- [0, range-1] if range is a positive integer
- [range+1, 0] if range is a negative integer
- Between min and max, inclusive, if range = [min,max] or [max,min]
how can replace randi instead randint?
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hi ;i have a question please help me , i want to replace randi instead randint in ofdm_basic code in r2017a , but i dont know how i must do it . please answer and help me . this is that line :
X=randint(1,Nused*Nframe,M); % bit: integer vector
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採用された回答
Jan
2018 年 1 月 2 日
編集済み: Jan
2018 年 1 月 2 日
Do you mean randint from the Communications Toolbox or is this from Octave code?
I assume, for both the answer is the same: It depends on what M is.
% X = randint(1, Nused*Nframe, M)
if length(M) == 1
if M > 0
Range = [0, M-1];
else
Range = [M+1, 0];
end
else
Range = [min(M), max(M)];
end
X = randi(Range, 1, Nused*Nframe);
I do not have this toolbox, but I guess that you should find something like this, if you look into the randint function:
edit randint
その他の回答 (5 件)
Geoff Hayes
2018 年 1 月 2 日
mohamad - which ofdm_basic code are you referring to? Is this something found on the MATLAB FileExchange or something else? From Communications System Toolbox Release Notes, it looks like randint has been removed from the Communications System Toolbox and is to be replaced with randi.
If we assume that the code
X=randint(1,Nused*Nframe,M)
creates a 1x(Nused*Nframe) matrix with integers in the interval [0,M-1], then we can replace the above with
X = randi(M, 1, Nused*Nframe) - 1;
I think that will work...
4 件のコメント
Mochan Yang
2019 年 7 月 22 日
bitstream=randi(2,1,N)-1
1 件のコメント
Walter Roberson
2019 年 7 月 22 日
bitstream = randi([0 1], 1, N);
or
bitstream = rand(1,N) >= 0.5;
khalil nasri
2021 年 3 月 6 日
randn( )>0;
1 件のコメント
Walter Roberson
2021 年 3 月 6 日
Normal distribution has peak probability at 0 exactly. Does testing for strict greater than give 50% exactly?
HONG CHENG
2022 年 4 月 28 日
you can look at this link
Specifically
You just need to change the order of paramters
a=randint(3,4,[1,4]);
a=randi([1,4],3,4);
0 件のコメント
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