Fastest way to compute distance beetween each points in python

问题

In my project I need to compute euclidian distance beetween each points stored in an array. The entry array is a 2D numpy array with 3 columns which are the coordinates(x,y,z) and each rows define a new point.

I'm usualy working with 5000 - 6000 points in my test cases.

My first algorithm use Cython and my second numpy. I find that my numpy algorithm is faster than cython.

edit: with 6000 points :

numpy 1.76 s / cython 4.36 s

Here's my cython code:

cimport cython
from libc.math cimport sqrt
@cython.boundscheck(False)
@cython.wraparound(False)
cdef void calcul1(double[::1] M,double[::1] R):

  cdef int i=0
  cdef int max = M.shape[0]
  cdef int x,y
  cdef int start = 1

  for x in range(0,max,3):
     for y in range(start,max,3):

        R[i]= sqrt((M[y] - M[x])**2 + (M[y+1] - M[x+1])**2 + (M[y+2] - M[x+2])**2)
        i+=1  

     start += 1

M is a memory view of the initial entry array but flatten() by numpy before the call of the function calcul1(), R is a memory view of a 1D output array to store all the results.

Here's my Numpy code :

def calcul2(M):

     return np.sqrt(((M[:,:,np.newaxis] - M[:,np.newaxis,:])**2).sum(axis=0))

Here M is the initial entry array but transpose() by numpy before the function call to have coordinates(x,y,z) as rows and points as columns.

Moreover this numpy function is quite convinient because the array it returns is well organise. It's a n by n array with n the number of points and each points has a row and a column. So for example the distance AB is stored at the intersection index of row A and column B.

Here's how I call them (cython function):

cpdef test():

  cdef double[::1] Mf 
  cdef double[::1] out = np.empty(17998000,dtype=np.float64) # (6000² - 6000) / 2

  M = np.arange(6000*3,dtype=np.float64).reshape(6000,3) # Example array with 6000 points
  Mf = M.flatten() #because my cython algorithm need a 1D array
  Mt = M.transpose() # because my numpy algorithm need coordinates as rows

  calcul2(Mt)

  calcul1(Mf,out)

Am I doing something wrong here ? For my project both are not fast enough.

1: Is there a way to improve my cython code in order to beat numpy's speed ?

2: Is there a way to improve my numpy code to compute even faster ?

3: Or any other solutions, but it must be a python/cython (like parallel computing) ?

Thank you.

回答1:

Not sure where you are getting your timings, but you can use scipy.spatial.distance:

M = np.arange(6000*3, dtype=np.float64).reshape(6000,3)
np_result = calcul2(M)
sp_result = sd.cdist(M.T, M.T) #Scipy usage
np.allclose(np_result, sp_result)
>>> True

Timings:

%timeit calcul2(M)
1000 loops, best of 3: 313 µs per loop

%timeit sd.cdist(M.T, M.T)
10000 loops, best of 3: 86.4 µs per loop

Importantly, its also useful to realize that your output is symmetric:

np.allclose(sp_result, sp_result.T)
>>> True

An alternative is to only compute the upper triangular of this array:

%timeit sd.pdist(M.T)
10000 loops, best of 3: 39.1 µs per loop

Edit: Not sure which index you want to zip, looks like you may be doing it both ways? Zipping the other index for comparison:

%timeit sd.pdist(M)
10 loops, best of 3: 135 ms per loop

Still about 10x faster than your current NumPy implementation.

来源：https://stackoverflow.com/questions/37298710/fastest-way-to-compute-distance-beetween-each-points-in-python

标签

python

numpy

optimization

cython