Numba代码比纯Python慢 [英] Numba code slower than pure python

问题描述

我一直在努力加快粒子过滤器的重新采样计算.由于python有很多方法可以加快速度，因此尽管我会尝试所有方法.不幸的是，numba版本的运行速度非常慢.由于Numba应该会加快速度，因此我认为这是我的错误.

I've been working on speeding up a resampling calculation for a particle filter. As python has many ways to speed it up, I though I'd try them all. Unfortunately, the numba version is incredibly slow. As Numba should result in a speed up, I assume this is an error on my part.

我尝试了4种不同的版本:

I tried 4 different versions:

1. Numba
2. Python
3. 脾气暴躁
4. Cython

每个代码如下:

import numpy as np
import scipy as sp
import numba as nb
from cython_resample import cython_resample

@nb.autojit
def numba_resample(qs, xs, rands):
    n = qs.shape[0]
    lookup = np.cumsum(qs)
    results = np.empty(n)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results

def python_resample(qs, xs, rands):
    n = qs.shape[0]
    lookup = np.cumsum(qs)
    results = np.empty(n)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results

def numpy_resample(qs, xs, rands):
    results = np.empty_like(qs)
    lookup = sp.cumsum(qs)
    for j, key in enumerate(rands):
        i = sp.argmax(lookup>key)
        results[j] = xs[i]
    return results

#The following is the code for the cython module. It was compiled in a
#separate file, but is included here to aid in the question.
"""
import numpy as np
cimport numpy as np
cimport cython

DTYPE = np.float64

ctypedef np.float64_t DTYPE_t

@cython.boundscheck(False)
def cython_resample(np.ndarray[DTYPE_t, ndim=1] qs, 
             np.ndarray[DTYPE_t, ndim=1] xs, 
             np.ndarray[DTYPE_t, ndim=1] rands):
    if qs.shape[0] != xs.shape[0] or qs.shape[0] != rands.shape[0]:
        raise ValueError("Arrays must have same shape")
    assert qs.dtype == xs.dtype == rands.dtype == DTYPE

    cdef unsigned int n = qs.shape[0]
    cdef unsigned int i, j 
    cdef np.ndarray[DTYPE_t, ndim=1] lookup = np.cumsum(qs)
    cdef np.ndarray[DTYPE_t, ndim=1] results = np.zeros(n, dtype=DTYPE)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results
"""

if __name__ == '__main__':
    n = 100
    xs = np.arange(n, dtype=np.float64)
    qs = np.array([1.0/n,]*n)
    rands = np.random.rand(n)

    print "Timing Numba Function:"
    %timeit numba_resample(qs, xs, rands)
    print "Timing Python Function:"
    %timeit python_resample(qs, xs, rands)
    print "Timing Numpy Function:"
    %timeit numpy_resample(qs, xs, rands)
    print "Timing Cython Function:"
    %timeit cython_resample(qs, xs, rands)

这将导致以下输出:

Timing Numba Function:
1 loops, best of 3: 8.23 ms per loop
Timing Python Function:
100 loops, best of 3: 2.48 ms per loop
Timing Numpy Function:
1000 loops, best of 3: 793 µs per loop
Timing Cython Function:
10000 loops, best of 3: 25 µs per loop

知道为什么numba代码这么慢吗?我以为它至少可以与Numpy相提并论.

Any idea why the numba code is so slow? I assumed it would be at least comparable to Numpy.

注意:如果有人对如何加快Numpy或Cython代码示例的速度有任何想法，那也很好:)我的主要问题是关于Numba.

推荐答案

问题是numba无法理解lookup的类型.如果将print nb.typeof(lookup)放入方法中，则会看到numba将其视为对象，这很慢.通常，我只会在locals字典中定义lookup的类型，但是却遇到了一个奇怪的错误.相反，我只是创建了一个小包装程序，以便可以显式定义输入和输出类型.

The problem is that numba can't intuit the type of lookup. If you put a print nb.typeof(lookup) in your method, you'll see that numba is treating it as an object, which is slow. Normally I would just define the type of lookup in a locals dict, but I was getting a strange error. Instead I just created a little wrapper, so that I could explicitly define the input and output types.

@nb.jit(nb.f8[:](nb.f8[:]))
def numba_cumsum(x):
    return np.cumsum(x)

@nb.autojit
def numba_resample2(qs, xs, rands):
    n = qs.shape[0]
    #lookup = np.cumsum(qs)
    lookup = numba_cumsum(qs)
    results = np.empty(n)

    for j in range(n):
        for i in range(n):
            if rands[j] < lookup[i]:
                results[j] = xs[i]
                break
    return results

那我的时间是:

print "Timing Numba Function:"
%timeit numba_resample(qs, xs, rands)

print "Timing Revised Numba Function:"
%timeit numba_resample2(qs, xs, rands)

---

Timing Numba Function:
100 loops, best of 3: 8.1 ms per loop
Timing Revised Numba Function:
100000 loops, best of 3: 15.3 µs per loop

---

如果使用jit而不是autojit，您甚至可以走得更快一些:

---

You can go even a little faster still if you use jit instead of autojit:

@nb.jit(nb.f8[:](nb.f8[:], nb.f8[:], nb.f8[:]))

对我来说，将其从15.3微秒降低到12.5微秒，但仍然令人印象深刻，即autojit的性能如何.

For me that lowers it from 15.3 microseconds to 12.5 microseconds, but it's still impressive how well autojit does.

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