Vectorized way of calculating row-wise dot product

问题:

I want to calculate the row-wise dot product of two matrices of the same dimension as fast as possible. This is the way I am doing it:

import numpy as np
a = np.array([[1,2,3], [3,4,5]])
b = np.array([[1,2,3], [1,2,3]])
result = np.array([])
for row1, row2 in a, b:
    result = np.append(result, np.dot(row1, row2))
print result

and of course the output is:

[ 26.  14.]

回答1:

Check out numpy.einsum for another method:

In [52]: a
Out[52]: 
array([[1, 2, 3],
       [3, 4, 5]])

In [53]: b
Out[53]: 
array([[1, 2, 3],
       [1, 2, 3]])

In [54]: einsum('ij,ij->i', a, b)
Out[54]: array([14, 26])

Looks like einsum is a bit faster than inner1d:

In [94]: %timeit inner1d(a,b)
1000000 loops, best of 3: 1.8 us per loop

In [95]: %timeit einsum('ij,ij->i', a, b)
1000000 loops, best of 3: 1.6 us per loop

In [96]: a = random.randn(10, 100)

In [97]: b = random.randn(10, 100)

In [98]: %timeit inner1d(a,b)
100000 loops, best of 3: 2.89 us per loop

In [99]: %timeit einsum('ij,ij->i', a, b)
100000 loops, best of 3: 2.03 us per loop

回答2:

Played around with this and found inner1d the fastest:

The plot was created with perfplot (a small project of mine)

import numpy
from numpy.core.umath_tests import inner1d
import perfplot

perfplot.show(
    setup=lambda n: (numpy.random.rand(n, 3), numpy.random.rand(n, 3)),
    n_range=[2**k for k in range(1, 18)],
    kernels=[
        lambda data: numpy.sum(data[0] * data[1], axis=1),
        lambda data: numpy.einsum('ij, ij->i', data[0], data[1]),
        lambda data: inner1d(data[0], data[1])
        ],
    labels=['np.sum(a*b, axis=1)', 'einsum', 'inner1d'],
    logx=True,
    logy=True,
    xlabel='len(a), len(b)'
    )

回答3:

Straightforward way to do that is:

import numpy as np
a=np.array([[1,2,3],[3,4,5]])
b=np.array([[1,2,3],[1,2,3]])
np.sum(a*b, axis=1)

which avoids the python loop and is faster in cases like:

def npsumdot(x, y):
    return np.sum(x*y, axis=1)

def loopdot(x, y):
    result = np.empty((x.shape[0]))
    for i in range(x.shape[0]):
        result[i] = np.dot(x[i], y[i])
    return result

timeit npsumdot(np.random.rand(500000,50),np.random.rand(500000,50))
# 1 loops, best of 3: 861 ms per loop
timeit loopdot(np.random.rand(500000,50),np.random.rand(500000,50))
# 1 loops, best of 3: 1.58 s per loop

回答4:

You'll do better avoiding the append, but I can't think of a way to avoid the python loop. A custom Ufunc perhaps? I don't think numpy.vectorize will help you here.

import numpy as np
a=np.array([[1,2,3],[3,4,5]])
b=np.array([[1,2,3],[1,2,3]])
result=np.empty((2,))
for i in range(2):
    result[i] = np.dot(a[i],b[i]))
print result

EDIT

Based on this answer, it looks like inner1d might work if the vectors in your real-world problem are 1D.

from numpy.core.umath_tests import inner1d
inner1d(a,b)  # array([14, 26])

回答5:

I came across this answer and re-verified the results with Numpy 1.14.3 running in Python 3.5. For the most part the answers above hold true on my system, although I found that for very large matrices (see example below), all but one of the methods are so close to one another that the performance difference is meaningless.

For smaller matrices, I found that einsum was the fastest by a considerable margin, up to a factor of two in some cases.

My large matrix example:

import numpy as np
from numpy.core.umath_tests import inner1d
a = np.random.randn(100, 1000000)  # 800 MB each
b = np.random.randn(100, 1000000)  # pretty big.

def loop_dot(a, b):
    result = np.empty((a.shape[1],))
    for i, (row1, row2) in enumerate(zip(a, b)):
        result[i] = np.dot(row1, row2)
%timeit inner1d(a, b)
# 128 ms ± 523 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit np.einsum('ij,ij->i', a, b)
# 121 ms ± 402 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit np.sum(a*b, axis=1)
# 411 ms ± 1.99 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit loop_dot(a, b)  # note the function call took negligible time
# 123 ms ± 342 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

So einsum is still the fastest on very large matrices, but by a tiny amount. It appears to be a statistically significant (tiny) amount though!