Looping with two variables from a vector - matlab

I have a 30-vector, x where each element of x follows a standardised normal distribution.
So in Matlab,
I have:
for i=1:30;
x(i)=randn;
end;
Now I want to create 30*30=900 elements from vector, x to make a 900-vector, C defined as follows:
I am unable to do the loop for two variables (k and l) properly. I have:
for k=1:30,l=1:30;
C(k,l)=(1/30)*symsum((x(i))*(x(i-abs(k-l))),1,30+abs(k-l));
end
It says '??? Undefined function or method 'symsum' for input arguments of type
'double'.'
I hope to gain from this a 900-vector, C which I will then rewrite as a matrix. The reason I have using two indices k and l instead of one is because I eventually want these indices to denote the (k,l)-entry of such a matrix so it is important that that my 900-vector will be in the form of C = [ row 1 row 2 row 3 ... row 30 ] so I can use the reshape tool i.e.
C'=reshape(C,30,30)
Could anyone help me with the code for the summation and getting such a 900 vector.


Let's try to make this a bit efficient.
n = 30;
x = randn(n,1);
%# preassign C for speed
C = zeros(n);
%# fill only one half of C, since it's symmetric
for k = 2:n
for l = 1:k-1
%# shift the x-vector by |k-l| and sum it up
delta = k-l; %# k is always larger than l
C(k,l) = sum( x(1:end-delta).*x(1+delta:end) );
end
end
%# fill in the other half of C
C = C + C';
%# add the diagonal (where delta is 0, and thus each
%# element of x is multiplied with itself
C(1:n+1:end) = sum(x.^2);


It seems to me that you want a matrix C of 30x30 elements.
Given the formula that you provided I would do
x = randn(1,30)
C = zeros(30,30)
for k=1:30
for l=1:30
v = abs(k-l);
for i =1:30-v
C(k,l) = C(k,l) + x(i)*x(i+v);
end
end
end
if you actually need the vector you can obtain it from the matrix.

Related

How to create a mxn matrix with a specific rank in matlab?

I want to create a m by n matrix with rank k.
Like A is 8 × 8 with rank 5 or B is 4 × 6 with rank 4.
So I try to write a function in MATLAB like below.
My thought is:
generate an m by n zeros matrix
generate m by n matrix and convert it into reduced row echelon form
assign rank of 2.'s matrix to num
if num = k, then assign current matrix to the output
break the iteration
function output = check_rank(m,n,k)
while 1
output = zeros(m,n);
matrix = randi(20,m,n);
tmp = rref(matrix);
num = rank(tmp);
if (num == k)
output = matrix;
break;
end
disp(output);
end
A = check_rank(8,8,4)
The outcome is an infinite loop and all the answers are 6x6 zeros matrix:
Command Window Output
I have also tried method in the how to create a rank k matrix using matlab?
A = zeros(8,8);
for i = 1:4, A = A + randn(8,1) * randn(1,8); end
A
rank(A)
It can reach my goal, but I have no idea how it work successfully?
Thanks, #anonymous!

If you want to generate a random matrix with specified rank, you can try to build a user function like below
function [Y,rk] = fn(m,n,k)
P = orth(randn(m,k));
Q = orth(randn(n,k))';
Y = P*Q;
rk = rank(Y);
end
where P and Q are unitary matrices. Y is the generated matrix with random values, and rk helps you check the rank.
Example
>> [Y,rk] = fn(8,6,5)
Y =
3.8613e-02 7.5837e-03 -7.1011e-02 -7.0392e-02 -3.8519e-02 1.6612e-01
-3.1381e-02 -3.6287e-02 1.4888e-01 -7.6202e-02 -3.7867e-02 3.2707e-01
-1.9689e-01 2.2684e-01 1.2606e-01 -1.2657e-03 1.9724e-01 7.2793e-02
-1.2652e-01 7.7531e-02 1.3906e-01 3.1568e-02 1.8327e-01 -1.3804e-01
-2.6604e-01 -1.4345e-01 1.6961e-03 -9.7833e-02 5.9299e-01 -1.5765e-01
1.7787e-01 -3.5007e-01 3.8482e-01 -6.0741e-02 -2.1415e-02 -2.4317e-01
8.9910e-02 -2.5538e-01 -1.8029e-01 -7.0032e-02 -1.0739e-01 2.2188e-01
-3.4824e-01 3.7603e-01 2.8561e-02 2.6553e-02 2.4871e-02 6.8021e-01
rk = 5

You can easily use eye function:
I = eye(k);
M = zeros(m,n);
M(1:k, 1:k) = I;
The rank(M) is equal to k.

