我需要(并且找不到)MurmurHash的纯python(无c ++)实现,而且我太新手了。速度或内存使用量对我的项目无关紧要。
我发现了一个尝试here,但它限制为31位散列,我真的需要一个64位散列。
答案 0 :(得分:9)
这是未经测试的(抱歉!),但这是我提出的版本。 Python允许任意大的整数,所以我为前8个字节(或64位)创建了一个掩码,然后我(通过按位AND)应用于所有可能产生大于64位的整数的算术结果。 也许其他人可以评论一般方法和可能的字节序问题等等。
def bytes_to_long(bytes):
assert len(bytes) == 8
return sum((b << (k * 8) for k, b in enumerate(bytes)))
def murmur(data, seed):
m = 0xc6a4a7935bd1e995
r = 47
MASK = 2 ** 64 - 1
data_as_bytes = bytearray(data)
h = seed ^ ((m * len(data_as_bytes)) & MASK)
for ll in range(0, len(data_as_bytes), 8):
k = bytes_to_long(data_as_bytes[ll:ll + 8])
k = (k * m) & MASK
k = k ^ ((k >> r) & MASK)
k = (k * m) & MASK
h = (h ^ k)
h = (h * m) & MASK
l = len(data_as_bytes) & 7
if l >= 7:
h = (h ^ (data_as_bytes[6] << 48))
if l >= 6:
h = (h ^ (data_as_bytes[5] << 40))
if l >= 5:
h = (h ^ (data_as_bytes[4] << 32))
if l >= 4:
h = (h ^ (data_as_bytes[3] << 24))
if l >= 3:
h = (h ^ (data_as_bytes[4] << 16))
if l >= 2:
h = (h ^ (data_as_bytes[4] << 8))
if l >= 1:
h = (h ^ data_as_bytes[4])
h = (h * m) & MASK
h = h ^ ((h >> r) & MASK)
h = (h * m) & MASK
h = h ^ ((h >> r) & MASK)
return h
答案 1 :(得分:5)
这是MurmurHash3 32的纯Python实现,它只对字符串进行哈希处理,但可以很容易地适应字节数组。这是算法Java version的端口。
def murmur3_x86_32(data, seed = 0):
c1 = 0xcc9e2d51
c2 = 0x1b873593
length = len(data)
h1 = seed
roundedEnd = (length & 0xfffffffc) # round down to 4 byte block
for i in range(0, roundedEnd, 4):
# little endian load order
k1 = (ord(data[i]) & 0xff) | ((ord(data[i + 1]) & 0xff) << 8) | \
((ord(data[i + 2]) & 0xff) << 16) | (ord(data[i + 3]) << 24)
k1 *= c1
k1 = (k1 << 15) | ((k1 & 0xffffffff) >> 17) # ROTL32(k1,15)
k1 *= c2
h1 ^= k1
h1 = (h1 << 13) | ((h1 & 0xffffffff) >> 19) # ROTL32(h1,13)
h1 = h1 * 5 + 0xe6546b64
# tail
k1 = 0
val = length & 0x03
if val == 3:
k1 = (ord(data[roundedEnd + 2]) & 0xff) << 16
# fallthrough
if val in [2, 3]:
k1 |= (ord(data[roundedEnd + 1]) & 0xff) << 8
# fallthrough
if val in [1, 2, 3]:
k1 |= ord(data[roundedEnd]) & 0xff
k1 *= c1
k1 = (k1 << 15) | ((k1 & 0xffffffff) >> 17) # ROTL32(k1,15)
k1 *= c2
h1 ^= k1
# finalization
h1 ^= length
# fmix(h1)
h1 ^= ((h1 & 0xffffffff) >> 16)
h1 *= 0x85ebca6b
h1 ^= ((h1 & 0xffffffff) >> 13)
h1 *= 0xc2b2ae35
h1 ^= ((h1 & 0xffffffff) >> 16)
return h1 & 0xffffffff
答案 2 :(得分:5)
修复Nikolas版本中的错误
def bytes_to_long(bytes):
assert len(bytes) == 8
return sum((b << (k * 8) for k, b in enumerate(bytes)))
def murmur64(data, seed = 19820125):
m = 0xc6a4a7935bd1e995
r = 47
MASK = 2 ** 64 - 1
data_as_bytes = bytearray(data)
h = seed ^ ((m * len(data_as_bytes)) & MASK)
off = len(data_as_bytes)/8*8
for ll in range(0, off, 8):
k = bytes_to_long(data_as_bytes[ll:ll + 8])
k = (k * m) & MASK
k = k ^ ((k >> r) & MASK)
k = (k * m) & MASK
h = (h ^ k)
h = (h * m) & MASK
l = len(data_as_bytes) & 7
if l >= 7:
h = (h ^ (data_as_bytes[off+6] << 48))
if l >= 6:
h = (h ^ (data_as_bytes[off+5] << 40))
if l >= 5:
h = (h ^ (data_as_bytes[off+4] << 32))
if l >= 4:
h = (h ^ (data_as_bytes[off+3] << 24))
if l >= 3:
h = (h ^ (data_as_bytes[off+2] << 16))
if l >= 2:
h = (h ^ (data_as_bytes[off+1] << 8))
if l >= 1:
h = (h ^ data_as_bytes[off])
h = (h * m) & MASK
h = h ^ ((h >> r) & MASK)
h = (h * m) & MASK
h = h ^ ((h >> r) & MASK)
return h
答案 3 :(得分:5)
MurmurHash3的另一个纯Python实现,完全兼容并可由mmh3包装器替换,但仍限于32位Murmur3: https://github.com/wc-duck/pymmh3
答案 4 :(得分:0)
我建议更改bytes_to_long
函数,以便可以使用少于8个字节的值:
def bytes_to_long(bytes):
length = len(bytes)
if length < 8:
extra = 8 - length
bytes = b'\000' * extra + bytes
assert len(bytes) == 8
return sum((b << (k * 8) for k, b in enumerate(bytes)))
这会使bytearray
填充Null字节,以便断言静止图像可以正常工作(您现在可以将其删除),但允许转换较小的值。