C ++：使用MPI的gatherv来连接不同长度的向量 [英] C++: Using MPI's gatherv to concatenate vectors of differing lengths

问题描述

因此我试图在C ++中的MPI进程之间复制不同长度的向量，即在所有节点上获取向量并将它们连接到节点0上的新向量。

我有以下代码，它不会返回我的预期，驱使我疯狂，并进一步下降线。

代码是这里（缩写）：

  //预先汇总所有numfrags以使_numFrag 
 // numfrags是局部大小_fragLoc 
 // _ numFrag是numfrags的总和
 
 MPI :: COMM_WORLD.Barrier（）; 
 cout<< _myid<< local numFrag =< _fragLoc.size（）< endl; 
 MPI :: COMM_WORLD.Barrier（）; 
 for（unsigned i = 0; i <_fragLoc.size（）; ++ i）cout < fragloc（<< i<）=< _fragLoc [i]< endl; 
 MPI :: COMM_WORLD.Barrier（）; 
 
 vector< int> outVector（_numFrag）; 
 int displ [_numprocs]; 
 
 if（_myid == 0）{
 double sum = 0; 
 for（int i = 0; i <_numprocs; ++ i）{
 displ [i] = sum; 
 cout<< _myid<< ：<< i<< ：<< sum< endl; 
 sum + = numfrags [i]; 
} 
} 
 
 MPI :: COMM_WORLD.Barrier（）; MPI :: COMM_WORLD.Gatherv（& _fragLoc [0]，numfrags [_myid]，MPI :: INT，& outVector [0]，& numfrags [0] 0）; 
 
 MPI :: COMM_WORLD.Barrier（）; 
 
 if（_myid == 0）{
 cout< X numFrag =< _numFrag<< endl; 
 for（unsigned i = 0; i <_numFrag; ++ i）cout < outVector（<< i<<）=< outVector [i]< endl; 
} 
  

举一个简单的例子，我有一个四节点运行。这里是作为伪代码的变量输入：

  int _numprocs = 4; 
 vector< int> numfrags = {0,1,0,1}; 
 vector< int> _fragLoc< node 0> = {}; 
 vector< int> _fragLoc< node 1> = {12}; 
 vector< int> _fragLoc< node 2> = {} 
 vector< int> _fragLoc< node 3> = {37}; 
 int _numFrag = 2; 
  

输出为：

  2local numFrag = 0 
 3local numFrag = 1 
 0local numFrag = 0 
 1local numFrag = 1 
 fragloc（0）= 12 
 fragloc （0）= 37 
 0：0：0 
 0：1：0 
 0：2：1 
 0：3：1 
 0： 
 X numFrag = 2 
 outVector（0）= 0 
 outVector（1）= 0 
  

但我预计个人fragLoc的一起放在outVector，这不会发生。任何建议？当我完成调试时，我会清除障碍。

解决方案

据我所知，如预期。

  #include< iostream> 
 #include< mpi.h> 
 #include< vector> 
 
 using namespace std; 
 
 int main（int argc，char ** argv）{
 
 MPI :: Init（argc，argv）; 
 int _myid = MPI :: COMM_WORLD.Get_rank（）; 
 int _numprocs = MPI :: COMM_WORLD.Get_size（）; 
 
 vector< int> _fragLoc; 
 switch（_myid）{
 case 0：break; 
 case 1：_fragLoc.push_back（12）;打破; 
 case 2：break; 
 case 3：_fragLoc.push_back（37）;打破; 
} 
 
 int locNumFrag = _fragLoc.size（）; 
 cout<< _myid<< local numFrag =< locNumFrag<< endl; 
 
 MPI :: COMM_WORLD.Barrier（）; // for printing 
 
 vector< int> numfrags（_numprocs）; 
 MPI :: COMM_WORLD.Allgather（& locNumFrag，1，MPI :: INT，& numfrags [0]，1，MPI :: INT）; 
 
 int _numFrag = 0; 
 for（int i = 0; i <_numprocs; i ++）
 _numFrag + = numfrags [i]; 
 
 for（unsigned i = 0; i <_fragLoc.size（）; ++ i）
 cout< fragloc（<< i<）=< _fragLoc [i]< endl; 
 
 MPI :: COMM_WORLD.Barrier（）; // for printing 
 
 vector< int> outVector（_numFrag）; 
 int displ [_numprocs]; 
 
 if（_myid == 0）{
 double sum = 0; 
 for（int i = 0; i <_numprocs; ++ i）{
 displ [i] = sum; 
 cout<< _myid<< ：<< i<< ：<< sum< endl; 
 sum + = numfrags [i]; 
} 
} 
 
 MPI :: COMM_WORLD.Gatherv（& _fragLoc [0]，numfrags [_myid]，MPI :: INT，& outVector [0] ; numfrags [0]，& displ [0]，MPI :: INT，0）; 
 
 if（_myid == 0）{
 cout< X numFrag =< _numFrag<< endl; 
 for（unsigned i = 0; i <_numFrag; ++ i）cout < outVector（<< i<）=< outVector [i]< endl; 
} 
 
 MPI :: Finalize（）; 
 return 0; 
} 
  

运行

  $ mpirun -np 4 ./gatherv 
 0local numFrag = 0 
 1local numFrag = 1 
 fragloc（0）= 12 
 2local numFrag = 0 
 3local numFrag = 1 
 fragloc（0）= 37 
 0：0：0 
 0：1：0 
 0：2：1 $ b b 0：3：1 
 X numFrag = 2 
 outVector（0）= 12 
 outVector（1）= 37 
  

so I'm trying to copy over vectors of different lengths between MPI processes in C++, namely taking vectors on all of the nodes and concatenating them into a new vector on node 0.

