generate array of Different random floating numbers in a specific range

Dear all, I am going to generate an array of Different random floating numbers in a specific range( such as range of 1, 100) in MATLAB. randperm is creating integer, but I want float numbers , preferable with 1 decimal point. Like 20.1 , 34.2 , 20.2, 27.4 Any help is really appreciated . Thanks

[Oleg Komarov]

Create randperm and divide by 100.

Answer 1

[Image Analyst]

If you want 150 numbers in the range of 1-100, try this (small modification of my prior code):

numberOfElementsNeeded = 150;
while true % Loop until we get the required number of elements.
    % Create an array of what should be more than enough numbers with 1 decimal
    % place.
    % We use more than enough in case there are duplicates that we need to remove.
    % It was specified that they all need to be different so there's a
    % possibility that some may need to be removed (if there are duplicates).
    m = double(int32(rand(3*numberOfElementsNeeded,1) * 1000)) / 10;
    % Find out if any happen to be identical.
    mUnique = unique(m);
    % unique() sorts them.  Randomize them to undo the sort
    finalArray = m(randperm(length(mUnique)));
    % Grab only the number of elements that we need.
    numElementsToTake = min([numberOfElementsNeeded length(finalArray)]);
    % See if we have enough.
    if numElementsToTake == numberOfElementsNeeded
        % If we have enough, then break out of the loop.
        break;
    end
    % If we didn't get enough, try again.
end
% Take just numberOfElementsNeeded of them
finalArray = finalArray(1:numberOfElementsNeeded);
% Print out to command window
finalArray

I just changed the line before the for loop to

numberOfElementsNeeded = 150;

and added the line

finalArray = finalArray(1:numberOfElementsNeeded);

after the loop.

Answer 2

[Image Analyst]

I think this code I wrote is fairly robust and flexible:

numberOfElementsNeeded = 100;
while true % Loop until we get the required number of elements.
    % Create an array of what should be more than enough numbers with 1 decimal
    % place.
    % We use more than enough in case there are duplicates that we need to remove.
    % It was specified that they all need to be different so there's a
    % possibility that some may need to be removed (if there are duplicates).
    m = double(int32(rand(3*numberOfElementsNeeded,1) * 1000)) / 10;
    % Find out if any happen to be identical.
    mUnique = unique(m);
    % unique() sorts them.  Randomize them to undo the sort
    finalArray = m(randperm(length(mUnique)));
    % Grab only the number of elements that we need.
    numElementsToTake = min([numberOfElementsNeeded length(finalArray)]);
    % See if we have enough.
    if numElementsToTake == numberOfElementsNeeded
        % If we have enough, then break out of the loop.
        break;
    end
    % If we didn't get enough, try again.
end
% Print out to command window
finalArray

[Image Analyst]

P.S. I hope this wasn't your homework! If so, ignore it.

[saharsahar]

No this is not any homw work, Actually I am working on research. Thanks for consideration.

[Md. Al-Imran Abir]

@Image Analyst Hello! Has there been any new function added for this in later versions? I am asking as I couldn't found any similar thread from recent times.

[Image Analyst]

How about

format long g
% Define parameters.
minValue = 1;
maxValue = 1000;
numberToGet = 9;
% Generate a set of floating point numbers.
r = minValue + (maxValue - minValue) * rand(1, numberToGet)

r = 1×9

602.200629477027 254.508319839586 586.850933385989 413.251448689176 336.006749585505 637.57299397731 540.719970607523 896.675054360822 919.776384436408

% Round values to  nearest .1.
r = round(r, 1)

r = 1×9

602.2 254.5 586.9 413.3 336 637.6 540.7 896.7 919.8

[Md. Al-Imran Abir]

I think this might no't ensure that there won't be any repetition though I didn't get any repitition while I ran it several times. I am not sure about it as I don't know anything about the underlying algorithm but I think when I will run it for thousands of times in loops, there might be some repitition.

[Image Analyst]

There may be repeats after the array is rounded to the nearest 10th. To avoid that use randperm with 10 times the max

format long g
% Define parameters.
minValue = 1;
maxValue = 1000;
numberToGet = 9;
% Generate a set of floating point numbers.
r = minValue + randperm((maxValue - minValue), numberToGet)

r = 1×9

299 955 426 922 86 582 207 757 191

% Round values to  nearest .1.
r = round(r/10, 1)

r = 1×9

29.9 95.5 42.6 92.2 8.6 58.2 20.7 75.7 19.1

Answer 3

[Image Analyst]

Another way that is not quite as random and not quite as robust or flexible as my first answer:

m = double(randperm(1000))/10;
finalArray = m(1:100)

Answer 4

[saharsahar]

• Thanks Image Analyst, These two methods works well when the total number of those numbers to be generated will be less than the integers in the range , But if I want to create for example 150 random floating numbers (with 1 decimal point) in a range(1:100) , it will not work, and I think it is beacuse of presence of randperm. Any idea is really appreciated.

Answer 5

[saharsahar]

That is exactly what I want. Many Thanks. However, I'm curious to know where the upper bound of the range (in this example 100) is mentioned in the code? I may change the upper bound to other numbers such as 25. I really appreciate it.

Answer 6

[Bruno Luong]

" (1, 100) in MATLAB. randperm is creating integer, but I want float numbers , preferable with 1 decimal point. Like 20.1 , 34.2 , 20.2, 27.4 "

lo = 1;
up = 100;
resolution = 0.1;

loi = ceil(lo/resolution)

loi = 10

upi = floor(up/resolution)

upi = 1000

n = 50; % length of your random vector
x = ((loi-1) + randperm((upi-loi)+1, n)) * resolution

x = 1×50

87.2000 25.8000 25.1000 82.5000 95.3000 83.9000 39.7000 84.7000 42.3000 23.0000 80.9000 10.6000 30.8000 67.5000 48.2000 7.2000 88.0000 34.6000 48.3000 92.2000 23.3000 30.9000 51.1000 11.2000 82.1000 80.8000 97.3000 69.9000 94.3000 50.1000