dawei
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,性能如此之高.
但是,即使我传入@numba.jit(nopython = True,parallel = True,nogil = True),它也没有使用所有CPU内核.
有没有办法使用numba @jit来使用所有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,dtype = np.int32)
c = np.ndarray(SIZE,dtype = np.int32)
@numba.jit(nopython = True,nogil = True)
def add(a,b,c):
for i in range(SIZE):
c[i] = a[i] + b[i]
start = time.time()
add(a,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使用所有核心.您必须了解numba使用一些启发式方法来使代码并行执行,有时这些启发式方法根本找不到任何在代码中并行化的内容.目前有一个pull request,如果无法使其“并行”,它会发出警告.所以它更像是“请尽可能并行执行”参数而不是“强制执行并行执行”.
但是,如果您确实知道可以并行化代码,则可以始终手动使用线程或进程.只是改编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,b)
timefunc(correct,"numba (1 thread)",func_nb,"numba (%d threads)" % nthreads,func_nb_mt,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核心(4核心机器因此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(很多)加速是加法是带宽受限的操作.这意味着从数组加载元素并将结果放在结果数组中需要花费更多的时间而不是实际添加.
在这些情况下,由于并行执行,您甚至可以看到减速！
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需要大量时间时,您才会看到并行执行会有很大改进. numba文档中的示例是这样的：
def func_np(a,b):
"""
Control function using Numpy.
"""
return np.exp(2.1 * a + 3.2 * b)
@jit('void(double[:],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)随着线程数量而缩放,因为两次乘法,一次加法和一次对math.exp的调用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,"numba (2 threads)",func_nb_mt2,"numba (3 threads)",func_nb_mt3,"numba (4 threads)",func_nb_mt4,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