I have the following code, which does not return what I expected, driving me crazy, and causing trouble further down the line.

The code is this (abbreviated):

//previously summed all of numfrags to make _numFrag
//numfrags is a vector of the local sizes of _fragLoc
//_numFrag is the total of numfrags

MPI::COMM_WORLD.Barrier();
cout << _myid << "local numFrag = " << _fragLoc.size() << endl;
MPI::COMM_WORLD.Barrier();
for (unsigned i = 0; i < _fragLoc.size(); ++i)      cout << "fragloc(" << i << ") = " << _fragLoc[i] << endl;
MPI::COMM_WORLD.Barrier();

vector<int> outVector (_numFrag);
int displ[_numprocs]; 

if (_myid == 0) {
    double sum = 0;
    for (int i = 0; i < _numprocs; ++i) {
        displ[i] = sum;
        cout << _myid << " : " << i << " : " << sum << endl;
        sum += numfrags[i];
    }
}

MPI::COMM_WORLD.Barrier(); MPI::COMM_WORLD.Gatherv(&_fragLoc[0], numfrags[_myid], MPI::INT, &outVector[0], &numfrags[0], &displ[0], MPI::INT,0);

MPI::COMM_WORLD.Barrier();

if (_myid == 0) {
    cout << "X numFrag = " << _numFrag << endl;
    for (unsigned i = 0; i < _numFrag; ++i) cout << "outVector(" << i << ") = " << outVector[i] << endl;
}

Giving a simple example, I have a four-node run. Here are the variable inputs as pseudocode:

int _numprocs = 4;
vector<int> numfrags = {0,1,0,1};
vector<int> _fragLoc <node 0> = {};
vector<int> _fragLoc <node 1> = {12};
vector<int> _fragLoc <node 2> = {};
vector<int> _fragLoc <node 3> = {37};
int _numFrag = 2;

The output is:

2local numFrag = 0
3local numFrag = 1
0local numFrag = 0
1local numFrag = 1
fragloc(0) = 12
fragloc(0) = 37
0 : 0 : 0
0 : 1 : 0
0 : 2 : 1
0 : 3 : 1
0: after stage 2
X numFrag = 2
outVector(0) = 0
outVector(1) = 0

But I expected the individual fragLoc's to be put together into outVector and this isn't happening. Any advice? I'll clean up the barriers when I'm done debugging.

解决方案

As far as I can tell, the code above works as expected.

#include <iostream>
#include <mpi.h>
#include <vector>

using namespace std;

int main(int argc, char **argv) {

    MPI::Init(argc, argv);
    int _myid = MPI::COMM_WORLD.Get_rank();
    int _numprocs = MPI::COMM_WORLD.Get_size();

    vector<int> _fragLoc;
    switch(_myid) {
        case 0: break;
        case 1: _fragLoc.push_back(12); break;
        case 2: break;
        case 3: _fragLoc.push_back(37); break;
    }

    int locNumFrag = _fragLoc.size();
    cout << _myid << "local numFrag = " << locNumFrag << endl;

    MPI::COMM_WORLD.Barrier();  // for printing

    vector<int> numfrags(_numprocs);
    MPI::COMM_WORLD.Allgather(&locNumFrag, 1, MPI::INT, &numfrags[0], 1, MPI::INT);

    int _numFrag = 0;
    for (int i=0; i<_numprocs; i++) 
        _numFrag += numfrags[i];

    for (unsigned i = 0; i < _fragLoc.size(); ++i)
        cout << "fragloc(" << i << ") = " << _fragLoc[i] << endl;

    MPI::COMM_WORLD.Barrier(); // for printing

    vector<int> outVector (_numFrag);
    int displ[_numprocs]; 

    if (_myid == 0) {
        double sum = 0;
        for (int i = 0; i < _numprocs; ++i) {
            displ[i] = sum;
            cout << _myid << " : " << i << " : " << sum << endl;
            sum += numfrags[i];
        }
    }

    MPI::COMM_WORLD.Gatherv(&_fragLoc[0], numfrags[_myid], MPI::INT, &outVector[0], &numfrags[0], &displ[0], MPI::INT,0);

    if (_myid == 0) {
        cout << "X numFrag = " << _numFrag << endl;
        for (unsigned i = 0; i < _numFrag; ++i) cout << "outVector(" << i << ") = " << outVector[i] << endl;
    }

    MPI::Finalize();
    return 0;
}

Running gives

$ mpirun -np 4 ./gatherv
0local numFrag = 0
1local numFrag = 1
fragloc(0) = 12
2local numFrag = 0
3local numFrag = 1
fragloc(0) = 37
0 : 0 : 0
0 : 1 : 0
0 : 2 : 1
0 : 3 : 1
X numFrag = 2
outVector(0) = 12
outVector(1) = 37

这篇关于C ++：使用MPI的gatherv来连接不同长度的向量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！