matlab中rand函数的用法

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2024年5月23日发(作者：)

一、matlab中的rand函数（用于产生随机数）

均匀分布的随机数或矩阵

语法

Y = rand(n)

Y = rand(m,n)

Y = rand([m n])

Y = rand(m,n,p,...)

Y = rand([m ])

Y = rand(size(A))

rand

s = rand('state')

描述

rand函数产生由在(0, 1)之间均匀分布的随机数组成的数组。

Y = rand(n) 返回一个n x n的随机矩阵。如果n不是数量，则返回错误信息。

Y = rand(m,n) 或 Y = rand([m n]) 返回一个m x n的随机矩阵。

Y = rand(m,n,p,...) 或 Y = rand([m ]) 产生随机数组。

Y = rand(size(A)) 返回一个和A有相同尺寸的随机矩阵。

1，rand(3)*-2 rand（3）是一个3*3的随机矩阵（数值范围在0~1之间）

然后就是每个数乘上-2

2 ，用matlab随机产生60个1到365之间的正数 1+fix（365*rand（1，

60））；

3，用rand函数随机取100个从-1到1的数x1，x2，...，x = rand(1,100) * 2

– 1

二、使用中应该注意的问题：

rand产生的是0到1(不包括1)的随机数.

matlab的rand函数生的是伪随机数,即由种子递推出来的,相同的种子,生成相同

的随机数.

matlab刚运行起来时,种子都为初始值,因此每次第一次执行rand得到的随机数都

是相同的.

1.多次运行,生成相同的随机数方法:

用rand('state',S)设定种子

S为35阶向量，最简单的设为0就好

例:

rand('state',0);rand(10)

2. 任何生成相同的随机数方法:

试着产生和时间相关的随机数,种子与当前时间有关.

rand('state',sum(100*clock))

即:

rand('state',sum(100*clock)) ;rand(10)

只要执行rand('state',sum(100*clock)) ;的当前计算机时间不现,生成的随机值就不

现.

也就是如果时间相同,生成的随机数还是会相同.

在你计算机速度足够快的情况下,试运行一下:

rand('state',sum(100*clock));A=rand(5,5);rand('state',sum(100*clock));B=rand(5,5);

A和B是相同.

所以建议再增加一个随机变量,变成:

rand('state',sum(100*clock)*rand(1));

.

有兴趣的可以查阅

Petr Savicky

Institute of Computer Science

Academy of Sciences of CR

Czech Republic

savicky@

September 16, 2006

Abstract

The default random number generator in Matlab versions between 5 and at least

7.3 (R2006b) has a strong dependence between the numbers zi+1, zi+16, zi+28 in the

generated sequence. In particular, there is no index i such that the inequalities

zi+1 < 1/4, 1/4 zi+16 < 1/2, and 1/2 zi+28 are satisfied simultaneously. This

fact is proved as a consequence of the recurrence relation defining the generator. A

random sequence satisfies the inequalities with probability 1/32. Another example

demonstrating the dependence is a simple function f with values −1 and 1, such that

the correlation between f(zi+1, zi+16) and sign(zi+28 − 1/2) is at least 0.416, while it

should be zero.

A simple distribution on three variables that closely approximates the joint

distribution of zi+1, zi+16, zi+28 is described. The region of zero density in the

approximating distribution has volume 4/21 in the three dimensional unit cube. For

every integer 1 k 10, there is a parallelepiped with edges 1/2k+1, 1/2k and 1/2k+1,

where the density of the distribution is 2k. Numerical simulation confirms that the

distribution of the original generator matches the approximation within small random

error corresponding to the sample size.

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