Python并行化执行详细解析

目录

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>

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)

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