This question already has answers here:
Generate a random number with max, min and mean (average) in Matlab
(6 answers)
Closed 9 years ago.
I want to generate a N dimensional column vector in matlab, with mean 0.5 ( variance is ok to adjust ) , but I want all numbers to be positive, does anyone know how to do it?
You can try this:
E.g. create a vector of 1000 random values drawn from a normal distribution with a mean of 0.5 and a standard deviation of 5.
a = 5;
b = 0.5;
y = a.*randn(1000,1) + b;
To make it positive, then you can delete all the numbers that are negative or zero and generate some more until you got n positive numbers.
Check out here for more info.
It depends on the distribution that you want. The rand(v) function generates a uniform random distribution (range [0,1] I believe although I'm not sure if either 0 or 1 are theoretically possible values) in an array with dimensions v.
So if you want a 3x4x5x6 array, you would do
myRandArray = rand([3 4 5 6]);
If you want the upper value to be larger, you could do
myRandArray = maxVal * rand([3 4 5 6]);
Where maxVal is the largest value. And if you want a range minVal to maxVal, then do
myRandArray = minVal + (maxVal - minVal) * rand([3 4 5 6]);
For other distributions (like randn for normal distribution) you can make adjustments to the above, obviously. If you want a "truncated normal distribution" - you may need to start with too many values:
dims = [3 4 5 6];
n = prod( dims );
tooMany = 0.5 + randn(2 * n); % since you want mean = 0.5
tooMany(tooMany < 0) = [];
if numel( tooMany ) > n
myRandArray = reshape(tooMany(1:n), dims);
end
You can obviously improve on this, but it's a general idea.
For example, you could generate only n values, see how many fail (say m), then generate another m * n / (n - m), and repeat as needed until you have exactly n.
Note that the mean of the final distribution is no longer 0.5 since we cut off the tail. A 'normal distribution' cannot remain 'normal' if you exclude certain values. But then you didn't specify what distribution you wanted...
Related
I know how to generate random numbers in a certain range in Matlab. What i am trying to do now is generate random numbers in a range where there is more chance of getting certain ones.
For example: how could i use Matlab to generate random numbers between 0 and 2, where 50% of them will be less than 0.5?
To get numbers between 0 and 2 I would use (2-0)*rand+0. How can i do this but get a certain percentage of the numbers generated to be less than 0.5? Is there a way to do this using the rand function?
Here is a suggestion:
N = 10; % how many random numbers to generate
bounds = [0 0.5 1 2]; % define the ranges
prob = cumsum([0.5 0.3 0.2]); % define the probabilities
% pick a random range with probability from 'prob':
s = size(bounds,2)-cumsum(bsxfun(#lt,rand(N,1),prob),2);
% pick a random number in this range:
b = rand(1,N).*(bounds(s(:,end)+1)-bounds(s(:,end)))+bounds(s(:,end))
Here we have a probability of prob(k) to draw a number between bounds(k) to bounds(k+1). Basically we first draw a range with defined probability, and then draw another number from the range. So we are interested only in b, but need s on the way (mainly for creating a lot of numbers in a vectorized manner).
so we get:
b =
Columns 1 through 5
0.5297 0.15791 0.88636 0.34822 0.062666
Columns 6 through 10
0.065076 0.54618 0.0039101 0.21155 0.82779
Or, for N = 100000 we can draw:
so we can see how the values are distributed between the 3 ranges in bounds.
You can use a multinomial distribution to draw the ranges, and then compute the random numbers. Here's how:
N = 10;
bounds = [0 0.5 1 2]; % define the ranges
d = diff(bounds);
% pick a N random ranges from a multinomial distribution:
s = mnrnd(N,[0.5 0.3 0.2]);
% pick a random number in this range:
b = rand(1,N).*repelem(d,s)+repelem(bounds(1:end-1),s)
so you get s:
s =
50 39 11
that says you take 50 values from the first range, 39 from the second, and so on...
And you got the result in b:
b =
Columns 1 through 5
0.28212 0.074551 0.18166 0.035787 0.33316
Columns 6 through 10
0.12404 0.93468 1.9808 1.4522 1.6955
So basically it works the same as the first method I posted here, but it may be more accurate and/or readable. Also, I didn't test which method is faster.
I'm trying to generate a 2-by-6 matrix of random numbers based on their density function, for example
f(x)= 2x-4 for 2 < x < 3; 0 otherwise
So from what I understand I have to find the cumulative density function first, x2-4x, and then I have to invert it so that I can use the rand function.
This is that part I do not understand, how do I get the inverted function
Try something similar to this method: https://stackoverflow.com/a/13914141/1011724.
However, your PDF is continuous so you need to adjust it slightly. The cumsum part becomes your CDF and the sum(r >= ... part becomes a definite integral from 0 to rand (which is just your CDF since it evaluates to 0 at x==0) so (ignoring your limits) you get
X = #(x)x.^2 - 4x
To generate a random matrix go X(rand(2,6))
To account for your limits you can just multiply the entire function by x > 2 & x < 3 but also if it's greater than 3 then although the PDF is 0, the CDF should still be 32 - 4 =5
X_limited = #(x)(x.^2 - 4x ).*(x > 2 & x < 3) + (x>=3)*5
If you plot a graph of (x > 2 & x < 3) you will see it is a rectangular function between 2 and 3 and so multiplying by it makes anything outside of that window 0 but leaves anything inside the window unchanged. Similarly, x >= 3 is a step function start at x == 3 and thus it adds 5 to any values higher than 3 and since the windowing function will make sure the first term is zero when x is greater then 3, this step function ensures a value of 5 for all x greater than 3.
