(некоторые из) Следующие ответы взяты из Обновите свои навыки Python: Изучение словаря :
Когда мы создаем словарь, каков его начальный размер?
Как видно из исходного кода:
/* 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
Представьте, что мы обновили много пар ключ-значение, я полагаю, нам нужно получить внешнюю хеш-таблицу.Я предполагаю, что нам нужно пересчитать хеш-функцию, чтобы адаптировать размер новой большей хеш-таблицы, сохраняя при этом своего рода логику с предыдущей хеш-таблицей ....
CPython проверяет размер хеш-таблицы каждый разраз мы добавим ключ.Если таблица заполнена на две трети, она изменит размер хеш-таблицы на GROWTH_RATE (которая в настоящее время установлена на 3) и вставит все элементы:
/* 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)
The USABLE_FRACTION - это две трети, которые я упомянул выше:
/* 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)
Кроме того, вычисление индекса:
i = (size_t)hash & mask;
, где маска HASH_TABLE_SIZE-1.
Вот какрассматриваются хеш-коллизии:
perturb >>= PERTURB_SHIFT;
i = (i*5 + perturb + 1) & mask;
Объясняется в исходном коде :
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).