Question

令a和b为具有8位整数（0-255）的相同大小的向量。我想计算那些矢量不同的比特数，即通过串联这些数字的二进制表示形成的矢量之间的汉明距离。例如：

a = [127,255]
b= [127,240]

使用numpy库

np.bitwise_xor(a,b)
# Output: array([ 0, 15])

我现在需要的是二进制表示上述数组的每个元素，并在数组的所有元素中计数1的数量。以上示例将给出汉明距离0 + 4 = 4.在Python中任何快速而优雅的解决方案？

Answer 1

方法＃1：我们可以将它们广播成二进制位＆amp;计算不同位的数量，如此 -

def hamming_distance(a, b):
    r = (1 << np.arange(8))[:,None]
    return np.count_nonzero( (a & r) != (b & r) )

示例运行 -

In [144]: a = [127,255]
     ...: b = [127,240]
     ...: 

In [145]: hamming_distance(a, b)
Out[145]: 4

方法＃2：使用bitwise-xor操作，我们可以找出a和b之间不同二进制位的数量 -

def hamming_distance_v2(a, b):
    r = (1 << np.arange(8))[:,None]
    return np.count_nonzero((np.bitwise_xor(a,b) & r) != 0)

Answer 2

如果要在程序执行期间多次调用距离函数，可以使用预先计算的位计数表来获得一些速度。这是汉明距离函数的另一个版本：

# _nbits[k] is the number of 1s in the binary representation of k for 0 <= k < 256.
_nbits = np.array(
      [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], dtype=np.uint8)


def hamming_distance1(a, b):
    c = np.bitwise_xor(a, b)
    n = _nbits[c].sum()
    return n

在下文中，a和b是在问题评论中给出的长度为32的Python列表。 divakar_hamming_distance()和divakar_hamming_distance_v2()来自@Divakar的回答。

以下是@Divakar职能的时间安排：

In [116]: %timeit divakar_hamming_distance(a, b)
The slowest run took 5.57 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 11.3 µs per loop

In [117]: %timeit divakar_hamming_distance_v2(a, b)
The slowest run took 5.35 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 10.3 µs per loop

hamming_distance1(a, b)有点快：

In [118]: %timeit hamming_distance1(a, b)
The slowest run took 6.04 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 7.42 µs per loop

在我的计算机上，初始化_nbits大约需要11μs，因此如果只调用一次该函数，则使用hamming_distance1没有任何好处。如果你打三次或更多次，那么性能会有净增长。

如果输入已经是numpy数组，那么所有函数都要快得多：

In [119]: aa = np.array(a)

In [120]: bb = np.array(b)

In [121]: %timeit divakar_hamming_distance_v2(aa, bb)
The slowest run took 8.22 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.72 µs per loop

In [122]: %timeit hamming_distance1(aa, bb)
The slowest run took 12.67 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.77 µs per loop

当然，如果您在计算汉明距离之前总是这样做，那么进行转换的时间必须包含在整体时间中。但是，如果您编写生成a和b的代码以利用先前的numpy，那么在计算汉明距离时，您可能已将它们作为numpy数组。

（我还对8位值之间的预计汉明距离的二维数组进行了实验 - 一个形状为（256,256）的数组 - 但初始化成本更高，性能增益也很小。）

Answer 3

也许不是最有效的方法，但最简单的imo是将你的ouptut数组转换为二进制形式的字符串，然后将所有转换回字符的字符的总和...

import numpy as np

output = np.random.randint(0,63,10)
hamming = ['{:b}'.format(x).count('1') for x in output]

Answer 4

sum(bin(x).count("1") for x in np.bitwise_xor(a,b))

获得整数数组的汉明距离的最快方法

4 个答案: