generate array of Different random floating numbers in a specific range

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saharsahar
saharsahar 2012-2-4
评论： Image Analyst 2023-8-27
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

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Image Analyst
Image Analyst 2012-2-4
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.

更多回答（5 个）

Image Analyst
Image Analyst 2012-2-4
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 个评论
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Md. Al-Imran Abir
Md. Al-Imran Abir 2023-8-27
编辑：Md. Al-Imran Abir 2023-8-27
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 2023-8-27
Ran in:
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 2012-2-4
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 2012-2-4
  • 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 2012-2-4
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 2023-8-27
编辑：Bruno Luong 2023-8-27
Ran in:
" (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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