Now you just need to generate random numbers in whatever your range is. Assuming it's between 0 and 5
x = rand(2,6)*5
X_limited(x)
I have a vector of certain size and I want to reshape it into a square matrix. Here is an example: Let's say the vector is of size 784. Then I would create a matrix of size 28x28. In Matlab I would do it with the following command:
reshape(x,28,28)
Of course it can be possible that it is not possible to have an exact square matrix. In this case the matrix should as squarish as possible.
How can I do this calculation? That means how can I calculate the values a and b in reshape(x,a,b)?
Start with a equal to the square root of numel(x) rounded down. If that number doesn't divide numel(x), subtract 1 and try again. That way you end with a equal to the closest integer to sqrt(x) (from below) that divides numel(x). b would then be numel(x)/a, but you can simply use [] as the third argument to reshape:
a = floor(sqrt(numel(x)));
while mod(x,a)
a = a-1;
end
result = reshape(x,a,[]);
Example:
x = 1:20;
gives
result =
1 5 9 13 17
2 6 10 14 18
3 7 11 15 19
4 8 12 16 20
One possible approach:
x = rand(1, 784);
divisors = find(rem(numel(x), 1:numel(x)) == 0);
[~, idx] = min(abs(divisors - sqrt(numel(x))));
x = reshape(x, divisors(idx), numel(x) / divisors(idx));
Let me explain:
Suppose you have a vector named x:
x = rand(1, 784);
First, you find the divisors of the size of x:
divisors = find(rem(numel(x), 1:numel(x)) == 0);
Then, you proceed to choose the divisor which is closest to the square root of x's size:
[~, idx] = min(abs(divisors - sqrt(numel(x))));
Finally, you reshape x using that divisor (and the corresponding multiple):
x = reshape(x, divisors(idx), numel(x) / divisors(idx));
It is not a simple problem to find closest factors of an integer. You need to use the MATLAB answers to the question Input an integer, find the two closest integers which, when multiplied, equal the input. From that question if you use the answer that provides the function findIntegerFactorsCloseToSquarRoot, you can use the following code to reshape.
[a, b] = findIntegerFactorsCloseToSquarRoot(numel(x));
reshape(x, a, b);
I suggest you to first check whether the number is prime or not by isprime(784).
Then you can use prime_factors = factor(784) to get the integer factorization of the number. (Depending on the MATLAB version you may use ifactor(784))
The rest needs just a little more work on prime_factors.
Problem : How do I use a continuous map - The Link1: Bernoulli Shift Map to model binary sequence?
Concept :
The Dyadic map also called as the Bernoulli Shift map is expressed as x(k+1) = 2x(k) mod 1. In Link2: Symbolic Dynamics, explains that the Bernoulli Map is a continuous map and is used as the Shift Map. This is explained further below.
A numeric trajectory can be symbolized by partitioning into appropriate regions and assigning it with a symbol. A symbolic orbit is obtained by writing down the sequence of symbols corresponding to the successive partition elements visited by the point in its orbit. One can learn much about the dynamics of the system by studying its symbolic orbits. This link also says that the Bernoulli Shift Map is used to represent symbolic dynamics.
Question :
How is the Bernoulli Shift Map used to generate the binary sequence? I tried like this, but this is not what the document in Link2 explains. So, I took the numeric output of the Map and converted to symbols by thresholding in the following way:
x = rand();
y = mod(2* x,1) % generate the next value after one iteration
y =
0.3295
if y >= 0.5 then s = 1
else s = 0
where 0.5 is the threshold value, called the critical value of the Bernoulli Map.
I need to represent the real number as fractions as explained here on Page 2 of Link2.
Can somebody please show how I can apply the Bernoulli Shift Map to generate symbolized trajectory (also called time series) ?
Please correct me if my understanding is wrong.
How do I convert a real valued numeric time series into symbolized i.e., how do I use the Bernoulli Map to model binary orbit /time series?
You can certainly compute this in real number space, but you risk hitting precision problems (depending on starting point). If you're interested in studying orbits, you may prefer to work in a rational fraction representation. There are more efficient ways to do this, but the following code illustrates one way to compute a series derived from that map. You'll see the period-n definition on page 2 of your Link 2. You should be able to see from this code how you could easily work in real number space as an alternative (in that case, the matlab function rat will recover a rational approximation from your real number).
[EDIT] Now with binary sequence made explicit!
