我想切割原始矩阵的边缘,并想知道是否有更快的方法。因为我需要多次使用相同的位置和position_u运行selectEdge函数,这意味着许多图表的索引不会改变?是否有可能生成一个可以解决所有问题的映射矩阵?
非常感谢
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.]]
对于后一种选择邻居边缘的实现来说,它会更慢
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
答案 0 :(得分:3)
您可以使用"advanced integer indexing"消除双for-loop
:
X, Y = positions[:,None], positions_u[None,:]
selectedMat[X, Y] += originalMat[X, Y]
selectedMat[Y, X] += originalMat[Y, X]
例如,
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_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)
这是一个timeit基准测试(使用IPython):
In [89]: %timeit selectEdge(positions, positions_u, originalMat, selectedMat)
100 loops, best of 3: 4.44 ms per loop
In [90]: %timeit alt_selectEdge(positions, positions_u, originalMat, selectedMat)
10000 loops, best of 3: 104 µs per loop