I have generated random numbers in MATLAB within a range using the below:
N=10000;
n=3000;
c=randperm(N,n);
I need another set of random numbers within the same range 1:N and of the same size n but excluding the values in c.
Any ideas?
You can use again randperm excluding the integers of array c to crate array d with same length n:
ok = 1:N;
ok(c) = [];
d=ok(randperm(numel(ok),n));
Related
I would like to generate an array which contains all ordered samples of length k taken from a set of n elements {a_1,...,a_n}, that is all the k-tuples (x_1,...,x_k) where each x_j can be any of the a_i (repetition of elements is allowed), and whose total number is n^k.
Is there a built-in function in Matlab to obtain it?
I have tried to write a code that iteratively uses the datasample function, but I couldn't get what desired so far.
An alternative way to get all the tuples is based on k-base integer representation.
If you take the k-base representation of all integers from 0 to n^k - 1, it gives you all possible set of k indexes, knowing that these indexes start at 0.
Now, implementing this idea is quite straightforward. You can use dec2base if k is lower than 10:
X = A(dec2base(0:(n^k-1), k)-'0'+1));
For k between 10 and 36, you can still use dec2base but you must take care of letters as there is a gap in ordinal codes between '9' and 'A':
X = A(dec2base(0:(n^k-1), k)-'0'+1));
X(X>=17) = X(X>=17)-7;
Above 36, you must use a custom made code for retrieving the representation of the integer, like this one. But IMO you may not need this as 2^36 is quite huge.
What you are looking for is ndgrid: it generates the grid elements in any dimension.
In the case k is fixed at the moment of coding, get all indexes of all elements a this way:
[X_1, ..., X_k] = ndgrid(1:n);
Then build the matrix X from vector A:
X = [A(X_1(:)), ..., A(X_k(:))];
If k is a parameter, my advice would be to look at the code of ndgrid and adapt it in a new function so that the output is a matrix of values instead of storing them in varargout.
What about this solution, I don't know if it's as fast as yours, but do you think is correct?
function Y = ordsampwithrep(X,K)
%ordsampwithrep Ordered samples with replacement
% Generates an array Y containing in its rows all ordered samples with
% replacement of length K with elements of vector X
X = X(:);
nX = length(X);
Y = zeros(nX^K,K);
Y(1,:) = datasample(X,K)';
k = 2;
while k < nX^K +1
temprow = datasample(X,K)';
%checknew = find (temprow == Y(1:k-1,:));
if not(ismember(temprow,Y(1:k-1,:),'rows'))
Y(k,:) = temprow;
k = k+1;
end
end
end
I have a 102-by-102 matrix. I want to select square sub-matrices of orders from 2 up to 8 using random column numbers. Here is what I have done so far.
matt is the the original matrix of size 102-by-102.
ittr = 30
cols = 3;
for i = 1:ittr
rr = randi([2,102], cols,1);
mattsub = matt([rr(1) rr(2) rr(3)], [rr(1) rr(2) rr(3)]);
end
I have to extract matrices of different orders from 2 to 8. Using the above code I would have to change the mattsub line every time I change cols. I believe it is possible to do with another loop inside but cannot figure out how. How can I do this?
There is no need to extract elements of a vector and concatenate them, just use the vector to index a matrix.
Instead of :
mattsub = matt([rr(1) rr(2) rr(3)], [rr(1) rr(2) rr(3)]);
Use this:
mattsub = matt(rr, rr);
Defining a set of random sizes is pretty easy using the randi function. Once this is done, they can be projected along your iterations number N using arrayfun. Within the iterations, the randperm and sort functions can be used in order to build the random indexers to the original matrix M.
Here is the full code:
% Define the starting parameters...
M = rand(102);
N = 30;
% Retrieve the matrix rows and columns...
M_rows = size(M,1);
M_cols = size(M,2);
% Create a vector of random sizes between 2 and 8...
sizes = randi(7,N,1) + 1;
% Generate the random submatrices and insert them into a vector of cells...
subs = arrayfun(#(x)M(sort(randperm(M_rows,x)),sort(randperm(M_cols,x))),sizes,'UniformOutput',false);
This can work on any type of matrix, even non-squared ones.
You don't need another loop, one is enough. If you use randi to get a random integer as size of your submatrix, and then use those to get random column and row indices you can easily get a random submatrix. Do note that the ouput is a cell, as the submatrices won't all be of the same size.
N=102; % Or substitute with some size function
matt = rand(N); % Initial matrix, use your own
itr = 30; % Number of iterations
mattsub = cell(itr,1); % Cell for non-uniform output
for ii = 1:itr
X = randi(7)+1; % Get random integer between 2 and 7
colr = randi(N-X); % Random column
rowr = randi(N-X); % random row
mattsub{ii} = matt(rowr:(rowr+X-1),colr:(colr+X-1));
end
I want to create two random integers on the interval [1,n] which are guaranteed to be different from each other. I feel like
ri(1)=randi([1 n]);
ri(2)=randi([1 n]);
while ri(1)==ri(2)
ri(2)=randi([1 n]);
end
is not really the smoothest thing you can do.