% start at some point on period-n orbit
period = 6;
num = 3;
den = 2^period-1;
% compute for this many steps of the sequence
num_steps = 20;
% for each step
for n = 1:num_steps
% * 2
num = num * 2;
% mod 1
if num >= den
num = num - den;
end
% simplify rational fraction
g = gcd(num, den);
if g > 1
num = num / g;
den = den / g;
end
% recover 8-bit binary representation
bits = 8;
q = 2^bits;
x = num / den * q;
b = dec2bin(x, bits);
% display
fprintf('%4i / %4i == 0.%s\n', num, den, b);
end
Ach... for completeness, here's the real-valued version. Pure mathematicians should look away now.
% start at some point on period-n orbit
period = 6;
num = 3;
den = 2^period-1;
% use floating point approximation
x = num / den;
% compute for this many steps of the sequence
num_steps = 20;
% for each step
for n = 1:num_steps
% apply map
x = mod(x*2, 1);
% display
[num, den] = rat(x);
fprintf('%i / %i\n', num, den);
end
And, for extra credit, why is this implementation fast but daft? (HINT: try setting num_steps to 50)...
% matlab vectorised version
period = 6;
num = 3;
den = 2^period-1;
x = zeros(1, num_steps);
x(1) = num / den;
y = filter(1, [1 -2], x);
[a, b] = rat(mod(y, 1));
disp([a' b']);
OK, this is supposed to be an answer, not a question, so let's answer my own questions...
It's fast because it uses Matlab's built-in (and highly optimised) filter function to handle the iteration (that is, in practice, the iteration is done in C rather than in M-script). It's always worth remembering filter in Matlab, I'm constantly surprised by how it can be turned to good use for applications that don't look like filtering problems. filter cannot do conditional processing, however, and does not support modulo arithmetic, so how do we get away with it? Simply because this map has the property that whole periods at the input map to whole periods at the output (because the map operation is multiply by an integer).
It's daft because it very quickly hits the aforementioned precision problems. Set num_steps to 50 and watch it start to get wrong answers. What's happening is the number inside the filter operation is getting to be so large (order 10^14) that the bit we actually care about (the fractional part) is no longer representable in the same double-precision variable.
This last bit is something of a diversion, which has more to do with computation than maths - stick to the first implementation if your interest lies in symbol sequences.
If you only want to deal with rational type of output, you'll first have to convert the starting term of your series into a rational number if it is not. You can do that with:
[N,D] = rat(x0) ;
Once you have a numerator N and a denominator D, it is very easy to calculate the series x(k+1)=mod(2*x(k), 1) , and you don't even need a loop.
for the part 2*x(k), it means all the Numerator(k) will be multiplied by successive power of 2, which can be done by matrix multiplication (or bsxfun for the lover of the function):
so 2*x(k) => in Matlab N.*(2.^(0:n-1)) (N is a scalar, the numerator of x0, n is the number of terms you want to calculate).
The Mod1 operation is also easy to translate to rational number: mod(x,1)=mod(Nx,Dx)/Dx (Nx and Dx being the numerator and denominator of x.
If you do not need to simplify the denominator, you could get all the numerators of the series in one single line:
xn = mod( N.*(2.^(0:n-1).'),D) ;
but for visual comfort, it is sometimes better to simplify, so consider the following function:
function y = dyadic_rat(x0,n)
[N,D] = rat(x0) ; %// get Numerator and Denominator of first element
xn = mod( N.*(2.^(0:n-1).'),D) ; %'// calculate all Numerators
G = gcd( xn , D ) ; %// list all "Greatest common divisor"
y = [xn./G D./G].' ; %'// output simplified Numerators and Denominators
If I start with the example given in your wiki link (x0=11/24), I get:
>> y = dyadic_rat(11/24,8)
y =
11 11 5 2 1 2 1 2
24 12 6 3 3 3 3 3
If I start with the example given by Rattus Ex Machina (x0=3/(2^6-1)), I also get the same result:
>> y = dyadic_rat(3/63,8)
y =
1 2 4 8 16 11 1 2
21 21 21 21 21 21 21 21
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Generate random number with given probability matlab
I need to create a column vector with random assignments of the number 1, 2 and 3. However i need to be able to control the percentage occurrence of each oif these 3 numbers.
For example, i have a 100 x 1 column vector and I want 30 of the number 1, 50 of the number 2 and 20 of the number 3, in a random assignments.
I am not sure whether you can do that with rand or randi function.
May be you can write a small module something like this :
bit1 = 1 * ones(1,20);
bit2 = 2 * ones(1,50);
bit3 = 3 * ones(1,30);
bits = [bit1 bit2 bit3];
randbits = bits(:, randperm(length(bits)))
You can do it using the CDF (cumulative destribution function) of the percentage of each number.
pdf = [ 30 50 20 ]/100; % the prob. distribution fun. of the samples
cdf = cumsum( pdf );
% I assume here all entries of the PDF are positive and sum(pdf)==1
% If this is not the case, you may normalize pdf to sum to 1.
The sampling itself
n = 100; % number of samples required
v = rand(n,1); % uniformly samples
tmp = bsxfun( #le, v, cdf );
[~, r] = max( tmp, [], 2 );
As observed by #Dan (see comment below), last line can be replaced with
r = numel(pdf) + 1 - sum( tmp, 2 );
The vector r is a random vector of integers 1,2,3 and should satisfy the desired pdf