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
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 vector of numbers (temperatures), and I am using the MATLAB function mink to extract the 5 smallest numbers from the vector to form a new variable. However, the numbers extracted using mink are automatically ordered from lowest to largest (of those 5 numbers). Ideally, I would like to retain the sequence of the numbers as they are arranged in the original vector. I hope my problem is easy to understand. I appreciate any advice.
The function mink that you use was introduced in MATLAB 2017b. It has (as Andras Deak mentioned) two output arguments:
[B,I] = mink(A,k);
The second output argument are the indices, such that B == A(I).
To obtain the set B but sorted as they appear in A, simply sort the vector of indices I:
B = A(sort(I));
For example:
>> A = [5,7,3,1,9,4,6];
>> [~,I] = mink(A,3);
>> A(sort(I))
ans =
3 1 4
For older versions of MATLAB, it is possible to reproduce mink using sort:
function [B,I] = mink(A,k)
[B,I] = sort(A);
B = B(1:k);
I = I(1:k);
Note that, in the above, you don't need the B output, your ordered_mink can be written as follows
function B = ordered_mink(A,k)
[~,I] = sort(A);
B = A(sort(I(1:k)));
Note: This solution assumes A is a vector. For matrix A, see Andras' answer, which he wrote up at the same time as this one.
First you'll need the corresponding indices for the extracted values from mink using its two-output form:
[vals, inds] = mink(array);
Then you only need to order the items in val according to increasing indices in inds. There are multiple ways to do this, but they all revolve around sorting inds and using the corresponding order on vals. The simplest way is to put these vectors into a matrix and sort the rows:
sorted_rows = sortrows([inds, vals]); % sort on indices
and then just extract the corresponding column
reordered_vals = sorted_rows(:,2); % items now ordered as they appear in "array"
A less straightforward possibility for doing the sorting after the above call to mink is to take the sorting order of inds and use its inverse to reverse-sort vals:
reverse_inds = inds; % just allocation, really
reverse_inds(inds) = 1:numel(inds); % contruct reverse permutation
reordered_vals = vals(reverse_inds); % should be the same as previously
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 have two lists of 2-dimensional points given as M x 2 - and N x 2 - matrices, respectively, with M and N possibly being very large.
What is the fastest way to determine how many of them are equal?
I am not sure whether you want to count repetitive entries, but if not you could use intersect or some quite intuitive algorithm based on sorting (see below). I would not prefer a nested-loop version...
function test_compareVecs()
%% create some random data
N = 31415;
M1 = 100000;
M2 = 200000;
vec = rand(N,2);
v1 = [rand(M1-N,2); vec];
v2 = [rand(M2-N,2); vec];
v1 = v1(randperm(M1),:);
v2 = v2(randperm(M2),:);
%% intersect
disp('intersect:');
tic
s = size(intersect(v1,v2,'rows'),1);
toc;
s
%% alternative approach
disp('alternative approach:');
tic;
s = compareVecs(v1,v2);
toc;
s
end
function s = compareVecs(v1,v2)
%% create help vector
help_vec = [[v1,zeros(size(v1,1),1)]; ...
[v2,ones(size(v2,1),1)]];
%% sort by first column
% note: for some reason "sortrows(help_vec,1)" is slower
hash_vec = help_vec(:,1); % dummy hash
[~,sidx] = sort(hash_vec);
help_vec = help_vec(sidx,:);
%% diff + compare
help_vec = diff(help_vec);
s = sum(help_vec(:,1) == 0 & ...
help_vec(:,2) == 0 & ...
help_vec(:,3) ~= 0);
end
Result
intersect:
Elapsed time is 0.145717 seconds.
s = 31415
alternative approach:
Elapsed time is 0.048084 seconds.
s = 31415
Compute all pair-wise distances with pdist2 and then count pairs with zero distance. If the coordinates are float values, you may want to use a tolerance instead of comparing against zero:
%// Data:
M = 10;
N = 8;
listM = randi(10,M,2)-1;
listN = randi(10,N,2)-1;
tol = 1e-6;
%// Distance matrix:
d = pdist2(listM, listN);
%// Count:
count = sum(d(:)<tol);
This should work irrespective of the order of the points in each list, or their lengths. It is a hash-table/dictionary solution that should be fast but with memory demand linear with the lengths of the lists. Please, note that the syntax below may not be perfect, but a quick reference to the main data structures mentioned should make corrections trivial.
(1) populate a dictionary-like containers.Map, in a way that the key is a unique function of the points, e.g. num2str(M(i,1))'-'num2str(M(i,2)).
(2) Then, go over all elements of the second list, create the key just as in (1) and check if it exists. If it does, set map(key)=1 else set it to 0. In the end, all the keys consisting of common points will have 1s stored, and the rest will be zeros.
(3) Finalize by summing over the values of the map (something like sum(map.values())) which should give you the total number of unique intersections among the two sets, irrespective of the order these points appear in each list.
OBS: if you don't want to count just unique intersections but all repeated points, in (2), rather than making map(key)=1, add 1 to map(key). The rest is the same.
I am trying to randomly disperse different numbers in MATLAB array:
I have two 3's, four 2's and I want to randomly populate ones vector (size 10,1).
End result look something like this:
A = [1;3;1;2;3;2;2;1;1;2;1;1]
Then I want to fix the values in A but add more random elements but I can only replace with higher numbers:
For example, to the matrix above I will randomly add two more 2's and two more 3's giving something like this
A= [3;3;2;2;3;2;2;2;1;2;1;3]
M = [3;3;2;2;2;2];
M(end+1:end+4) = 1;
M=M(randperm(10))
The second half of your question needs a lot of clarification.
First part
You can use randsample for that:
A = ones(1,12); %// original values
v = [3 3 2 2 2 2]; %// values to "disperse" in A
ind_replace = randsample(1:numel(A), numel(v)); %// index of entries to be replaced
A(ind_replace) = v;
If you don't have randsample (which is part of the Statistics Toolbox), use randperm and select the first few elements:
ind_replace = randperm(numel(A));
ind_replace = ind_replace(1:numel(v));
A(ind) = v;
Second part
To only replace entries which equal 1:
v = [2 2 3 3]; %// values to "disperse" among the 1 values in A
ind_ones = find(A==1); %// index of entries which equal one
ind_replace = randsample(1:numel(ind_ones), numel(v)); %// index within the above
%// Or: ind_replace = randperm(numel(ind_ones));
%// ind_replace = ind_replace(1:numel(v));
A(ind_ones(ind_replace)) = v;
Note this generalizes the first part, that is, it can also be used when all entries of A equal 1.