Reputation: 513

How to vectorize finding max value in numpy array with if statement?

My Setup: Python 2.7.4.1, Numpy MKL 1.7.1, Windows 7 x64, WinPython

Context:

I tried to implement the Sequential Minimal Optimization algorithm for solving SVM. I use maximal violating pair approach.

The problem:

In working set selection procedure i want to find maximum value of gradient and its index for elements which met some condition, y[i]*alpha[i]<0 or y[i]*alpha[i]

#y - array of -1 and 1
y=np.array([-1,1,1,1,-1,1])
#alpha- array of floats in range [0,C]
alpha=np.array([0.4,0.1,1.33,0,0.9,0])
#grad - array of floats
grad=np.array([-1,-1,-0.2,-0.4,0.4,0.2])


GMaxI=float('-inf')       
GMax_idx=-1        
n=alpha.shape[0] #usually n=100000
C=4
B=[0,0,C]
for i in xrange(0,n):
    yi=y[i]  #-1 or 1
    alpha_i=alpha[i]

    if (yi * alpha_i< B[yi+1]): # B[-1+1]=0 B[1+1]=C
        if( -yi*grad[i]>=GMaxI):
            GMaxI= -yi*grad[i]
            GMax_idx = i

This procedure is called many times (~50000) and profiler shows that this is the bottleneck. It is possible to vectorize this code?

Edit 1: Add some small exemplary data

Edit 2: I have checked solution proposed by hwlau , larsmans and Mr E. Only solutions proposed Mr E is correct. Below sample code with all three answers:

import numpy as np
y=np.array([   -1,  -1,   -1,  -1,   -1,   -1,   -1,   -1])
alpha=np.array([0,  0.9,  0.4, 0.1, 1.33,    0,  0.9,    0])
grad=np.array([-3, -0.5,   -1,  -1, -0.2,   -4, -0.4, -0.3])
C=4
B=np.array([0,0,C])

#hwlau - wrong index and value
filter = (y*alpha < C*0.5*(y+1)).astype('float')
GMax_idx = (filter*(-y*grad)).argmax()
GMax = -y[GMax_idx]*grad[GMax_idx]

print GMax_idx,GMax

#larsmans - wrong index
neg_y_grad = (-y * grad)[y * alpha < B[y + 1]]
GMaxI = np.max(neg_y_grad)
GMax_ind = np.argmax(neg_y_grad)

print GMax_ind,GMaxI

#Mr E - correct result
BY = np.take(B, y+1)
valid_mask = (y * alpha < BY)
values = -y * grad
values[~valid_mask] = np.min(values) - 1.0

GMaxI = values.max()
GMax_idx = values.argmax()
print GMax_idx,GMaxI

Output (GMax_idx, GMaxI)
0 -3.0 
3 -0.2
4 -0.2

Conclusions

After checking all solutions, the fastest one (2x-6x) is solution proposed by @ali_m. However it requires to install some python packages: numba and all its prerequisites.

I have some trouble to use numba with class methods, so I create global functions which are autojited with numba, my solution look something like this:

from numba import autojit  

@autojit
def FindMaxMinGrad(A,B,alpha,grad,y):
    '''
    Finds i,j indices with maximal violatin pair scheme
    A,B - 3 dim arrays, contains bounds A=[-C,0,0], B=[0,0,C]
    alpha - array like, contains alpha coeficients
    grad - array like, gradient
    y - array like, labels
    '''
    GMaxI=-100000
    GMaxJ=-100000

    GMax_idx=-1
    GMin_idx=-1

    for i in range(0,alpha.shape[0]):

        if (y[i] * alpha[i]< B[y[i]+1]):
            if( -y[i]*grad[i]>GMaxI):
                GMaxI= -y[i]*grad[i]
                GMax_idx = i

        if (y[i] * alpha[i]> A[y[i]+1]):
            if( y[i]*grad[i]>GMaxJ):
                GMaxJ= y[i]*grad[i]
                GMin_idx = i

    return (GMaxI,GMaxJ,GMax_idx,GMin_idx)  

class SVM(object):

    def working_set(self,....):
        FindMaxMinGrad(.....)

