PythonPP+lambda:示例

    直接上代码。 python PP 下载， 可以到官网下载 Python PP 下载 

 

    lambda 能做的， 基本上普通函数都可以做到。 lambda 的主要作用是简化表达式， 并且似乎尤其适合于表达科学计算公式。 结合 map, reduce 等函数，可以获得更强大而“优雅”的表达能力。 这里的“优雅”并不一定表示易理解性，而常常是因为一句短小精炼的代码使得人眼前一亮。   

    以下代码演示 lambda 的用法及 python 并发编程。    

　

import pp
from multiprocessing import Process, Pool, Pipe

def addNFunc(n):
    return lambda x:x+n

def fac(n):
    if n <= 1:
        return 1
    return reduce(lambda x,y: x*y, map(addNFunc(1), range(n)))

def computingFacOfRange(begin, end):
    for i in map(addNFunc(begin), range(end-begin+1)):
        print i, '! =', fac(i)

def another():
    print 'another quick job'
 
def  QuadraticSum(n, m):
    ''' compute 1^m + 2^m + ... + n^m '''
    if n <= 1:
        return 1
    return reduce(lambda x,y: x+y**m, map(addNFunc(1), range(n)))

def expressQS(n,m):
    if n <= 1:
        return '1'
    return reduce(lambda x,y: "%s+%s" %(str(x),str(y)+'^'+str(m)), map(addNFunc(1), range(n)))

def obtainQuadraticSum(n, m):
    return "%s=%s" % (expressQS(n, m), QuadraticSum(n,m))

def obtainQuadraticSumByPipe(conn, n, m):
    conn.send(obtainQuadraticSum(n, m))

def printQuadraticSum(n, m):
    print obtainQuadraticSum(n, m)

def testQuadraticSum():
    for n in [1,2,3,4,5]:
        for m in [-1,1,2,3]:
            printQuadraticSum(n, m)
                  
def usingPP():
    ''' using pp module for python concurrent programming '''
    nworkers = 10
    ppservers = ()
    job_server = pp.Server(nworkers, ppservers=ppservers)

    print "Starting pp with", job_server.get_ncpus(), "workers"

    jobs = []
    for i in range(nworkers):
        jobs.append(job_server.submit(computingFacOfRange, (i*10+1, (i+1)*10), (fac,addNFunc,)))

    jobs.append(job_server.submit(another, ()))

    for job in jobs:
        job()
    
    job_server.print_stats()
    

def usingMultiprocess():
    ''' using multiprocessing module for python concurrent programming '''
    num = 100
    processes = []

    print '************ using original process ***********'
    input_conn, output_conn = Pipe()
    for m in [-1,1,2,3]:
        p = Process(target=obtainQuadraticSumByPipe, args=(input_conn, num,m,))
        p.start()
        print output_conn.recv()

    print '------------- using Pool -------------'
    pool = Pool(processes=4)
    for m in [-1,1,2,3]:
        pool.apply(printQuadraticSum, (num,m))
       

if __name__ == '__main__':
    testQuadraticSum()
    usingPP()
    usingMultiprocess()

 

作者：@琴水玉

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