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
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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
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
Image Analyst
2012-2-4
P.S. I hope this wasn't your homework! If so, ignore it.
saharsahar
2012-2-4
Md. Al-Imran Abir
2023-8-27
@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.
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)
% Round values to nearest .1.
r = round(r, 1)
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.
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)
% Round values to nearest .1.
r = round(r/10, 1)
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)
" (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)
upi = floor(up/resolution)
n = 50; % length of your random vector
x = ((loi-1) + randperm((upi-loi)+1, n)) * resolution
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