Why is numpy's einsum faster than numpy's built in functions?

Question:

Lets start with three arrays of dtype=np.double. Timings are performed on a intel CPU using numpy 1.7.1 compiled with icc and linked to intel’s mkl. A AMD cpu with numpy 1.6.1 compiled with gcc without mkl was also used to verify the timings. Please note the timings scale nearly linearly with system size and are not due to the small overhead incurred in the numpy functions if statements these difference will show up in microseconds not milliseconds:

arr_1D=np.arange(500,dtype=np.double)
large_arr_1D=np.arange(100000,dtype=np.double)
arr_2D=np.arange(500**2,dtype=np.double).reshape(500,500)
arr_3D=np.arange(500**3,dtype=np.double).reshape(500,500,500)

First lets look at the np.sum function:

np.all(np.sum(arr_3D)==np.einsum('ijk->',arr_3D))
True

%timeit np.sum(arr_3D)
10 loops, best of 3: 142 ms per loop

%timeit np.einsum('ijk->', arr_3D)
10 loops, best of 3: 70.2 ms per loop

Powers:

np.allclose(arr_3D*arr_3D*arr_3D,np.einsum('ijk,ijk,ijk->ijk',arr_3D,arr_3D,arr_3D))
True

%timeit arr_3D*arr_3D*arr_3D
1 loops, best of 3: 1.32 s per loop

%timeit np.einsum('ijk,ijk,ijk->ijk', arr_3D, arr_3D, arr_3D)
1 loops, best of 3: 694 ms per loop

Outer product:

np.all(np.outer(arr_1D,arr_1D)==np.einsum('i,k->ik',arr_1D,arr_1D))
True

%timeit np.outer(arr_1D, arr_1D)
1000 loops, best of 3: 411 us per loop

%timeit np.einsum('i,k->ik', arr_1D, arr_1D)
1000 loops, best of 3: 245 us per loop

All of the above are twice as fast with np.einsum. These should be apples to apples comparisons as everything is specifically of dtype=np.double. I would expect the speed up in an operation like this:

np.allclose(np.sum(arr_2D*arr_3D),np.einsum('ij,oij->',arr_2D,arr_3D))
True

%timeit np.sum(arr_2D*arr_3D)
1 loops, best of 3: 813 ms per loop

%timeit np.einsum('ij,oij->', arr_2D, arr_3D)
10 loops, best of 3: 85.1 ms per loop

Einsum seems to be at least twice as fast for np.inner, np.outer, np.kron, and np.sum regardless of axes selection. The primary exception being np.dot as it calls DGEMM from a BLAS library. So why is np.einsum faster that other numpy functions that are equivalent?

The DGEMM case for completeness:

np.allclose(np.dot(arr_2D,arr_2D),np.einsum('ij,jk',arr_2D,arr_2D))
True

%timeit np.einsum('ij,jk',arr_2D,arr_2D)
10 loops, best of 3: 56.1 ms per loop

%timeit np.dot(arr_2D,arr_2D)
100 loops, best of 3: 5.17 ms per loop

The leading theory is from @sebergs comment that np.einsum can make use of SSE2, but numpy’s ufuncs will not until numpy 1.8 (see the change log). I believe this is the correct answer, but have not been able to confirm it. Some limited proof can be found by changing the dtype of input array and observing speed difference and the fact that not everyone observes the same trends in timings.

Asked By: Daniel

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Answers:

I think these timings explain what’s going on:

a = np.arange(1000, dtype=np.double)
%timeit np.einsum('i->', a)
100000 loops, best of 3: 3.32 us per loop
%timeit np.sum(a)
100000 loops, best of 3: 6.84 us per loop

a = np.arange(10000, dtype=np.double)
%timeit np.einsum('i->', a)
100000 loops, best of 3: 12.6 us per loop
%timeit np.sum(a)
100000 loops, best of 3: 16.5 us per loop

a = np.arange(100000, dtype=np.double)
%timeit np.einsum('i->', a)
10000 loops, best of 3: 103 us per loop
%timeit np.sum(a)
10000 loops, best of 3: 109 us per loop

