generate array of Different random floating numbers in a specific range

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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

Accepted Answer

Image Analyst
Image Analyst on 4 Feb 2012
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.

More Answers (5)

Image Analyst
Image Analyst on 4 Feb 2012
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
  6 Comments
Md. Al-Imran Abir
Md. Al-Imran Abir on 27 Aug 2023
Edited: Md. Al-Imran Abir on 27 Aug 2023
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
Image Analyst on 27 Aug 2023
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

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Image Analyst
Image Analyst on 4 Feb 2012
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)

saharsahar
saharsahar on 4 Feb 2012
  • 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.

saharsahar
saharsahar on 4 Feb 2012
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.

Bruno Luong
Bruno Luong on 27 Aug 2023
Edited: Bruno Luong on 27 Aug 2023
" (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

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