python - 为稀疏的 64 位无符号整数创建最小完美哈希

就个人而言，我只是使用CMPH gperf生成一个表，或者为大量键生成一个表，然后完成它。

如果您必须在 Python 中执行此操作，那么我发现这篇博客文章包含一些 Python 2 代码，这些代码在使用中间表将字符串键转换为最小完美哈希方面非常有效。

根据您的要求调整帖子中的代码，在 0.35 秒内为 50k 项生成最小完美哈希：

>>> import random
>>> testdata = {random.randrange(2**64): random.randrange(2**64)
...             for __ in range(50000)}  # 50k random 64-bit keys
>>> import timeit
>>> timeit.timeit('gen_minimal_perfect_hash(testdata)', 'from __main__ import  gen_minimal_perfect_hash, testdata', number=10)
3.461486832005903

我所做的更改：

我切换到 Python 3，遵循 Python 样式指南并使代码更 Pythonic
我将 64 位无符号整数转换为 8 字节字节串int.to_bytes()
我没有存储负数，而是使用标志来区分中间表中的直接输入值和散列输入值

改编后的代码：

# Easy Perfect Minimal Hashing
# By Steve Hanov. Released to the public domain.
# Adapted to Python 3 best practices and 64-bit integer keys by Martijn Pieters
#
# Based on:
# Edward A. Fox, Lenwood S. Heath, Qi Fan Chen and Amjad M. Daoud,
# "Practical minimal perfect hash functions for large databases", CACM, 35(1):105-121
# also a good reference:
# Compress, Hash, and Displace algorithm by Djamal Belazzougui,
# Fabiano C. Botelho, and Martin Dietzfelbinger
from itertools import count, groupby


def fnv_hash_int(value, size, d=0x811c9dc5):
    """Calculates a distinct hash function for a given 64-bit integer.

    Each value of the integer d results in a different hash value. The return
    value is the modulus of the hash and size.

    """
    # Use the FNV algorithm from http://isthe.com/chongo/tech/comp/fnv/
    # The unsigned integer is first converted to a 8-character byte string.
    for c in value.to_bytes(8, 'big'):
        d = ((d * 0x01000193) ^ c) & 0xffffffff

    return d % size


def gen_minimal_perfect_hash(dictionary, _hash_func=fnv_hash_int):
    """Computes a minimal perfect hash table using the given Python dictionary.

    It returns a tuple (intermediate, values). intermediate and values are both
    lists. intermediate contains the intermediate table of indices needed to
    compute the index of the value in values; a tuple of (flag, d) is stored, where
    d is either a direct index, or the input for another call to the hash function.
    values contains the values of the dictionary.

    """
    size = len(dictionary)

    # Step 1: Place all of the keys into buckets
    buckets = [[] for __ in dictionary]
    intermediate = [(False, 0)] * size
    values = [None] * size

    for key in dictionary:
        buckets[_hash_func(key, size)].append(key)

    # Step 2: Sort the buckets and process the ones with the most items first.
    buckets.sort(key=len, reverse=True)
    # Only look at buckets of length greater than 1 first; partitioned produces
    # groups of buckets of lengths > 1, then those of length 1, then the empty
    # buckets (we ignore the last group).
    partitioned = (g for k, g in groupby(buckets, key=lambda b: len(b) != 1))
    for bucket in next(partitioned, ()):
        # Try increasing values of d until we find a hash function
        # that places all items in this bucket into free slots
        for d in count(1):
            slots = {}
            for key in bucket:
                slot = _hash_func(key, size, d=d)
                if values[slot] is not None or slot in slots:
                    break
                slots[slot] = dictionary[key]
            else:
                # all slots filled, update the values table; False indicates
                # these values are inputs into the hash function
                intermediate[_hash_func(bucket[0], size)] = (False, d)
                for slot, value in slots.items():
                    values[slot] = value
                break

    # The next group is buckets with only 1 item. Process them more quickly by
    # directly placing them into a free slot.
    freelist = (i for i, value in enumerate(values) if value is None)

    for bucket, slot in zip(next(partitioned, ()), freelist):
        # These are 'direct' slot references
        intermediate[_hash_func(bucket[0], size)] = (True, slot)
        values[slot] = dictionary[bucket[0]]

    return (intermediate, values)


def perfect_hash_lookup(key, intermediate, values, _hash_func=fnv_hash_int):
    "Look up a value in the hash table defined by intermediate and values"
    direct, d = intermediate[_hash_func(key, len(intermediate))]
    return values[d if direct else _hash_func(key, len(values), d=d)]

