问题描述
我知道对于很大的矩阵,单个元素访问的numpy速度很慢.代码的以下部分大约需要7-8分钟才能运行.矩阵的大小约为3000 * 3000
I came to know that numpy is slow for individual element accesses for a very big matrix. The following part of the code takes about 7-8 minutes to run. Size of the Matrix is about 3000*3000
import numpy as np
................
................
ArrayLength=len(Coordinates)
AdjMatrix=np.zeros((len(Angles),len(Angles)))
for x in range(0, Arraylength):
for y in range(x+1, Arraylength-x):
distance=Distance(Coordinates[x],Coordinates[y)
if(distance<=radius)
AdjMatrix[x][y]=distance
AdjMatrix[y][x]=distance
我基本上正在尝试为包含约3000个节点的图构造一个邻接矩阵.有人可以帮我做这种麻木的方式吗?还是其他选择?
I am basically trying to construct an adjacency matrix for a graph that consists of about 3000 nodes. Can someone help me in doing this numpy way? Or any alternatives?
这是Distance()函数
Here is the Distance() function
Def Distance(p1,p2):
distance=np.sqrt(np.square(p1[0]-p2[0])+np.square(p1[1]-p2[1]))
return distance
顺便说一下,我将坐标作为元组传递.如p [0] = x坐标和p [1] = y坐标.
By the way I am passing coordinates as tuples.. As in p[0]=x-coordinate and p[1]= y- coordinate.
推荐答案
可以发布Distance()
函数吗?如果是常用功能,scipy.spatial.distance.cdist
可以非常快速地计算距离矩阵:
Can you post the Distance()
function? If it's common function, scipy.spatial.distance.cdist
can calculate the distance matrix very quickly:
修改:
您确实可以使用pdist
,这是一个示例:
You can use pdist
indeed, here is an example:
from scipy.spatial.distance import pdist, squareform
coordinates = [(0.0, 0), (1.0, 2.0), (-1.0, 0.5), (3.1, 2.1)]
dist = squareform(pdist(coordinates))
print dist
输出:
[[ 0. 2.23606798 1.11803399 3.74432905]
[ 2.23606798 0. 2.5 2.1023796 ]
[ 1.11803399 2.5 0. 4.40113622]
[ 3.74432905 2.1023796 4.40113622 0. ]]
如果要屏蔽某些数据:
dist[dist > 3.0] = 0
print dist
输出:
[[ 0. 2.23606798 1.11803399 0. ]
[ 2.23606798 0. 2.5 2.1023796 ]
[ 1.11803399 2.5 0. 0. ]
[ 0. 2.1023796 0. 0. ]]
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