我正在使用MPI4PY将n / p列分散到两个输入数据过程中。但是,我无法发送所需的列。为了对最终注释中的结果进行报告,我必须对代码进行哪些更改?
矩阵为:
[1, 2, 3, 4]
[5, 6, 7, 8]
[9, 10, 11, 12]
[13, 14, 15, 16]
然后,n = 4并且p = 2。每个过程将分别具有2列。
这是我的代码:
# Imports
from mpi4py import MPI
import numpy as np
comm = MPI.COMM_WORLD
size = comm.Get_size()
rank = comm.Get_rank()
rows = 4
num_columns = rows/size
data=None
if rank == 0:
data = np.matrix([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]])
recvbuf = np.empty((rows, int(num_columns)), dtype='int')
comm.Scatterv(data, recvbuf, root=0)
print('Rank: ',rank, ', recvbuf received:\n ',recvbuf)
我得到以下输出:
Rank: 0 , recvbuf received:
[[1 2]
[3 4]
[5 6]
[7 8]]
Rank: 1 , recvbuf received:
[[ 9 10]
[11 12]
[13 14]
[15 16]]
我想获取以下输出:
Rank: 0 , recvbuf received:
[[1 2]
[5 6]
[9 10]
[13 14]]
Rank: 1 , recvbuf received:
[[ 3 4]
[7 8]
[11 12]
[15 16]]
我认为这段代码可以满足你的需求。这里的问题是Scatterv根本不关心numpy数组的形状,它只考虑包含值的线性内存块。因此,最简单的方法是事先将数据处理为正确的顺序。请注意,这send_data
是一维数组,但这并不重要,因为Scatterv不在乎。在另一端,已经定义了recvbuf的形状,Scatterv只是从接收到的1D输入中填充它。
# Imports
from mpi4py import MPI
import numpy as np
comm = MPI.COMM_WORLD
size = comm.Get_size()
rank = comm.Get_rank()
rows = 4
num_cols = rows/size
send_data=None
if rank == 0:
data = np.matrix([[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]])
# Split into sub-arrays along required axis
arrs = np.split(data, size, axis=1)
# Flatten the sub-arrays
raveled = [np.ravel(arr) for arr in arrs]
# Join them back up into a 1D array
send_data = np.concatenate(raveled)
recvbuf = np.empty((rows, int(num_cols)), dtype='int')
comm.Scatterv(send_data, recvbuf, root=0)
print('Rank: ',rank, ', recvbuf received:\n ',recvbuf)
非常感谢您,您的回答很完美!