更有效的计算numpy距离的方法？

Question

我有一个关于如何尽可能快地计算numpy距离的问题,

def getR1(VVm,VVs,HHm,HHs):
    t0=time.time()
    R=VVs.flatten()[numpy.newaxis,:]-VVm.flatten()[:,numpy.newaxis]
    R*=R
    R1=HHs.flatten()[numpy.newaxis,:]-HHm.flatten()[:,numpy.newaxis]
    R1*=R1
    R+=R1
    del R1
    print "R1\t",time.time()-t0, R.shape, #11.7576191425 (108225, 10500) 
    print numpy.max(R) #4176.26290975
    # uses 17.5Gb ram
    return R


def getR2(VVm,VVs,HHm,HHs):
    t0=time.time()
    precomputed_flat = numpy.column_stack((VVs.flatten(), HHs.flatten()))
    measured_flat = numpy.column_stack((VVm.flatten(), HHm.flatten()))
    deltas = precomputed_flat[None,:,:] - measured_flat[:, None, :]
    #print time.time()-t0, deltas.shape # 5.861109972 (108225, 10500, 2)
    R = numpy.einsum('ijk,ijk->ij', deltas, deltas)
    print "R2\t",time.time()-t0,R.shape, #14.5291359425 (108225, 10500)
    print numpy.max(R) #4176.26290975
    # uses 26Gb ram
    return R


def getR3(VVm,VVs,HHm,HHs):
    from numpy.core.umath_tests import inner1d
    t0=time.time()
    precomputed_flat = numpy.column_stack((VVs.flatten(), HHs.flatten()))
    measured_flat = numpy.column_stack((VVm.flatten(), HHm.flatten()))
    deltas = precomputed_flat[None,:,:] - measured_flat[:, None, :]
    #print time.time()-t0, deltas.shape # 5.861109972 (108225, 10500, 2)
    R = inner1d(deltas, deltas)
    print "R3\t",time.time()-t0, R.shape, #12.6972110271 (108225, 10500)
    print numpy.max(R) #4176.26290975
    #Uses 26Gb
    return R


def getR4(VVm,VVs,HHm,HHs):
    from scipy.spatial.distance import cdist
    t0=time.time()
    precomputed_flat = numpy.column_stack((VVs.flatten(), HHs.flatten()))
    measured_flat = numpy.column_stack((VVm.flatten(), HHm.flatten()))
    R=spdist.cdist(precomputed_flat,measured_flat, 'sqeuclidean') #.T
    print "R4\t",time.time()-t0, R.shape, #17.7022118568 (108225, 10500)
    print numpy.max(R) #4176.26290975
    # uses 9 Gb ram
    return R

def getR5(VVm,VVs,HHm,HHs):
    from scipy.spatial.distance import cdist
    t0=time.time()
    precomputed_flat = numpy.column_stack((VVs.flatten(), HHs.flatten()))
    measured_flat = numpy.column_stack((VVm.flatten(), HHm.flatten()))
    R=spdist.cdist(precomputed_flat,measured_flat, 'euclidean') #.T
    print "R5\t",time.time()-t0, R.shape, #15.6070930958 (108225, 10500)
    print numpy.max(R) #64.6240118667
    # uses only 9 Gb ram
    return R

def getR6(VVm,VVs,HHm,HHs):
    from scipy.weave import blitz
    t0=time.time()
    R=VVs.flatten()[numpy.newaxis,:]-VVm.flatten()[:,numpy.newaxis]
    blitz("R=R*R") # R*=R
    R1=HHs.flatten()[numpy.newaxis,:]-HHm.flatten()[:,numpy.newaxis]
    blitz("R1=R1*R1") # R1*=R1
    blitz("R=R+R1") # R+=R1
    del R1
    print "R6\t",time.time()-t0, R.shape, #11.7576191425 (108225, 10500) 
    print numpy.max(R) #4176.26290975
    return R

导致以下时间:

R1  11.7737319469 (108225, 10500) 4909.66881791
R2  15.1279799938 (108225, 10500) 4909.66881791
R3  12.7408981323 (108225, 10500) 4909.66881791
R4  17.3336868286 (10500, 108225) 4909.66881791
R5  15.7530870438 (10500, 108225) 70.0690289494
R6  11.670968771 (108225, 10500) 4909.66881791

虽然最后一个给出sqrt((VVm-VVs)^ 2 +(HHm-HHs)^ 2),而其他给出(VVm-VVs)^ 2 +(HHm-HHs)^ 2,这不是很重要,因为在我的代码中另外进一步,我为每个i取最小值R [i,:],并且sqrt不会影响最小值,(如果我对距离感兴趣,我只需要取sqrt(值),而不是在整个阵列上执行sqrt,因此实际上没有时间差异.

