在mpi4py中用运算符作为每个元素的求和

2024-09-30 00:32:25 发布

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我正在编写mpi python代码。例如,四个过程的数据如下:

data on procs0: [1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0]
data on procs1: [0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 0, 0]
data on procs2: [0, 0, 0, 0, 0, 0, 7, 8, 9, 0, 0, 0]
data on procs3: [0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 11, 12]

我想使用mpi4py库中的reduce函数来减少procs0上的数据,结果是:

^{pr2}$

如何使用mpi4py lib函数来实现它?在

编辑时间: 以上是一个简单的特例,不能使用集合,请看下面的另一个例子:

data on procs0: [1,0,0,0,0,0]
data on procs1: [0,2,0,0,0,0]
data on procs2: [0,0,0,3,0,0]
data on procs3: [0,0,0,0,4,5]

理想的结果必须是:

result on procs0: [1,2,0,3,4,5]

Tags: 数据函数代码reducedataon过程lib
2条回答

从你的问题中我不确定你想要的是数据的总和还是最大值,我用mpi Reduce函数写了一个简单的例子,它计算总和。在

#!/usr/bin/env python
import numpy as np
from mpi4py import MPI
comm = MPI.COMM_WORLD

comm.Barrier()
t_start = MPI.Wtime()

# this array lives on each processor
data = np.zeros(5)
for i in xrange(comm.rank, len(data), comm.size):
    # set data in each array that is different for each processor
    data[i] = i

# print out the data arrays for each processor
print '[%i]'%comm.rank, data
comm.Barrier()

# the 'totals' array will hold the sum of each 'data' array
if comm.rank==0:
    # only processor 0 will actually get the data
    totals = np.zeros_like(data)
else:
    totals = None

# use MPI to get the totals 
comm.Reduce(
    [data, MPI.DOUBLE],
    [totals, MPI.DOUBLE],
    op = MPI.SUM,
    root = 0
)

# print out the 'totals'
# only processor 0 actually has the data
print '[%i]'%comm.rank, totals

comm.Barrier()
t_diff = MPI.Wtime() - t_start
if comm.rank==0: print t_diff

将此代码保存在文件reduce_test.py中并使用命令mpirun -np 3 ./reduce_test.py运行它,在我的机器上得到以下输出:

^{pr2}$

请注意,将调用comm.Reduce中的参数op = MPI.SUM更改为op = MPI.MAX将计算最大值,而不是求和。在

将list comprehension与zip一起使用,存储每个列的最大值

In [1]: procs0=[1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0]    
In [2]: procs1=[0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 0, 0]
In [3]: procs2=[0, 0, 0, 0, 0, 0, 7, 8, 9, 0, 0, 0]
In [4]: procs3=[0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 11, 12]

In [5]: [max(i) for i in zip(procs0, procs1, procs2, procs3)]
Out[5]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

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