前言：
并行编程比程序编程困难，除非正常编程需要创建大量数据，计算耗时太长，物理行为模拟困难
例子：N体问题
物理前提：
- 牛顿定律
- 时间离散运动方程
普通计算方法
import numpy as np
import time
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
Ns = [2**i for i in range(1,10)]
runtimes = []
def remove_i(x,i):
"从所有粒子中去除本粒子"
shape = (x.shape[0]-1,)+x.shape[1:]
y = np.empty(shape,dtype=float)
y[:i] = x[:i]
y[i:] = x[i+1:]
return y
def a(i,x,G,m):
"计算加速度"
x_i = x[i]
x_j = remove_i(x,i)
m_j = remove_i(m,i)
diff = x_j - x_i
mag3 = np.sum(diff**2,axis=1)**1.5
result = G * np.sum(diff * (m_j / mag3)[:,np.newaxis],axis=0)
return result
def timestep(x0,v0,G,m,dt):
N = len(x0)
x1 = np.empty(x0.shape,dtype=float)
v1 = np.empty(v0.shape,dtype=float)
for i in range(N):
a_i0 = a(i,x0,G,m)
v1[i] = a_i0 * dt + v0[i]
x1[i] = a_i0 * dt**2 + v0[i] * dt + x0[i]
return x1,v1
def initial_cond(N,D):
x0 = np.array([[1,1,1],[10,10,10]])
v0 = np.array([[10,10,1],[0,0,0]])
m = np.array([10,10])
return x0,v0,m
def stimulate(N,D,S,G,dt):
fig = plt.figure()
ax = Axes3D(fig)
x0,v0,m = initial_cond(N,D)
for s in range(S):
x1,v1 = timestep(x0,v0,G,m,dt)
x0,v0 = x1,v1
t = 0
for i in x0:
ax.scatter(i[0],i[1],i[2],label=str(s*dt),c=["black","green","red"][t])
t += 1
t = 0
plt.show()
start = time.time()
stimulate(2,3,3000,9.8,1e-3)
stop = time.time()
runtimes.append(stop - start)效果图
Python 并行化执行
首先我们给出一个可以用来写自己的并行化程序的，额，一串代码
import datetime
import multiprocessing as mp
def accessional_fun():
f = open("accession.txt","r")
result = float(f.read())
f.close()
return result
def final_fun(name, param):
result = 0
for num in param:
result += num + accessional_fun() * 2
return {name: result}
if __name__ == '__main__':
start_time = datetime.datetime.now()
num_cores = int(mp.cpu_count())
print("你使用的计算机有: " + str(num_cores) + " 个核，当然了，Intel 7 以上的要除以2")
print("如果你使用的 Python 是 32 位的，注意数据量不要超过两个G")
print("请你再次检查你的程序是否已经改成了适合并行运算的样子")
pool = mp.Pool(num_cores)
param_dict = {'task1': list(range(10, 300)),
'task2': list(range(300, 600)),
'task3': list(range(600, 900)),
'task4': list(range(900, 1200)),
'task5': list(range(1200, 1500)),
'task6': list(range(1500, 1800)),
'task7': list(range(1800, 2100)),
'task8': list(range(2100, 2400)),
'task9': list(range(2400, 2700)),
'task10': list(range(2700, 3000))}
results = [pool.apply_async(final_fun, args=(name, param)) for name, param in param_dict.items()]
results = [p.get() for p in results]
end_time = datetime.datetime.now()
use_time = (end_time - start_time).total_seconds()
print("多进程计算 共消耗: " + "{:.2f}".format(use_time) + " 秒")
print(results)运行结果：如下：
accession.txt 里的内容是2.5 这就是一个累加的问题，每次累加的时候都会读取文件中的2.5
如果需要运算的问题是类似于累加的问题，也就是可并行运算的问题，那么才好做出并行运算的改造
再举一个例子
import math
import time
import multiprocessing as mp
def final_fun(name, param):
result = 0
for num in param:
result += math.cos(num) + math.sin(num)
return {name: result}
if __name__ == '__main__':
start_time = time.time()
num_cores = int(mp.cpu_count())
print("你使用的计算机有: " + str(num_cores) + " 个核，当然了，Intel 7 以上的要除以2")
print("如果你使用的 Python 是 32 位的，注意数据量不要超过两个G")
print("请你再次检查你的程序是否已经改成了适合并行运算的样子")
pool = mp.Pool(num_cores)
param_dict = {'task1': list(range(10, 3000000)),
'task2': list(range(3000000, 6000000)),
'task3': list(range(6000000, 9000000)),
'task4': list(range(9000000, 12000000)),