问题仍然存在:为什么第一个解决方案是最好的,(第二个和第三个解决方案速度较慢的原因是因为增量= ...需要5.8秒,(这也是为什么这两个方法需要26Gb)),为什么sqeuclidean比欧几里德慢吗？

sqeuclidean应该做(VVm-VVs)^ 2 +(HHm-HHs)^ 2,而我认为它做了不同的事情.任何人都知道如何找到该方法的源代码(C或底部的任何内容)？我认为它确实是sqrt((VVm-VVs)^ 2 +(HHm-HHs)^ 2)^ 2(唯一的原因我能想到为什么它会慢于(VVm-VVs)^ 2 +(HHm-HHs) ^ 2 - 我知道这是一个愚蠢的原因,任何人都有一个更合乎逻辑的原因？)

既然我对C一无所知,我怎么用scipy.weave来内联呢？这个代码是否可以编译,就像你使用python一样？或者我需要安装特殊的东西吗？

编辑:好吧,我尝试使用scipy.weave.blitz,(R6方法),这稍微快一些,但我认为有人比我知道更多的C仍然可以提高这个速度？我只是采用了形式为+ = b或*=的行,然后查看了它们在C中的含义,并将它们放在闪电战语句中,但我想如果我将行与flatten和newaxis放在一起C也是,它应该更快,但我不知道我怎么能这样做(知道C的人可能会解释？).现在,闪电战和我的第一种方法之间的差异不足以真正由C vs numpy引起我猜？

我猜其他方法,比如deltas = ...也可以更快,当我把它放在C？

Answer 1

每当你有乘法和求和时,尝试使用其中一个点积函数或np.einsum.由于您要预先分配数组,而不是为水平和垂直坐标设置不同的数组,因此将它们堆叠在一起:

precomputed_flat = np.column_stack((svf.flatten(), shf.flatten()))
measured_flat = np.column_stack((VVmeasured.flatten(), HHmeasured.flatten()))
deltas = precomputed_flat - measured_flat[:, None, :]

从这里开始,最简单的是:

dist = np.einsum('ijk,ijk->ij', deltas, deltas)

您也可以尝试以下方法:

from numpy.core.umath_tests import inner1d
dist = inner1d(deltas, deltas)

当然还有SciPy的空间模块cdist:

from scipy.spatial.distance import cdist
dist = cdist(precomputed_flat, measured_flat, 'euclidean')

编辑 我无法在如此大的数据集上运行测试,但这些时间相当有启发性:

len_a, len_b = 10000, 1000

a = np.random.rand(2, len_a)
b =  np.random.rand(2, len_b)
c = np.random.rand(len_a, 2)
d = np.random.rand(len_b, 2)

In [3]: %timeit a[:, None, :] - b[..., None]
10 loops, best of 3: 76.7 ms per loop

In [4]: %timeit c[:, None, :] - d
1 loops, best of 3: 221 ms per loop

对于上面较小的数据集,通过在内存中以不同方式排列数据,我可以稍微加快您的方法并与之scipy.spatial.distance.cdist匹配inner1d:

precomputed_flat = np.vstack((svf.flatten(), shf.flatten()))
measured_flat = np.vstack((VVmeasured.flatten(), HHmeasured.flatten()))
deltas = precomputed_flat[:, None, :] - measured_flat

import scipy.spatial.distance as spdist
from numpy.core.umath_tests import inner1d

In [13]: %timeit r0 = a[0, None, :] - b[0, :, None]; r1 = a[1, None, :] - b[1, :, None]; r0 *= r0; r1 *= r1; r0 += r1
10 loops, best of 3: 146 ms per loop

In [14]: %timeit deltas = (a[:, None, :] - b[..., None]).T; inner1d(deltas, deltas)
10 loops, best of 3: 145 ms per loop

In [15]: %timeit spdist.cdist(a.T, b.T)
10 loops, best of 3: 124 ms per loop

In [16]: %timeit deltas = a[:, None, :] - b[..., None]; np.einsum('ijk,ijk->jk', deltas, deltas)
10 loops, best of 3: 163 ms per loop