Question

我正在尝试计算两组D维向量之间的成对欧氏距离，并尝试优化以下天真实现的运行时性能。基本上下面的代码返回描述符（AxD）中每一行之间的成对平方欧几里德距离，并且码本（BxD）中的每一行返回结果（AxB）矩阵。

从python转向基于cython的实现，我看到了一个很好的改进。我是cython的新手，并且想知道是否可以进一步优化此实现以使性能更接近纯c或基于java的实现。

pairwise_dist.pyx

import numpy as np
cimport numpy as np
import math
import cython
np.import_array()
DTYPE = np.float64
ctypedef np.float64_t DTYPE_t

@cython.boundscheck(False)
@cython.wraparound(False)

def _pdist(   np.ndarray[DTYPE_t, ndim=2] codebook,
              np.ndarray[DTYPE_t, ndim=2] descriptors,
              np.ndarray[DTYPE_t, ndim=2] result  ):

            cdef int i, j, k
            cdef int codebook_size = codebook.shape[0]
            cdef int n_descriptors = descriptors.shape[0]
            cdef int n_dimensions = codebook.shape[1]

            cdef float dist, diff

            for i in xrange(n_descriptors):
                for j in xrange(codebook_size):
                    dist = 0.0
                    for k in xrange(n_dimensions):
                        diff = codebook[j,k] - descriptors[i,k]
                        dist = dist + diff*diff
                    result[i,j] = dist

test.py：

import time
import numpy
CODEBOOK_SIZE = 4096
CODEBOOK_DIM = 64
DESC_DIM = 64
DESC_SIZE = 4000

codebook = numpy.random.randint(256,size=(CODEBOOK_SIZE,CODEBOOK_DIM))
descriptors = numpy.random.randint(256,size=(DESC_SIZE,DESC_DIM))
codebook = codebook.astype(numpy.float64)
descriptors = descriptors.astype(numpy.float64)

result = numpy.zeros((descriptors.shape[0],codebook.shape[0]),codebook.dtype)
from pairwise_dist import _pdist

stime = time.time()
_pdist(codebook,descriptors,result)
etime = time.time()
print "Time taken: "+ str(stime-etime)

build.py：

from distutils.core import setup
from Cython.Build import cythonize

setup(
    name = "pairwise_dist",
    ext_modules = cythonize('pairwise_dist.pyx'),
)

python build_cython.py build_ext --inplace

Answer 1

您是否出于教育原因使用Cython？否则，我建议使用scipy.spatial模块

import numpy as np
from scipy.spatial.distance import pdist, squareform

object_1 = [0.2, 4.5, 198, 0.003]
object_2 = [0.3, 2.0, 999, 0.001]
object_3 = [0.1, 9.2, 321, 0.023]
list_of_objects = [object_1, object_2, object_3]

# make a 4x3 matrix from list of objects
X = np.array(list_of_objects)

#calculate pairwise distances
distances = pdist(X)

#make a square matrix from result
distances_as_2d_matrix = squareform(distances)

print distances
print distances_as_2d_matrix

pdist支持多个指标，欧几里德只有一个。

用cython提高成对欧氏距离的运行时性能

1 个答案: