python的简单MapReduce实现：计算π

# 外面的正方形面积
As = (2r)(2r) or 4r*r
# 里面的圆的面积
Ac = pi*r*r

pi = Ac / (r*r)
As = 4r*r
r*r = As / 4
pi = 4 * Ac / As

#coding=utf-8
import random
import multiprocessing
from multiprocessing import Process

class MapReduce(object):
    
    def __init__(self, map_func, reduce_func, workers_num=None):
        self.map_func = map_func
        self.reduce_func = reduce_func
        self.workers_num = workers_num
        if not workers_num:
            workers_num = multiprocessing.cpu_count()*2
        self.pool = multiprocessing.Pool(workers_num)

    def __call__(self, inputs):
        map_result = self.pool.map(self.map_func, inputs)
        reduce_result = self.reduce_func(map_result)
        return reduce_result

def calculator(*args):
    print multiprocessing.current_process().name,' processing'
    points, circle_round = args[0]
    points_in_circle = 0
    for i in range(points):
	    # 这里其实只取了1/4圆
        x = random.random()*circle_round
        y = random.random()*circle_round
        if (x**2 + y**2) < circle_round**2:
            points_in_circle += 1
    return points_in_circle

def count_circle_points(points_list):
    return sum(points_list)

if __name__ == '__main__':
    # 半径
    CIRCLE_ROUND = 10
    # 总点数
    POINTS = 10000000
    # 总进程数
    WORKERS_NUM = 10

    map_reduce = MapReduce(calculator, count_circle_points, WORKERS_NUM)
    inputs = [(POINTS/WORKERS_NUM, CIRCLE_ROUND)] * WORKERS_NUM
    all_points_in_circle = map_reduce(inputs)
    ac_as = float(all_points_in_circle)/POINTS
    print 'pi approach to:%7f'%(4*ac_as)