Generating a Pseudo-random sequence of plus/minus 1 integers

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被撕碎了的回忆 2021-01-23 06:24

Can anybody help me create a simple pseudo-random sequence of +-1 integers with length 1000 using Matlab?

I.e. a sequence such as

-1 -1 1 1 -1 -1 1 -1 -         


        
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  • 2021-01-23 06:26

    Thanks for these many helpful answers. I figure this topic might be general enough it may well deserve a comparison.

    In my setup ( Windows8.4 x64 i74820k cpu and with R2014a) the fastest version is consistently:

    x=2*round(rand(L,1))-1;
    

    Being half an order of magnitude faster than the slowest solution. Hope this helps.

    comparison: figure comparing execution times for pseudo-random sign generation

    code:

    L=[];
    for expon=0:6
        for mant=1:9
        L=cat(1,L,mant*power(10,expon));
        end
    end
    clear expon mant
    
    t1=zeros(length(L),1);
    x=2*round(rand(L(1),1))-1;
    for li=1:length(L)
        tic,
        x=2*round(rand(L(li),1))-1;
        t1(li)=toc;
    end
    
    t2=zeros(length(L),1);
    x=(rand(L(1),1)>0.5)*2-1;
    for li=1:length(L)
        tic,
        x=(rand(L(li),1)>0.5)*2-1;
        t2(li)=toc;
    end
    
    t3=zeros(length(L),1);
    x=(randi([0,1],L(1),1)>0.5)*2-1;
    for li=1:length(L)
        tic,
        x=(randi([0,1],L(li),1)>0.5)*2-1;
        t3(li)=toc;
    end
    
    t4=zeros(length(L),1);
    x=rand(L(1),1);ind=x>=0.5;x(ind)=1;x(~ind)=-1;
    for li=1:length(L)
        tic,
        x=rand(L(li),1);
        ind=x>=0.5;
        x(ind)=1;
        x(~ind)=-1;
        t4(li)=toc;
    end
    
    t5=zeros(length(L),1);
    x=sign(randn(L(1),1));
    for li=1:length(L)
        tic,
        x=sign(randn(L(li),1));
        x(x==0)=1;
        t5(li)=toc;
    end
    
    t6=zeros(length(L),1);
    vec = [-1 1];
    x=vec(randi(numel(vec),L(1),1));
    for li=1:length(L)
        tic,
        x=vec(randi(numel(vec),L(li),1));
        t6(li)=toc;
    end
    
    t7=zeros(length(L),1);
    x=2*randi(2,L(1),1)-3;
    for li=1:length(L)
        tic,
        x=2*randi(2,L(li),1)-3;
        t7(li)=toc;
    end
    
    t8=zeros(length(L),1);
    x=randsample([-1 1],L(1),true);
    for li=1:length(L)
        tic,
        x=randsample([-1 1],L(li),true);
        t8(li)=toc;
    end
    
    clear x vec ind li
    
    figure,
    loglog(L,[t1 t2 t3 t4 t5 t6 t7 t8],'.-','linewidth',2)
    grid on 
    grid minor
    title('Generating pseudo-random sequence +1/-1')
    ylabel('Exec. Time [s]')
    xlabel('Output Vector Length')
    T{1}='x=2*round(rand(L(1),1))-1';
    T{2}='x=(rand(L(1),1)>0.5)*2-1';
    T{3}='x=(randi([0,1],L(1),1)>0.5)*2-1';
    T{4}='x=rand(L(1),1);ind=x>=0.5;x(ind)=1;x(~ind)=-1';
    T{5}='x=sign(randn(L(1),1))';
    T{6}='vec=[-1 1];x=vec(randi(numel(vec),L(1),1))';
    T{7}='x=2*randi(2,L(1),1)-3';
    T{8}='x=randsample([-1 1],L(1),true)';
    legend(T,'location','northwest')
    
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  • 2021-01-23 06:40

    Simply user randsrc function.

    It will generate random sequences of 1 and -1.

    For example

    out = randsrc(2,3)

    out =

    -1    -1    -1
     1    -1     1
    
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  • 2021-01-23 06:43

    Some alternatives:

    x = 2*randi(2, 1000, 1)-3; %// generate 1 and 2 values, and transform to -1 and 1
    x = 2*(rand(1, 1000, 1)<=.5)-1; %// similar to Rayryeng's answer but in one step
    x = randsample([-1 1], 1000, true); %// sample with replacement from the set [-1 1]
    
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  • 2021-01-23 06:47

    You can try generating a random sequence of floating point numbers from [0,1] and any values less than 0.5 set to -1, and anything larger set to 1:

    x = rand(1000,1);
    ind = x >= 0.5;
    x(ind) = 1;
    x(~ind) = -1;
    

    Another suggestion I have is to use the sign function combined with randn so that we can generate both positive and negative numbers. sign generates values that are either -1, 0, 1 depending on the sign of the input. If the input is negative, the output is -1, +1 when positive and 0 when 0. You could do an additional check where any values that are output to 0, set them to -1 or 1:

    x = sign(randn(1000,1));
    x(x == 0) = 1;
    

    One more (inspired by Luis Mendo) would be to have a vector of [-1,1] and use randi to generate a sequence of either 1 or 2, then use this and sample into this vector:

    vec = [-1 1];
    x = vec(randi(numel(vec), 1000, 1));
    

    This code can be extended where vec can be anything you want, and we can sample from any element in vec to produce a random sequence of values (observation made by Luis Mendo. Thanks!).

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  • 2021-01-23 06:52
    x = rand(N,1);
    y = sign(x-0.5);
    
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