So you basically have an almost constant 3us overhead when calling np.sum over np.einsum, so they basically run as fast, but one takes a little longer to get going. Why could that be? My money is on the following:

a = np.arange(1000, dtype=object)
%timeit np.einsum('i->', a)
Traceback (most recent call last):
...
TypeError: invalid data type for einsum
%timeit np.sum(a)
10000 loops, best of 3: 20.3 us per loop

Not sure what is going on exactly, but it seems that np.einsum is skipping some checks to extract type specific functions to do the multiplications and additions, and is going directly with * and + for standard C types only.


The multidimensional cases are not different:

n = 10; a = np.arange(n**3, dtype=np.double).reshape(n, n, n)
%timeit np.einsum('ijk->', a)
100000 loops, best of 3: 3.79 us per loop
%timeit np.sum(a)
100000 loops, best of 3: 7.33 us per loop

n = 100; a = np.arange(n**3, dtype=np.double).reshape(n, n, n)
%timeit np.einsum('ijk->', a)
1000 loops, best of 3: 1.2 ms per loop
%timeit np.sum(a)
1000 loops, best of 3: 1.23 ms per loop

So a mostly constant overhead, not a faster running once they get down to it.

Answered By: Jaime

First off, there’s been a lot of past discussion about this on the numpy list. For example, see:
http://numpy-discussion.10968.n7.nabble.com/poor-performance-of-sum-with-sub-machine-word-integer-types-td41.html
http://numpy-discussion.10968.n7.nabble.com/odd-performance-of-sum-td3332.html

Some of boils down to the fact that einsum is new, and is presumably trying to be better about cache alignment and other memory access issues, while many of the older numpy functions focus on a easily portable implementation over a heavily optimized one. I’m just speculating, there, though.


However, some of what you’re doing isn’t quite an “apples-to-apples” comparison.

In addition to what @Jamie already said, sum uses a more appropriate accumulator for arrays

For example, sum is more careful about checking the type of the input and using an appropriate accumulator. For example, consider the following:

In [1]: x = 255 * np.ones(100, dtype=np.uint8)

In [2]: x
Out[2]:
array([255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255, 255,
       255, 255, 255, 255, 255, 255, 255, 255, 255], dtype=uint8)

Note that the sum is correct:

In [3]: x.sum()
Out[3]: 25500

While einsum will give the wrong result:

In [4]: np.einsum('i->', x)
Out[4]: 156

But if we use a less limited dtype, we’ll still get the result you’d expect:

In [5]: y = 255 * np.ones(100)

In [6]: np.einsum('i->', y)
Out[6]: 25500.0
Answered By: Joe Kington

Now that numpy 1.8 is released, where according to the docs all ufuncs should use SSE2, I wanted to double check that Seberg’s comment about SSE2 was valid.

To perform the test a new python 2.7 install was created- numpy 1.7 and 1.8 were compiled with icc using standard options on a AMD opteron core running Ubuntu.

This is the test run both before and after the 1.8 upgrade:

import numpy as np
import timeit

arr_1D=np.arange(5000,dtype=np.double)
arr_2D=np.arange(500**2,dtype=np.double).reshape(500,500)
arr_3D=np.arange(500**3,dtype=np.double).reshape(500,500,500)

print 'Summation test:'
print timeit.timeit('np.sum(arr_3D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print timeit.timeit('np.einsum("ijk->", arr_3D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print '----------------------n'


print 'Power test:'
print timeit.timeit('arr_3D*arr_3D*arr_3D',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print timeit.timeit('np.einsum("ijk,ijk,ijk->ijk", arr_3D, arr_3D, arr_3D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print '----------------------n'


print 'Outer test:'
print timeit.timeit('np.outer(arr_1D, arr_1D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print timeit.timeit('np.einsum("i,k->ik", arr_1D, arr_1D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print '----------------------n'


print 'Einsum test:'
print timeit.timeit('np.sum(arr_2D*arr_3D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print timeit.timeit('np.einsum("ij,oij->", arr_2D, arr_3D)',
                      'import numpy as np; from __main__ import arr_1D, arr_2D, arr_3D',
                      number=5)/5
print '----------------------n'

