Python 用numpy数组索引加速求边矩阵
我想切割原始矩阵的边缘,想知道有没有更快的方法。因为我需要在相同的位置和位置上多次运行selectEdge函数,这意味着索引对于许多图不会改变?是否有可能生成一个可以修复所有问题的映射矩阵 多谢各位Python 用numpy数组索引加速求边矩阵,python,algorithm,numpy,matrix,indexing,Python,Algorithm,Numpy,Matrix,Indexing,我想切割原始矩阵的边缘,想知道有没有更快的方法。因为我需要在相同的位置和位置上多次运行selectEdge函数,这意味着索引对于许多图不会改变?是否有可能生成一个可以修复所有问题的映射矩阵 多谢各位 def selectEdge(positions, positions_u, originalMat, selectedMat): """ select Edge by neighbors of all points many to many m positions
def selectEdge(positions, positions_u, originalMat, selectedMat):
""" select Edge by neighbors of all points
many to many
m positions
n positions
would have m*n edges
update selectedMat
"""
for ele in positions:
for ele_u in positions_u:
selectedMat[ele][ele_u] += originalMat[ele][ele_u]
selectedMat[ele_u][ele] += originalMat[ele_u][ele]
return selectedMat
我只需要上三角矩阵,因为它是对称的
def test_selectEdge(self):
positions, positions_u = np.array([0,1,5,7]), np.array([2,3,4,6])
originalMat, selectedMat = np.array([[1.0]*8]*8), np.array([[0.0]*8]*8)
selectedMat = selectEdge(positions, positions_u, originalMat, selectedMat)
print 'position, positions_u'
print positions, positions_u
print 'originalMat', originalMat
print 'selectedMat', selectedMat
这是我的测试结果
position, positions_u
[0 1 5 7] [2 3 4 6]
originalMat
[[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]
[ 1. 1. 1. 1. 1. 1. 1. 1.]]
selectedMat
[[ 0. 0. 1. 1. 1. 0. 1. 0.]
[ 0. 0. 1. 1. 1. 0. 1. 0.]
[ 1. 1. 0. 0. 0. 1. 0. 1.]
[ 1. 1. 0. 0. 0. 1. 0. 1.]
[ 1. 1. 0. 0. 0. 1. 0. 1.]
[ 0. 0. 1. 1. 1. 0. 1. 0.]
[ 1. 1. 0. 0. 0. 1. 0. 1.]
[ 0. 0. 1. 1. 1. 0. 1. 0.]]
对于后一种选择相邻边的实现,速度会更慢
def selectNeighborEdges(originalMat, selectedMat, relation):
""" select Edge by neighbors of all points
one to many
Args:
relation: dict, {node1:[node i, node j,...], node2:[node i, node j, ...]}
update selectedMat
"""
for key in relation:
selectedMat = selectEdge([key], relation[key], originalMat, selectedMat)
return selectedMat
您可以使用以下方法消除双
for循环
:
比如说,
import numpy as np
def selectEdge(positions, positions_u, originalMat, selectedMat):
for ele in positions:
for ele_u in positions_u:
selectedMat[ele][ele_u] += originalMat[ele][ele_u]
selectedMat[ele_u][ele] += originalMat[ele_u][ele]
return selectedMat
def alt_selectEdge(positions, positions_u, originalMat, selectedMat):
X, Y = positions[:,None], positions_u[None,:]
selectedMat[X, Y] += originalMat[X, Y]
selectedMat[Y, X] += originalMat[Y, X]
return selectedMat
N, M = 100, 50
positions = np.random.choice(np.arange(N), M, replace=False)
positions_u = np.random.choice(np.arange(N), M, replace=False)
originalMat = np.random.random((N, N))
selectedMat = np.zeros_like(originalMat)
首先检查selectEdge
和alt\u selectEdge
是否返回相同的结果:
expected = selectEdge(positions, positions_u, originalMat, selectedMat)
result = alt_selectEdge(positions, positions_u, originalMat, selectedMat)
assert np.allclose(expected, result)
下面是一个timeit基准测试(使用IPython):
您可以使用以下方法消除双
for循环
:
比如说,
import numpy as np
def selectEdge(positions, positions_u, originalMat, selectedMat):
for ele in positions:
for ele_u in positions_u:
selectedMat[ele][ele_u] += originalMat[ele][ele_u]
selectedMat[ele_u][ele] += originalMat[ele_u][ele]
return selectedMat
def alt_selectEdge(positions, positions_u, originalMat, selectedMat):
X, Y = positions[:,None], positions_u[None,:]
selectedMat[X, Y] += originalMat[X, Y]
selectedMat[Y, X] += originalMat[Y, X]
return selectedMat
N, M = 100, 50
positions = np.random.choice(np.arange(N), M, replace=False)
positions_u = np.random.choice(np.arange(N), M, replace=False)
originalMat = np.random.random((N, N))
selectedMat = np.zeros_like(originalMat)
首先检查selectEdge
和alt\u selectEdge
是否返回相同的结果:
expected = selectEdge(positions, positions_u, originalMat, selectedMat)
result = alt_selectEdge(positions, positions_u, originalMat, selectedMat)
assert np.allclose(expected, result)
下面是一个timeit基准测试(使用IPython):