我正在使用 numbas @jit 装饰器在 python 中添加两个 numpy 数组。如果我使用 @jit 与 python 相比，性能是如此之高。

然而，即使我传入 @numba.jit(nopython = True, parallel = True, nogil = True)，它也没有利用所有 CPU 内核。

有什么方法可以通过 numba @jit 使用所有 CPU 内核。

这是我的代码:

import time                                                
import numpy as np                                         
import numba                                               

SIZE = 2147483648 * 6                                      

a = np.full(SIZE, 1, dtype = np.int32)                     

b = np.full(SIZE, 1, dtype = np.int32)                     

c = np.ndarray(SIZE, dtype = np.int32)                     

@numba.jit(nopython = True, parallel = True, nogil = True) 
def add(a, b, c):                                          
    for i in range(SIZE):                                  
        c[i] = a[i] + b[i]                                 

start = time.time()                                        
add(a, b, c)                                               
end = time.time()                                          

print(end - start)                                        

最佳答案

您可以将 parallel=True 传递给任何 numba jitted 函数，但这并不意味着它总是利用所有内核。您必须了解 numba 使用一些启发式方法使代码并行执行，有时这些启发式方法根本找不到代码中的任何内容来并行化。当前有一个 pull request这样，如果无法使其“平行”，它就会发出警告。所以它更像是一个“如果可能请让它并行执行”参数而不是“强制并行执行”。

但是，如果您真的知道可以并行化代码，则始终可以手动使用线程或进程。只是调整 example of using multi-threading from the numba docs :

#!/usr/bin/env python
from __future__ import print_function, division, absolute_import

import math
import threading
from timeit import repeat

import numpy as np
from numba import jit

nthreads = 4
size = 10**7  # CHANGED

# CHANGED
def func_np(a, b):
    """
    Control function using Numpy.
    """
    return a + b

# CHANGED
@jit('void(double[:], double[:], double[:])', nopython=True, nogil=True)
def inner_func_nb(result, a, b):
    """
    Function under test.
    """
    for i in range(len(result)):
        result[i] = a[i] + b[i]

def timefunc(correct, s, func, *args, **kwargs):
    """
    Benchmark *func* and print out its runtime.
    """
    print(s.ljust(20), end=" ")
    # Make sure the function is compiled before we start the benchmark
    res = func(*args, **kwargs)
    if correct is not None:
        assert np.allclose(res, correct), (res, correct)
    # time it
    print('{:>5.0f} ms'.format(min(repeat(lambda: func(*args, **kwargs),
                                          number=5, repeat=2)) * 1000))
    return res

def make_singlethread(inner_func):
    """
    Run the given function inside a single thread.
    """
    def func(*args):
        length = len(args[0])
        result = np.empty(length, dtype=np.float64)
        inner_func(result, *args)
        return result
    return func

def make_multithread(inner_func, numthreads):
    """
    Run the given function inside *numthreads* threads, splitting its
    arguments into equal-sized chunks.
    """
    def func_mt(*args):
        length = len(args[0])
        result = np.empty(length, dtype=np.float64)
        args = (result,) + args
        chunklen = (length + numthreads - 1) // numthreads
        # Create argument tuples for each input chunk
        chunks = [[arg[i * chunklen:(i + 1) * chunklen] for arg in args]
                  for i in range(numthreads)]
        # Spawn one thread per chunk
        threads = [threading.Thread(target=inner_func, args=chunk)
                   for chunk in chunks]
        for thread in threads:
            thread.start()
        for thread in threads:
            thread.join()
        return result
    return func_mt


func_nb = make_singlethread(inner_func_nb)
func_nb_mt = make_multithread(inner_func_nb, nthreads)

a = np.random.rand(size)
b = np.random.rand(size)

correct = timefunc(None, "numpy (1 thread)", func_np, a, b)
timefunc(correct, "numba (1 thread)", func_nb, a, b)
timefunc(correct, "numba (%d threads)" % nthreads, func_nb_mt, a, b)

我突出显示了我更改的部分，其他所有内容都是从示例中逐字复制的。这利用了我机器上的所有内核(4 核机器因此 4 线程)但没有显示出显着的加速:

numpy (1 thread)       539 ms
numba (1 thread)       536 ms
numba (4 threads)      442 ms

在这种情况下，多线程缺乏(大量)加速是因为加法是一种带宽受限的操作。这意味着从数组中加载元素并将结果放入结果数组所花的时间比实际添加要多得多。

在这些情况下，您甚至可以看到由于并行执行而导致速度变慢!

只有当函数更复杂并且实际操作与加载和存储数组元素相比花费大量时间时，您才会看到并行执行的巨大改进。 numba 文档中的示例就是这样一个:

def func_np(a, b):
    """
    Control function using Numpy.
    """
    return np.exp(2.1 * a + 3.2 * b)

@jit('void(double[:], double[:], double[:])', nopython=True, nogil=True)
def inner_func_nb(result, a, b):
    """
    Function under test.
    """
    for i in range(len(result)):
        result[i] = math.exp(2.1 * a[i] + 3.2 * b[i])

这实际上(几乎)随线程数缩放，因为两次乘法、一次加法和一次调用 math.exp 比加载和存储结果要慢得多:

func_nb = make_singlethread(inner_func_nb)
func_nb_mt2 = make_multithread(inner_func_nb, 2)
func_nb_mt3 = make_multithread(inner_func_nb, 3)
func_nb_mt4 = make_multithread(inner_func_nb, 4)

a = np.random.rand(size)
b = np.random.rand(size)

correct = timefunc(None, "numpy (1 thread)", func_np, a, b)
timefunc(correct, "numba (1 thread)", func_nb, a, b)
timefunc(correct, "numba (2 threads)", func_nb_mt2, a, b)
timefunc(correct, "numba (3 threads)", func_nb_mt3, a, b)
timefunc(correct, "numba (4 threads)", func_nb_mt4, a, b)

结果:

numpy (1 thread)      3422 ms
numba (1 thread)      2959 ms
numba (2 threads)     1555 ms
numba (3 threads)     1080 ms
numba (4 threads)      797 ms

关于python - 如何使 numba @jit 使用所有 cpu 内核(并行化 numba @jit)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/45610292/