Take a random draw of all possible pairs of indices in Matlab

Consider a Matlab matrix B which lists all possible unordered pairs (without repetitions) from [1 2 ... n]. For example, if n=4,
B=[1 2;
1 3;
1 4;
2 3;
2 4;
3 4]
Note that B has size n(n-1)/2 x 2
I want to take a random draw of m rows from B and store them in a matrix C. Continuing the example above, I could do that as
m=2;
C=B(randi([1 size(B,1)],m,1),:);
However, in my actual case, n=371293. Hence, I cannot create B and, then, run the code above to obtain C. This is because storing B would require a huge amount of memory.
Could you advise on how I could proceed to create C, without having to first store B? Comments on a different question suggest to
Draw at random m integers between 1 and n(n-1)/2.
I=randi([1 n*(n-1)/2],m,1);
Use ind2sub to obtain C.
Here, I'm struggling to implement the second step.
Thanks to the comments below, I wrote this
n=4;
m=10;
coord=NaN(m,2);
R= randi([1 n^2],m,1);
for i=1:m
[cr, cc]=ind2sub([n,n],R(i));
if cr>cc
coord(i,1)=cc;
coord(i,2)=cr;
elseif cr<cc
coord(i,1)=cr;
coord(i,2)=cc;
end
end
coord(any(isnan(coord),2),:) = []; %delete NaN rows from coord
I guess there are more efficient ways to implement the same thing.

You can use the function named myind2ind in this post to take random rows of all possible unordered pairs without generating all of them.
function [R , C] = myind2ind(ii, N)
jj = N * (N - 1) / 2 + 1 - ii;
r = (1 + sqrt(8 * jj)) / 2;
R = N -floor(r);
idx_first = (floor(r + 1) .* floor(r)) / 2;
C = idx_first-jj + R + 1;
end
I=randi([1 n*(n-1)/2],m,1);
[C1 C2] = myind2ind (I, n);

If you look at the odds, for i=1:n-1, the number of combinations where the first value is equal to i is (n-i) and the total number of cominations is n*(n-1)/2. You can use this law to generate the first column of C. The values of the second column of C can then be generated randomly as integers uniformly distributed in the range [i+1, n]. Here is a code that performs the desired tasks:
clc; clear all; close all;
% Parameters
n = 371293; m = 10;
% Generation of C
R = rand(m,1);
C = zeros(m,2);
s = 0;
t = n*(n-1)/2;
for i=1:n-1
if (i<n-1)
ind_i = R>=s/t & R<(s+n-i)/t;
else % To avoid rounding errors for n>>1, we impose (s+n-i)=t at the last iteration (R<(s+n-i)/t=1 always true)
ind_i = R>=s/t;
end
C(ind_i,1) = i;
C(ind_i,2) = randi([i+1,n],sum(ind_i),1);
s = s+n-i;
end
% Display
C
Output:
C =
84333 266452
46609 223000
176395 328914
84865 94391
104444 227034
221905 302546
227497 335959
188486 344305
164789 266497
153603 354932
Good luck!

Performance of using a matrix as vector index

In my code I have a slow part of which the idea can be summarized in the following short example:
A = randi(10,5); %Random 5×5 matrix containing integers ranging from 0 to 10
B = rand(10,1); %Random 10×1 vector containing values ranging from 0 to 1
C = B(A); %New 5×5 matrix consisting of elements from B, indexed using A
In my case, the matrix A is sized 1000×1000, B is a 500×1 vector and C is also 1000×1000. Given that this 3rd line is in a for loop, where A is constant and B is updated every iteration, how can I further improve speed performance? According to the profile viewer 75% of code execution is at this single line. As expected, using a for loop for this operation is much slower (10x for a 1000×1000 matrix):
AA = A(:); %Convert matrix to vector
for k=1:length(AA) %Loop through this vector and use it as index
D(k) = B(AA(k));
end
E = reshape(D,5,5); %Reshape vector to matrix of 5x5
Any ideas to optimize this?
Edit: Script used to measure performance:
N = 1500;
A = randi(500,N);
AA = A(:);
D = zeros(N,N);
B = rand(500,1);
f1 = #() VectorIndex(A,B);
timeit(f1,1)
f2 = #() LoopIndex(AA,B,N);
timeit(f2,1)
function C = VectorIndex(A,B)
C = B(A);
end
function D = LoopIndex(AA,B,N)
D = zeros(N,N);
for k=1:length(AA)
D(k) = B(AA(k));
end
D = reshape(D,N,N);
end