'task5': list(range(12000000, 15000000)),
'task6': list(range(15000000, 18000000)),
'task7': list(range(18000000, 21000000)),
'task8': list(range(21000000, 24000000)),
'task9': list(range(24000000, 27000000)),
'task10': list(range(27000000, 30000000))}
results = [pool.apply_async(final_fun, args=(name, param)) for name, param in param_dict.items()]
results = [p.get() for p in results]
end_time = time.time()
use_time = end_time - start_time
print("多进程计算 共消耗: " + "{:.2f}".format(use_time) + " 秒")
result = 0
for i in range(0,10):
result += results[i].get("task"+str(i+1))
print(result)
start_time = time.time()
result = 0
for i in range(10,30000000):
result += math.cos(i) + math.sin(i)
end_time = time.time()
print("单进程计算 共消耗: " + "{:.2f}".format(end_time - start_time) + " 秒")
print(result)运行结果：
力学问题改进：
import numpy as np
import time
from mpi4py import MPI
from mpi4py.MPI import COMM_WORLD
from types import FunctionType
from matplotlib import pyplot as plt
from multiprocessing import Pool
def remove_i(x,i):
shape = (x.shape[0]-1,) + x.shape[1:]
y = np.empty(shape,dtype=float)
y[:1] = x[:1]
y[i:] = x[i+1:]
return y
def a(i,x,G,m):
x_i = x[i]
x_j = remove_i(x,i)
m_j = remove_i(m,i)
diff = x_j - x_i
mag3 = np.sum(diff**2,axis=1)**1.5
result = G * np.sum(diff * (m_j/mag3)[:,np.newaxis],axis=0)
return result
def timestep(x0,v0,G,m,dt,pool):
N = len(x0)
takes = [(i,x0,v0,G,m,dt) for i in range(N)]
results = pool.map(timestep_i,takes)
x1 = np.empty(x0.shape,dtype=float)
v1 = np.empty(v0.shape,dtype=float)
for i,x_i1,v_i1 in results:
x1[i] = x_i1
v1[i] = v_i1
return x1,v1
def timestep_i(args):
i,x0,v0,G,m,dt = args
a_i0 = a(i,x0,G,m)
v_i1 = a_i0 * dt + v0[i]
x_i1 = a_i0 * dt ** 2 +v0[i]*dt + x0[i]
return i,x_i1,v_i1
def initial_cond(N,D):
x0 = np.random.rand(N,D)
v0 = np.zeros((N,D),dtype=float)
m = np.ones(N,dtype=float)
return x0,v0,m
class Pool(object):
def __init__(self):
self.f = None
self.P = COMM_WORLD.Get_size()
self.rank = COMM_WORLD.Get_rank()
def wait(self):
if self.rank == 0:
raise RuntimeError("Proc 0 cannot wait!")
status = MPI.Status()
while True:
task = COMM_WORLD.recv(source=0,tag=MPI.ANY_TAG,status=status)
if not task:
break
if isinstance(task,FunctionType):
self.f = task
continue
result = self.f(task)
COMM_WORLD.isend(result,dest=0,tag=status.tag)
def map(self,f,tasks):
N = len(tasks)
P = self.P
Pless1 = P - 1
if self.rank != 0:
self.wait()
return
if f is not self.f:
self.f = f
requests = []
for p in range(1,self.P):
r = COMM_WORLD.isend(f,dest=p)
requests.append(r)
MPI.Request.waitall(requests)
results = []
for i in range(N):
result = COMM_WORLD.recv(source=(i%Pless1)+1,tag=i)
results.append(result)
return results
def __del__(self):
if self.rank == 0:
for p in range(1,self.p):
COMM_WORLD.isend(False,dest=p)
def simulate(N,D,S,G,dt):
x0,v0,m = initial_cond(N,D)
pool = Pool()
if COMM_WORLD.Get_rank()==0:
for s in range(S):
x1,v1 = timestep(x0,v0,G,m,dt,pool)
x0,v0 = x1,v1
else:
pool.wait()
if __name__ == '__main__':
simulate(128,3,300,1.0,0.001)
Ps = [1,2,4,8]
runtimes = []
for P in Ps:
start = time.time()
simulate(128,3,300,1.0,0.001)
stop = time.time()
runtimes.append(stop - start)
print(runtimes)到此这篇关于Python 并行化执行详细解析的文章就介绍到这了,更多相关Python 并行化执行内容请搜索编程网以前的文章或继续浏览下面的相关文章希望大家以后多多支持编程网！
