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

Question

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 -1 -1 1 1 1 1 -1 -1 -1 -1 1
I tried using this code below but this is the RANGE -1 to 1, which includes 0 values. I only want -1 and 1. Thanks
x = randi([-1 1],1000,1);

Answer 1

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!).

Answer 2

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]

Answer 3

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

Answer 4

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')

Answer 5

x = rand(N,1);
y = sign(x-0.5);