CODEWITHSUNDEEP
pythonperformancenumpy-einsumnumexpr
masouduut94
masouduut94

Reputation: 1122

How can I speed up the performance by using numpy einsum and numexpr in calculating kernel functions?

I am trying to define a few of famous kernels like RBF, hyperbolic tangent, Fourier and etc for svm.SVR method in sklearn library. I started working on rbf (I know there's a default kernel in svm for rbf but I need to define one to be able to customize it later), and found some useful link in here and chose this one:

def my_kernel(X,Y):
    K = np.zeros((X.shape[0],Y.shape[0]))
    for i,x in enumerate(X):
        for j,y in enumerate(Y):
            K[i,j] = np.exp(-1*np.linalg.norm(x-y)**2)
    return K

clf=SVR(kernel=my_kernel)

I used this one because I could use it for my train (with shape of [3850,4]) and test data (with shape of [1200,4]) which have different shapes. But the problem is that it's too slow and I have to wait so long for the results. I even used static-typing and memoryviews in cython, but its performance is not as good as the default svm rbf kernel. I also found this link which is about the same problem but working with numpy.einsum and numexpr.evaluate is a little bit confusing for me. It turns out that this was the best code in terms of speed performance:

from scipy.linalg.blas import sgemm

def app2(X, gamma, var):
    X_norm = -gamma*np.einsum('ij,ij->i',X,X)
    return ne.evaluate('v * exp(A + B + C)', {\
        'A' : X_norm[:,None],\
        'B' : X_norm[None,:],\
        'C' : sgemm(alpha=2.0*gamma, a=X, b=X, trans_b=True),\
        'g' : gamma,\
        'v' : var\
    })

This code just works for one input (X) and I couldn't find a way to modify it for my case (two inputs with two different sizes - The kernel function gets matrices with shape (m,n) and (l,n) and outputs (m,l) according to svm docs ). I guess I only need to replace the K[i,j] = np.exp(-1*np.linalg.norm(x-y)**2) from the first code in the second one to speed it up. Any helps would be appreciated.

Upvotes: 0

Views: 540

Answers (2)

serge-sans-paille
serge-sans-paille

Reputation: 2139

You can just pipe your original kernel through pythran

kernel.py:

import numpy as np

#pythran export my_kernel(float64[:,:], float64[:,:])
def my_kernel(X,Y):
    K = np.zeros((X.shape[0],Y.shape[0]))
    for i,x in enumerate(X):
        for j,y in enumerate(Y):
            K[i,j] = np.exp(-1*np.linalg.norm(x-y)**2)
    return K

Compilation step:

> pythran kernel.py

There's no rewriting step (you need to put the kernel in a separate file though) and the acceleration is significant: 19 times faster on my laptop, using

> python -m timeit -s 'from numpy.random import random; x = random((100,100)); y = random((100,100)); from svr_kernel import my_kernel as k' 'k(x,y)'

to gather timings.

Upvotes: 2

max9111
max9111

Reputation: 6482

Three possible variants

Variants 1 and 3 makes use of

(a-b)^2 = a^2 + b^2 - 2ab

as described here or here. But for special cases like a small second dimension Variant 2 is also OK. 

import numpy as np
import numba as nb
import numexpr as ne
from scipy.linalg.blas import sgemm

def vers_1(X,Y, gamma, var):
    X_norm = -gamma*np.einsum('ij,ij->i',X,X)
    Y_norm = -gamma*np.einsum('ij,ij->i',Y,Y)
    return ne.evaluate('v * exp(A + B + C)', {\
        'A' : X_norm[:,None],\
        'B' : Y_norm[None,:],\
        'C' : sgemm(alpha=2.0*gamma, a=X, b=Y, trans_b=True),\
        'g' : gamma,\
        'v' : var\
    })

#Maybe easier to read but slow, if X.shape[1] gets bigger
@nb.njit(fastmath=True,parallel=True)
def vers_2(X,Y):
    K = np.empty((X.shape[0],Y.shape[0]),dtype=X.dtype)
    for i in nb.prange(X.shape[0]):
        for j in range(Y.shape[0]):
            sum=0.
            for k in range(X.shape[1]):
                sum+=(X[i,k]-Y[j,k])**2
            K[i,j] = np.exp(-1*sum)
    return K

@nb.njit(fastmath=True,parallel=True)
def vers_3(A,B):
    dist=np.dot(A,B.T)

    TMP_A=np.empty(A.shape[0],dtype=A.dtype)
    for i in nb.prange(A.shape[0]):
        sum=0.
        for j in range(A.shape[1]):
            sum+=A[i,j]**2
        TMP_A[i]=sum

    TMP_B=np.empty(B.shape[0],dtype=A.dtype)
    for i in nb.prange(B.shape[0]):
        sum=0.
        for j in range(B.shape[1]):
            sum+=B[i,j]**2
        TMP_B[i]=sum

    for i in nb.prange(A.shape[0]):
        for j in range(B.shape[0]):
            dist[i,j]=np.exp((-2.*dist[i,j]+TMP_A[i]+TMP_B[j])*-1)
    return dist

Timings

gamma = 1.
var = 1.
X=np.random.rand(3850,4)
Y=np.random.rand(1200,4)

res_1=vers_1(X,Y, gamma, var)
res_2=vers_2(X,Y)
res_3=vers_3(X,Y)
np.allclose(res_1,res_2)
np.allclose(res_1,res_3)


%timeit res_1=vers_1(X,Y, gamma, var)
19.8 ms ± 615 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit res_2=vers_2(X,Y)
16.1 ms ± 938 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit res_3=vers_3(X,Y)
13.5 ms ± 162 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Upvotes: 1

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