上面生成了两个列表，每个列表有 50k 个条目；第一个表中的值是(boolean, integer)整数在该范围内的元组[0, tablesize)（理论上，这些值的范围最高可达 2^16，但如果需要 65k+ 次尝试为您的数据找到插槽排列，我会感到非常惊讶）。您的表大小小于 50k，因此当将其表示为 C 数组时，上述安排使得可以将此列表中的条目存储为 4 个字节（bool并且short为 3 个，但对齐规则添加一个字节填充）。

快速测试一下哈希表是否正确并再次产生正确的输出：

>>> tables = gen_minimal_perfect_hash(testdata)
>>> for key, value in testdata.items():
...     assert perfect_hash_lookup(key, *tables) == value
...

您只需要在 C 中实现查找功能：

该fnv_hash_int操作可以获取一个指向 64 位整数的指针，然后将该指针转换为一个 8 位值数组，并将索引递增 8 次以访问每个单独的字节；使用合适的函数来确保大端（网络）顺序。
您不需要0xffffffff在 C 中进行掩码，因为无论如何都会自动丢弃 C 整数值的溢出。
len(intermediate) == len(values) == len(dictionary)并且可以在常数中捕获。
假设 C99，将中间表存储为结构类型的数组，其中aflag为unsigned ；这只是 3 个字节，加上 1 个填充字节以在 4 字节边界上对齐。数组中的数据类型取决于输入字典中的值。booldshortvalues

如果您原谅我的 C 技能，这里有一个示例实现：

mph_table.h

#include "mph_generated_table.h"
#include <arpa/inet.h>
#include <stdint.h>

#ifndef htonll
// see https://stackoverflow.com/q/3022552
#define htonll(x) ((1==htonl(1)) ? (x) : ((uint64_t)htonl((x) & 0xFFFFFFFF) << 32) | htonl((x) >> 32))
#endif

uint64_t mph_lookup(uint64_t key);

mph_table.c

#include "mph_table.h"
#include <stdbool.h>
#include <stdint.h>

#define FNV_OFFSET 0x811c9dc5
#define FNV_PRIME 0x01000193

uint32_t fnv_hash_modulo_table(uint32_t d, uint64_t key) {
    d = (d == 0) ? FNV_OFFSET : d;
    uint8_t* keybytes = (uint8_t*)&key;
    for (int i = 0; i < 8; ++i) {
        d = (d * FNV_PRIME) ^ keybytes[i];
    }
    return d % TABLE_SIZE;
}

uint64_t mph_lookup(uint64_t key) {
    _intermediate_entry entry = 
        mph_tables.intermediate[fnv_hash_modulo_table(0, htonll(key))];
    return mph_tables.values[
        entry.flag ?
            entry.d : 
            fnv_hash_modulo_table((uint32_t)entry.d, htonll(key))];
}

这将依赖于生成的头文件，产生于：

from textwrap import indent

template = """\
#include <stdbool.h>
#include <stdint.h>

#define TABLE_SIZE %(size)s

typedef struct _intermediate_entry {
    bool flag;
    uint16_t d;
} _intermediate_entry;
typedef struct mph_tables_t {
    _intermediate_entry intermediate[TABLE_SIZE];
    uint64_t values[TABLE_SIZE];
} mph_tables_t;

static const mph_tables_t mph_tables = {
    {  // intermediate
%(intermediate)s
    },
    {  // values
%(values)s
    }
};
"""

tables = gen_minimal_perfect_hash(dictionary)
size = len(dictionary)
cbool = ['false, ', 'true,  ']
perline = lambda i: zip(*([i] * 10))
entries = (f'{{{cbool[e[0]]}{e[1]:#06x}}}' for e in tables[0])
intermediate = indent(',\n'.join([', '.join(group) for group in perline(entries)]), ' ' * 8)
entries = (format(v, '#018x') for v in tables[1])
values = indent(',\n'.join([', '.join(group) for group in perline(entries)]), ' ' * 8)

with open('mph_generated_table.h', 'w') as generated:
    generated.write(template % locals())

dictionary你的输入表在哪里。

用散列函数编译gcc -O3是内联的（循环展开），整个mph_lookup函数以 300 个 CPU 指令运行。循环遍历我生成的所有 50k 个随机密钥的快速基准测试表明，我的 2.9 GHz Intel Core i7 笔记本电脑每秒可以为这些密钥查找 5000 万个值（每个密钥 0.02 微秒）。

python - 为稀疏的 64 位无符号整数创建最小完美哈希

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