2.
     * GROWTH_RATE was set to used*2 in version 3.3.0
     * GROWTH_RATE was set to used*2 + capacity/2 in 3.4.0-3.6.0.
     */
    #define GROWTH_RATE(d) ((d)->ma_used*3)

这USABLE_FRACTION是我上面提到的三分之二：

    /* USABLE_FRACTION is the maximum dictionary load.
     * Increasing this ratio makes dictionaries more dense resulting in more
     * collisions.  Decreasing it improves sparseness at the expense of spreading
     * indices over more cache lines and at the cost of total memory consumed.
     *
     * USABLE_FRACTION must obey the following:
     *     (0 < USABLE_FRACTION(n) < n) for all n >= 2
     *
     * USABLE_FRACTION should be quick to calculate.
     * Fractions around 1/2 to 2/3 seem to work well in practice.
     */
    #define USABLE_FRACTION(n) (((n) << 1)/3)

此外，索引计算为：

i = (size_t)hash & mask;

面具在哪儿HASH_TABLE_SIZE-1。

处理哈希冲突的方法如下：

perturb >>= PERTURB_SHIFT;
i = (i*5 + perturb + 1) & mask;

在源代码中说明：

    The first half of collision resolution is to visit table indices via this
    recurrence:
        j = ((5*j) + 1) mod 2**i
    For any initial j in range(2**i), repeating that 2**i times generates each
    int in range(2**i) exactly once (see any text on random-number generation for
    proof).  By itself, this doesn't help much:  like linear probing (setting
    j += 1, or j -= 1, on each loop trip), it scans the table entries in a fixed
    order.  This would be bad, except that's not the only thing we do, and it's
    actually *good* in the common cases where hash keys are consecutive.  In an
    example that's really too small to make this entirely clear, for a table of
    size 2**3 the order of indices is:
        0 -> 1 -> 6 -> 7 -> 4 -> 5 -> 2 -> 3 -> 0 [and here it's repeating]
    If two things come in at index 5, the first place we look after is index 2,
    not 6, so if another comes in at index 6 the collision at 5 didn't hurt it.
    Linear probing is deadly in this case because there the fixed probe order
    is the *same* as the order consecutive keys are likely to arrive.  But it's
    extremely unlikely hash codes will follow a 5*j+1 recurrence by accident,
    and certain that consecutive hash codes do not.
    The other half of the strategy is to get the other bits of the hash code
    into play.  This is done by initializing a (unsigned) vrbl "perturb" to the
    full hash code, and changing the recurrence to:
        perturb >>= PERTURB_SHIFT;
        j = (5*j) + 1 + perturb;
        use j % 2**i as the next table index;
    Now the probe sequence depends (eventually) on every bit in the hash code,
    and the pseudo-scrambling property of recurring on 5*j+1 is more valuable,
    because it quickly magnifies small differences in the bits that didn't affect
    the initial index.  Note that because perturb is unsigned, if the recurrence
    is executed often enough perturb eventually becomes and remains 0.  At that
    point (very rarely reached) the recurrence is on (just) 5*j+1 again, and
    that's certain to find an empty slot eventually (since it generates every int
    in range(2**i), and we make sure there's always at least one empty slot).