One method is to use randperm so that you generate a random permutation of n values that are enumerated from 1 up to and including n, and only return the first two elements of the result:
ri = randperm(n, 2);
Older versions of MATLAB do not support calling randperm this way. Older versions only accept the one input variant, which by default returns the entire permutation of the n values. Therefore, you can call randperm using the one input version, then subset into the final result to return what you need:
ri = randperm(n);
ri = ri([1 2]);
Use randperm to create two unique values in range 1...n
out = randperm(n, 2)
out(1) = number 1
out(2) = number 2
If you wish to include 0's in your range. then:
out = randperm(n+1, 2);
out = out-1;
out(1) = number 1
out(2) = number 2
Here's another way:
ri(1) = randi([1 n]); % choose ri(1) uniformly from the set 1,...,n
ri(2) = randi([1 n-1]); % choose ri(2) uniformly from 1,...,n-1
ri(2) = ri(2) + (ri(2)>=ri(1)); % transform 1,...,n-1 into 1,...,ri(1)-1,ri(1)+1,...,n
I want to generate 5 different random variables, & I want also to satisfy other condition which is N(rand1,rand2) =0 where N is 10-by-10 matrix that contains 0s & 1s.
This is the code that I wrote , it generate different random number , but I want to satisfy the other condition.
nb_sources=5;
nb_Des=5;
rand_nb= randperm(n,n);
source = [rand_nb(1:nb_sources)] ;
distination= [rand_nb(nb_sources+1:nb_sources+nb_Des)] ;
Since you're interested only in N(r1,r2)=0, you need to enumerate all these elements of N (lets say from from 1 to 30), generate 5 random numbers as rand(30,5,1) and pick up the indices. E.g. something like this
Nelem = 5;
[I,J] = find(N==0);
ind = randperm(size(I,1));
Res=[I(ind(1:Nelem)),J(ind(1:Nelem))];
I am currently trying to code up a function to assign probabilities to a collection of vectors using a histogram count. This is essentially a counting exercise, but requires some finesse to be able to achieve efficiently. I will illustrate with an example:
Say that I have a matrix X = [x1, x2....xM] with N rows and M columns. Here, X represents a collection of M, N-dimensional vectors. IN other words, each of the columns of X is an N-dimensional vector.
As an example, we can generate such an X for M = 10000 vectors and N = 5 dimensions using:
X = randint(5,10000)
This will produce a 5 x 10000 matrix of 0s and 1s, where each column is represents a 5 dimensional vector of 1s and 0s.
I would like to assign a probability to each of these vectors through a basic histogram count. The steps are simple: first find the unique columns of X; second, count the number of times each unique column occurs. The probability of a particular occurrence is then the #of times this column was in X / total number of columns in X.
Returning to the example above, I can do the first step using the unique function in MATLAB as follows:
UniqueXs = unique(X','rows')'
The code above will return UniqueXs, a matrix with N rows that only contains the unique columns of X. Note that the transposes are due to weird MATLAB input requirements.
However, I am unable to find a good way to count the number of times each of the columns in UniqueX is in X. So I'm wondering if anyone has any suggestions?
Broadly speaking, I can think of two ways of achieving the counting step. The first way would be to use the find function, though I think this may be slow since find is an elementwise operation. The second way would be to call unique recursively as it can also provide the index of one of the unique columns in X. This should allow us to remove that column from X and redo unique on the resulting X and keep counting.
Ideally, I think that unique might already be doing some counting so the most efficient way would probably be to work without the built-in functions.
Here are two solutions, one assumes all values are either 0's or 1's (just like the example in your description), the other does not. Both codes should be very fast (more so the one with binary values), even on large data.
1) only zeros and ones
%# random vectors of 0's and 1's
x = randi([0 1], [5 10000]); %# RANDINT is deprecated, use RANDI instead
%# convert each column to a binary string
str = num2str(x', repmat('%d',[1 size(x,1)])); %'
%# convert binary representation to decimal number
num = (str-'0') * (2.^(size(s,2)-1:-1:0))'; %'# num = bin2dec(str);
%# count frequency of how many each number occurs
count = accumarray(num+1,1); %# num+1 since it starts at zero
%# assign probability based on count
prob = count(num+1)./sum(count);
2) any positive integer
%# random vectors with values 0:MAX_NUM
x = randi([0 999], [5 10000]);
%# format vectors as strings (zero-filled to a constant length)
nDigits = ceil(log10( max(x(:)) ));
frmt = repmat(['%0' num2str(nDigits) 'd'], [1 size(x,1)]);
str = cellstr(num2str(x',frmt)); %'
%# find unique strings, and convert them to group indices
[G,GN] = grp2idx(str);
%# count frequency of occurrence
count = accumarray(G,1);
%# assign probability based on count
prob = count(G)./sum(count);
Now we can see for example how many times each "unique vector" occurred:
>> table = sortrows([GN num2cell(count)])
table =
'000064850843749' [1] # original vector is: [0 64 850 843 749]
'000130170550598' [1] # and so on..
'000181606710020' [1]
'000220492735249' [1]
'000275871573376' [1]
'000525617682120' [1]
'000572482660558' [1]
'000601910301952' [1]
...
Note that in my example with random data, the vector space becomes very sparse (as you increase the maximum possible value), thus I wouldn't be surprised if all counts were equal to 1...