1 
from random import *
N = 100
gamma = 0.7
connect = zeros((N,N))

for i in range(N):
    for j in range(i+1):
        if random() < gamma:
            connect[i,j] = 1
            connect[j,i] = 1
        else:
            connect[i,j] = 0
            connect[j,i] = 0

我尝试做的是创建一个对称矩阵，填充零和一（概率为 0.7）。这是双 for 循环，非常低效......我将用 numpy 制作一些东西，我相信这可以大大加快速度？有谁知道如何进行？非常感谢！

4

1 回答 1

2

You could use the numpy random module to generate random vectors, and use those vectors to seed the matrix. For example:

import numpy as np

N = 100
gamma = 0.7
connect = np.zeros((N,N),dtype=np.int32)

for i in range(0,N):
        dval = np.diag((np.random.random_sample(size=(N-i))<gamma).astype(np.int32),i)
        connect += dval
        if (i>0):
                connect += dval.T

does this diagonally using numpy.diag, but you could do it row-wise to assemble the upper or lower triangular portion, then use addition to form the symmetrical matrix. I don't have a feeling for which might be faster.

EDIT: In fact this row wise version is about 5 times faster than the diagonal version, which I guess shouldn't be all that surprising given the memory access patterns it uses compared to diagonal assembly.

N = 100
gamma = 0.7
connect = np.zeros((N,N),dtype=np.int32)

for i in range(0,N):
    rval = (np.random.random_sample(size=(N-i))<gamma).astype(np.int32)
    connect[i,i:] = rval

connect += np.triu(connect,1).T

EDIT 2

This is even simpler and about 4 times faster than the row-wise version above. Here a triangular matrix is formed directly from a full matrix of weights, then added to its transpose to produce the symmetric matrix:

N = 100
gamma = 0.7
a=np.triu((np.random.random_sample(size=(N,N))<gamma).astype(np.int32))
connect = a + np.triu(a,1).T

On the Linux system I tested it on, version 1 takes about 6.5 milliseconds, version 2 takes about 1.5 milliseconds, version 3 takes about 450 microseconds.

于 2011-06-01T09:13:47.807 回答