Upvotes: 5

Answers (4)

ali_m

Reputation: 74154

You can probably do quite a lot better than plain vectorization if you use numba to JIT-compile your original code that used nested loops.

import numpy as np
from numba import autojit

@autojit
def jit_max_grad(y, alpha, grad, B):
    maxgrad = -inf
    maxind = -1
    for ii in xrange(alpha.shape[0]):
        if (y[ii] * alpha[ii] < B[y[ii] + 1]):
            g = -y[ii] * grad[ii]
            if g >= maxgrad:
                maxgrad = g
                maxind = ii
    return maxind, maxgrad

For comparison, here's Mr E's vectorized version:

def mr_e_max_grad(y, alpha, grad, B):

    BY = np.take(B, y+1)
    valid_mask = (y * alpha < BY)
    values = -y * grad
    values[~valid_mask] = np.min(values) - 1.0
    GMaxI = values.max()
    GMax_idx = values.argmax()
    return GMax_idx, GMaxI

Timing:

y = np.array([   -1,  -1,   -1,  -1,   -1,   -1,   -1,   -1])
alpha = np.array([0,  0.9,  0.4, 0.1, 1.33,    0,  0.9,    0])
grad = np.array([-3, -0.5,   -1,  -1, -0.2,   -4, -0.4, -0.3])
C = 4
B = np.array([0,0,C])

%timeit mr_e_max_grad(y, alpha, grad, B)
# 100000 loops, best of 3: 19.1 µs per loop

%timeit jit_max_grad(y, alpha, grad, B)
# 1000000 loops, best of 3: 1.07 µs per loop

Update: if you want to see what the timings look like on bigger arrays, it's easy to define a function that generates semi-realistic fake data based on your description in the question:

def make_fake(n, C=4):
    y = np.random.choice((-1, 1), n)
    alpha = np.random.rand(n) * C
    grad = np.random.randn(n)
    B = np.array([0,0,C])
    return y, alpha, grad, B

%%timeit y, alpha, grad, B = make_fake(100000, 4)
mr_e_max_grad(y, alpha, grad, B)
# 1000 loops, best of 3: 1.83 ms per loop

%%timeit y, alpha, grad, B = make_fake(100000, 4)
jit_max_grad(y, alpha, grad, B)
# 1000 loops, best of 3: 471 µs per loop

Upvotes: 3

Fred Foo

Reputation: 363487

If you change B from a list to a NumPy array, you can at least vectorize the yi * alpha_i< B[yi+1] and push the loop inwards:

GMaxI = float('-inf')      
GMax_idx = -1      
for i in np.where(y * alpha < B[y + 1])[0]:
    if -y[i] * grad[i] >= GMaxI:
        GMaxI= -y[i] * grad[i]
        GMax_idx = i

That should save a bit of time. Next up, you can vectorize -y[i] * grad[i]:

GMaxI = float('-inf')
GMax_idx = -1
neg_y_grad = -y * grad
for i in np.where(y * alpha < B[y + 1])[0]:
    if neg_y_grad[i] >= GMaxI:
        GMaxI= -y[i] * grad[i]
        GMax_idx = i

Finally, we can vectorize away the entire loop by using max and argmax on -y * grad, filtered by y * alpha < B[y + 1]:

neg_y_grad = (-y * grad)
GMaxI = np.max(neg_y_grad[y * alpha < B[y + 1]])
GMax_idx = np.where(neg_y_grad == GMaxI)[0][0]

Upvotes: 2

unsym

Reputation: 2200

Here you go:

y=np.array([-1,1,1,1,-1,1])
alpha=np.array([0.4,0.1,1.33,0,0.9,0])
grad=np.array([-1,-1,-0.2,-0.4,0.4,0.2])
C=4

filter = (y*alpha < C*0.5*(y+1)).astype('float')
GMax_idx = (filter*(-y*grad)).argmax()
GMax = -y[GMax_idx]*grad[GMax_idx]

No benchmark tried, but it is pure numerical and vectorized so it should be fast.