How to set upper and lower limit of random number

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Stephen john
Stephen john 2022-3-16
评论： Walter Roberson 2022-3-17
Hello everyone, I hope you are doing well. i have the following code. it generate a random number
I want to set the upper limit and lower limit for the values and generate a pattern.
for example i have value 350, then the lower limit will be 300 and upper limit will be 350, or any variable which define the upper and lower limit of the dataset., I have tried the following method but it does not work
levels=round(rand*498)+2;
Dataset=round(rand(1,1000)*(levels*1.1))+round(levels*0.5);
scatter(1:length(Dataset),Dataset)

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回答（2 个）

Voss
Voss 2022-3-16
lower_limit = 300; % to use your example values
upper_limit = 350;
% generate a uniformly distributed random *number* between lower_limit and
% upper_limit:
X = rand()*(upper_limit-lower_limit)+lower_limit
% generate a uniformly distributed pseudorandom *integer* between lower_limit
% and upper_limit:
X = randi(upper_limit-lower_limit+1)+lower_limit-1
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Walter Roberson
Walter Roberson 2022-3-17
Ran in:
control = randi([2 1000], 1, 1000);
filtered = nan(size(control));
idx = find(control == 300);
if ~isempty(idx)
%talking about what happens when 300 is first found, is only meaningful
%if 300 was found at all
filtered(idx(1)) = randi([250 300], 1, 1); %250 to 300 the first time
idx = idx(2:end); %remaining locations
end
sometimes = rand(size(idx)) <= .3; %sometimes means a 30% chance, right?
filtered(idx(sometimes)) = randi([250 350], 1, nnz(sometimes));
idx = find(control == 500);
sometimes = rand(size(idx)) <= 0.64; %sometimes means a 64% chance, right?
filtered(idx(sometimes)) = randi([450 550], 1, nnz(sometimes));
%now verify
idx = find(~isnan(filtered))
idx = 1×3
301 707 877
control(idx)
ans = 1×3
500 300 500
filtered(idx)
ans = 1×3
545 294 507

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David Hill
David Hill 2022-3-16
dataSet=50*rand(1,1000)+300;
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Stephen john
Stephen john 2022-3-17
@David Hill the numbers are generated between 1-1000, then the upper and lower limit of the numbers depend on the span.
David Hill
David Hill 2022-3-17
r=998*rand(1,1000)+2;
span=100;
for k=1:1000
R(k,:)=span*rand(1,50)+r(k)-50;%each row is 50 samples within the +-50 of each element in r
end

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