Create a variable number of terms in an anonymous function that outputs a vector

I'd like to create an anonymous function that does something like this:
n = 5;
x = linspace(-4,4,1000);
f = #(x,a,b,n) a(1)*exp(b(1)^2*x.^2) + a(2)*exp(b(2)^2*x.^2) + ... a(n)*exp(b(n)^2*x.^2);
I can do this as such, without passing explicit parameter n:
f1 = #(x,a,b) a(1)*exp(-b(1)^2*x.^2);
for j = 2:n
f1 = #(x,a,b) f1(x,a,b) + a(j)*exp(b(j)^2*x.^2);
end
but it seems, well, kind of hacky. Does someone have a better solution for this? I'd like to know how someone else would treat this.

Your hacky solution is definitely not the best, as recursive function calls in MATLAB are not very efficient, and you can quickly run into the maximum recursion depth (500 by default).
You can introduce a new dimension along which you can sum up your arrays a and b. Assuming that x, a and b are row vectors:
f = #(x,a,b,n) a(1:n)*exp((b(1:n).^2).'*x.^2)
This will use the first dimension as summing dimension: (b(1:n).^2).' is a column vector, which produces a matrix when multiplied by x (this is a dyadic product, to be precise). The resulting n * length(x) matrix can be multiplied by a(1:n), since the latter is a matrix of size [1,n]. This vector-matrix product will also perform the summation for us.
Mini-proof:
n = 5;
x = linspace(-4,4,1000);
a = rand(1,10);
b = rand(1,10);
y = 0;
for k=1:n
y = y + a(k)*exp(b(k)^2*x.^2);
end
y2 = a(1:n)*exp((b(1:n).^2).'*x.^2); %'
all(abs(y-y2))<1e-10
The last command returns 1, so the two are essentially identical.

Iterating over all integer vectors summing up to a certain value in MATLAB?

I would like to find a clean way so that I can iterate over all the vectors of positive integers of length, say n (called x), such that sum(x) == 100 in MATLAB.
I know it is an exponentially complex task. If the length is sufficiently small, say 2-3 I can do it by a for loop (I know it is very inefficient) but how about longer vectors?
Thanks in advance,

Here is a quick and dirty method that uses recursion. The idea is that to generate all vectors of length k that sum to n, you first generate vectors of length k-1 that sum to n-i for each i=1..n, and then add an extra i to the end of each of these.
You could speed this up by pre-allocating x in each loop.
Note that the size of the output is (n + k - 1 choose n) rows and k columns.
function x = genperms(n, k)
if k == 1
x = n;
elseif n == 0
x = zeros(1,k);
else
x = zeros(0, k);
for i = 0:n
y = genperms(n-i,k-1);
y(:,end+1) = i;
x = [x; y];
end
end
Edit
As alluded to in the comments, this will run into memory issues for large n and k. A streaming solution is preferable, which generates the outputs one at a time. In a non-strict language like Haskell this is very simple -
genperms n k
| k == 1 = return [n]
| n == 0 = return (replicate k 0)
| otherwise = [i:y | i <- [0..n], y <- genperms (n-i) (k-1)]
viz.
>> mapM_ print $ take 10 $ genperms 100 30
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,100]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,99]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,98]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,97]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,4,96]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,95]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,94]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,7,93]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,8,92]
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,9,91]
which runs virtually instantaneously - no memory issues to worry about.
In Python you could achieve something nearly as simple using generators and the yield keyword. In Matlab it is certainly possible, but I leave the translation up to you!

This is one possible method to generate all vectors at once (will give memory problems for moderately large n):
s = 10; %// desired sum
n = 3; %// number of digits
vectors = cell(1,n);
[vectors{:}] = ndgrid(0:s); %// I assume by "integer" you mean non-negative int
vectors = cell2mat(cellfun(#(c) reshape(c,1,[]), vectors, 'uni', 0).');
vectors = vectors(:,sum(vectors)==s); %// each column is a vector
Now you can iterate over those vectors:
for vector = vectors %// take one column at each iteration
%// do stuff with the vector
end
To avoid memory problems it is better to generate each vector as needed, instead of generating all of them initially. The following approach iterates over all possible n-vectors in one for loop (regardless of n), rejecting those vectors whose sum is not the desired value:
s = 10; %// desired sum
n = 3;; %// number of digits
for number = 0: s^n-1
vector = dec2base(number,s).'-'0'; %// column vector of n rows
if sum(vector) ~= s
continue %// reject that vector
end
%// do stuff with the vector
end