如果你要調用的距離函數多在您執行程序的一次執行期間,您可以通過使用預計算的位計數表獲得一些速度。這裏的(另一個)版本的漢明距離函數:
# _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
在下面,a
和b
在這個問題評論給定長度32的Python列表。 divakar_hamming_distance()
和divakar_hamming_distance_v2()
來自@ Divakar的回答。
這裏有定時@ Divakar的功能:
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)
是快了一點:
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
在我的電腦,初始化_nbits
大約需要11微秒,所以沒有優勢如果您只調用一次函數,則使用hamming_distance1
。如果你三次或更多次稱呼它,則表現有淨增益。
如果輸入已經numpy的陣列,所有的功能都顯著快:
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
當然,如果你總是這樣,你計算的漢明距離之前,做轉換的時候一定要包括在總體時間中。但是,如果您編寫生成a
和b
的代碼以便早日利用numpy,則在計算海明距離時,可能已將它們作爲numpy陣列。
(I也試驗了一下與預先計算的漢明距離的8個值之間的2-d陣列 - 具有形狀(256陣列,256) - 但初始化成本較高和性能增益小。)
那不'0 + 1'代替因爲254是除了在一個位全爲1,而255是全1? – Divakar
大概只需要一個標準的popcount配方,在陣列上播放它,然後對結果進行求和。您可以通過將數組的緩衝區視爲更大的dtype來獲得加速。 – user2357112
@Divakar這是我的錯誤。接得好。樣本數據中的數字更新爲240。 –