Numpy 1.7.1:

Summation test:
0.172988510132
0.0934836149216
----------------------

Power test:
1.93524689674
0.839519000053
----------------------

Outer test:
0.130380821228
0.121401786804
----------------------

Einsum test:
0.979052495956
0.126066613197

Numpy 1.8:

Summation test:
0.116551589966
0.0920487880707
----------------------

Power test:
1.23683619499
0.815982818604
----------------------

Outer test:
0.131808176041
0.127472200394
----------------------

Einsum test:
0.781750011444
0.129271841049

I think this is fairly conclusive that SSE plays a large role in the timing differences, it should be noted that repeating these tests the timings very by only ~0.003s. The remaining difference should be covered in the other answers to this question.

Answered By: Daniel

An update for numpy 1.21.2: Numpy’s native functions are faster than einsums in almost all cases. Only einsum’s outer variant and the sum23 test faster than the non-einsum versions.

If you can use numpy’s native functions, do that.

(Images created with perfplot, a project of mine.)

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Code to reproduce the plots:

import numpy
import perfplot


def setup1(n):
    return numpy.arange(n, dtype=numpy.double)


def setup2(n):
    return numpy.arange(n ** 2, dtype=numpy.double).reshape(n, n)


def setup3(n):
    return numpy.arange(n ** 3, dtype=numpy.double).reshape(n, n, n)


def setup23(n):
    return (
        numpy.arange(n ** 2, dtype=numpy.double).reshape(n, n),
        numpy.arange(n ** 3, dtype=numpy.double).reshape(n, n, n),
    )


def numpy_sum(a):
    return numpy.sum(a)


def einsum_sum(a):
    return numpy.einsum("ijk->", a)


perfplot.save(
    "sum.png",
    setup=setup3,
    kernels=[numpy_sum, einsum_sum],
    n_range=[2 ** k for k in range(10)],
)


def numpy_power(a):
    return a * a * a


def einsum_power(a):
    return numpy.einsum("ijk,ijk,ijk->ijk", a, a, a)


perfplot.save(
    "power.png",
    setup=setup3,
    kernels=[numpy_power, einsum_power],
    n_range=[2 ** k for k in range(9)],
)


def numpy_outer(a):
    return numpy.outer(a, a)


def einsum_outer(a):
    return numpy.einsum("i,k->ik", a, a)


perfplot.save(
    "outer.png",
    setup=setup1,
    kernels=[numpy_outer, einsum_outer],
    n_range=[2 ** k for k in range(13)],
)


def dgemm_numpy(a):
    return numpy.dot(a, a)


def dgemm_einsum(a):
    return numpy.einsum("ij,jk", a, a)


def dgemm_einsum_optimize(a):
    return numpy.einsum("ij,jk", a, a, optimize=True)


perfplot.save(
    "dgemm.png",
    setup=setup2,
    kernels=[dgemm_numpy, dgemm_einsum],
    n_range=[2 ** k for k in range(13)],
)


def dot_numpy(a):
    return numpy.dot(a, a)


def dot_einsum(a):
    return numpy.einsum("i,i->", a, a)


perfplot.save(
    "dot.png",
    setup=setup1,
    kernels=[dot_numpy, dot_einsum],
    n_range=[2 ** k for k in range(20)],
)


def sum23_numpy(data):
    a, b = data
    return numpy.sum(a * b)


def sum23_einsum(data):
    a, b = data
    return numpy.einsum("ij,oij->", a, b)


perfplot.save(
    "sum23.png",
    setup=setup23,
    kernels=[sum23_numpy, sum23_einsum],
    n_range=[2 ** k for k in range(10)],
)
Answered By: Nico Schlömer