我最近也走上了这条路;虽然听起来我的应用程序略有不同。我对在大量字符串上近似集合操作很感兴趣。
您确实做出了需要快速位向量的关键观察。根据您想要放入布隆过滤器的内容,您可能还需要考虑所使用的散列算法的速度。你可能会发现这个库很有用。您可能还想修改下面使用的随机数技术,该技术仅对您的密钥进行一次散列。
在非 Java 位数组实现方面:
我使用BitVector构建了我的布隆过滤器。我花了一些时间分析和优化库,并将我的补丁回馈给 Avi。转到该 BitVector 链接并向下滚动到 v1.5 中的致谢以查看详细信息。最后,我意识到性能不是这个项目的目标,并决定不使用它。
这是我躺在身边的一些代码。我可以把它放在 python-bloom 的谷歌代码上。欢迎提出建议。
from BitVector import BitVector
from random import Random
# get hashes from http://www.partow.net/programming/hashfunctions/index.html
from hashes import RSHash, JSHash, PJWHash, ELFHash, DJBHash
#
# ryan.a.cox@gmail.com / www.asciiarmor.com
#
# copyright (c) 2008, ryan cox
# all rights reserved
# BSD license: http://www.opensource.org/licenses/bsd-license.php
#
class BloomFilter(object):
def __init__(self, n=None, m=None, k=None, p=None, bits=None ):
self.m = m
if k > 4 or k < 1:
raise Exception('Must specify value of k between 1 and 4')
self.k = k
if bits:
self.bits = bits
else:
self.bits = BitVector( size=m )
self.rand = Random()
self.hashes = []
self.hashes.append(RSHash)
self.hashes.append(JSHash)
self.hashes.append(PJWHash)
self.hashes.append(DJBHash)
# switch between hashing techniques
self._indexes = self._rand_indexes
#self._indexes = self._hash_indexes
def __contains__(self, key):
for i in self._indexes(key):
if not self.bits[i]:
return False
return True
def add(self, key):
dupe = True
bits = []
for i in self._indexes(key):
if dupe and not self.bits[i]:
dupe = False
self.bits[i] = 1
bits.append(i)
return dupe
def __and__(self, filter):
if (self.k != filter.k) or (self.m != filter.m):
raise Exception('Must use bloom filters created with equal k / m paramters for bitwise AND')
return BloomFilter(m=self.m,k=self.k,bits=(self.bits & filter.bits))
def __or__(self, filter):
if (self.k != filter.k) or (self.m != filter.m):
raise Exception('Must use bloom filters created with equal k / m paramters for bitwise OR')
return BloomFilter(m=self.m,k=self.k,bits=(self.bits | filter.bits))
def _hash_indexes(self,key):
ret = []
for i in range(self.k):
ret.append(self.hashes[i](key) % self.m)
return ret
def _rand_indexes(self,key):
self.rand.seed(hash(key))
ret = []
for i in range(self.k):
ret.append(self.rand.randint(0,self.m-1))
return ret
if __name__ == '__main__':
e = BloomFilter(m=100, k=4)
e.add('one')
e.add('two')
e.add('three')
e.add('four')
e.add('five')
f = BloomFilter(m=100, k=4)
f.add('three')
f.add('four')
f.add('five')
f.add('six')
f.add('seven')
f.add('eight')
f.add('nine')
f.add("ten")
# test check for dupe on add
assert not f.add('eleven')
assert f.add('eleven')
# test membership operations
assert 'ten' in f
assert 'one' in e
assert 'ten' not in e
assert 'one' not in f
# test set based operations
union = f | e
intersection = f & e
assert 'ten' in union
assert 'one' in union
assert 'three' in intersection
assert 'ten' not in intersection
assert 'one' not in intersection
此外,就我而言,我发现为 BitVector 提供更快的 count_bits 函数很有用。将此代码放入 BitVector 1.5,它应该为您提供更高效的位计数方法:
def fast_count_bits( self, v ):
bits = (
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8 )
return bits[v & 0xff] + bits[(v >> 8) & 0xff] + bits[(v >> 16) & 0xff] + bits[v >> 24]